Class 12 Commerce Maths - Integrals

Evaluate :
$\displaystyle\int \sqrt{\dfrac{1-\cos 2x}{1+\cos 2x}}dx$

$I=\int \sqrt{\dfrac{1-\cos 2x}{1+\cos 2x}}dx$

$=\int \sqrt{\dfrac{2\sin^2x}{2\cos^2x}}dx$

$=\int \tan x dx=\log (\sec x)+C$

Evaluate : $\int_{-1}^{0} (e^{3x}+7x-5)dx$ .

$\int _{ -1 }^{ 0 }{ \left( { e }^{ 3x }+7x-5 \right) } dx$

$=\int _{ -1 }^{ 0 }{ { e }^{ 3x } } dx+7\int _{ -1 }^{ 0 }{ x } dx-5\int _{ -1 }^{ 0 } dx$

${ \left[ \dfrac { { e }^{ 3x } }{ 3 } \right] }_{ -1 }^{ 0 }+{ \left[ 7\dfrac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }-5{ \left[ x \right] }_{ -1 }^{ 0 }$

$=\dfrac { { e }^{ -3*0 } }{ 3 } -\dfrac { { e }^{ -3 } }{ 3 } +7\dfrac { x*0 }{ 2 } -7\dfrac { { \left( -1 \right) }^{ 2 } }{ 2 } -5\left( 0-\left( -1 \right) \right)$

$=\dfrac { 1 }{ 3 } -\dfrac { 1 }{ { 3e }^{ 3 } } -\dfrac { 7 }{ 2 } -5$

$=\dfrac { 2-21-30 }{ 6 } -\dfrac { 1}{ 3{ e }^{ 3 } }$

$=-\dfrac { 49 }{ 6 } -\dfrac { 1 }{ { 3e }^{ 3 } }$

$=\dfrac { -98{ e }^{ 3 }-2 }{ 6{ e }^{ 3 } }$

$\int_{0}^{1} \dfrac {xe^{x}}{(1 + x)^{2}}dx$ .

First, we need to find the indefinite integral.

$\int _0^1\dfrac{xe^x}{\left(1+x\right)^2}dx$

$\mathrm{Apply\:Integration\:By\:Parts:}\:u=xe^x,\:v'=\dfrac{1}{\left(1+x\right)^2}$

$=xe^x\left(-\dfrac{1}{1+x}\right)-\int \left(e^x+e^xx\right)\left(-\dfrac{1}{1+x}\right)dx$

$=-\dfrac{e^xx}{x+1}-\int \:-\dfrac{e^xx+e^x}{x+1}dx$

Now, we will evalaute the

$\int \:-\dfrac{e^xx+e^x}{x+1}dx$

$\int \:-\dfrac{e^xx+e^x}{x+1}dx=-e^x$

So,

$=-\dfrac{e^xx}{x+1}-\left(-e^x\right)$

$=-\dfrac{e^xx}{x+1}+e^x+C$

Now, we will evalaute the defnite integral

$\int _0^1\dfrac{xe^x}{\left(1+x\right)^2}dx=-\dfrac{e}{2}+e-1$

$=\dfrac{1}{2}\left(e-2\right)$

Find $\displaystyle\int \dfrac{\sin^6x}{\cos^8x} \ dx$

$I=\int \dfrac{\sin^6x}{\cos^8x} dx$

$I=\int \tan^6x \sec^2x \ dx$

Put

$\tan x =t$

$\sec^2x \ dx=dt$

Therefore,

$I=\int t^6 dt=\dfrac{t^7}{7}+C =\dfrac{\tan^7x}{7}+C$

Evaluate the integrals
$\int _{ 0 }^{ 2 }{ \cfrac { 1 }{ \left( { x }^{ 2 }+1 \right) } dx }$

Let

$x=tan\theta$

$\Rightarrow dx=sec^{2}\theta d\theta$

$\Rightarrow \int_{0}^{2}\dfrac {1}{sec^{2}\theta} sec ^{2}\theta d\theta$

$\Rightarrow \int _{0}^{2}\theta =\int _{0}^{2}tan^{-1}x$

$\Rightarrow tan^{-1}2-0$

$= tan^{-1}2$

Evaluate : $\displaystyle \int_{0}^{1}\dfrac {2x + 3}{5x^{2} + 1} dx$

$I=\displaystyle \int_{0}^{1}\dfrac{2x+3}{5x^2+1}dx=\int_{0}^{1}\dfrac{2x}{5x^2+1}dx+\int_{0}^{1}\dfrac{3}{5x^2+1}dx$

$\Rightarrow I=I_1+I_2$

$I_1=\displaystyle \int_{0}^{1}\dfrac{2x}{5x^2+1}dx$

Put

$5x^2+1=t\Rightarrow 5\times 2xdx=dt$ and limits will be

Lower limit

$x=0\Rightarrow t=1$ and Upper limit

$x=1\Rightarrow t=6$

$I_1=\displaystyle \dfrac{1}{5}\int_{1}^{6}\dfrac{1}{t}dt=\dfrac{1}{5}|\ln|t||_1^{6}=\dfrac{1}{5}|\ln 6-\ln 1|=\dfrac{1}{5}\ln 6$

$I_2=\displaystyle \int_{0}^{1}\dfrac{3}{5x^2+1}dx=\dfrac{3}{5}\int_{0}^{1}\dfrac{1}{x^2+\dfrac{1}{5}}dx$

Integral of

$\displaystyle \int_{a}^{b}\dfrac{1}{x^2+a^2}dx=\left | \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}\right |_{a}^{b}$

Here,

$a=\dfrac{1}{\sqrt{5}}$

$I_2=\dfrac{3}{5}\left |\sqrt{5}\tan^{-1}(\sqrt{5}x) \right |_{0}^{1}=\dfrac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$

$\therefore I= \dfrac{1}{5}\ln 6+\dfrac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$

Evaluate :
$\int \dfrac{2\cos x}{3 \sin^2x} \ dx$

$I=\int \dfrac{2\cos x}{3\sin^2x} dx$

Let

$\sin x =t \implies \cos x \ dx=dt$

Therefore,

$I=\int \dfrac{2 \ dt}{3t^2}$

$I=\dfrac{2}{3} \dfrac{t^{-1}}{(-1)}$

$I=-\dfrac{2}{3}(\sin x)^{-1}+C$

Evaluate :
$\displaystyle\int \dfrac{x^3-x^2+x-1}{x-1}dx$

Let

$I=\int \dfrac{x^3-x^2+x-1}{x-1}dx$

$=\int \dfrac{x^2(x-1)+1(x-1)}{x-1}dx$

$=\int (x^2+1) \ dx=\dfrac{x^3}{3}+x+C$

$\int { \dfrac { 3x }{ 3x-1 } dx }$

$\int {\dfrac{{3x}}{{3x - 1}}\,\,dx}$

$= \int {\dfrac{{\left( {3x - 1} \right) + 1}}{{3x - 1}}} \,dx$

$= \int {1\,dx + \int {\dfrac{{dx}}{{3x - 1}}} }$

$= x + \dfrac{1}{3}\log \left( {3x - 1} \right) + c$

Evaluate :
$\int \dfrac{dx}{x(x^5+3)}$

$I=\int \dfrac{dx}{x(x^5+3)}$

$I=\int \dfrac{x^4}{x^5(x^5+3)}dx$

Put

$t=x^5$

$5x^4 \ dx=dt$

Therefore,

$I=\int \dfrac{dt}{5t(t+3)}$

$=\dfrac{1}{5} \int \dfrac{1}{3}[\dfrac{1}{t}-\dfrac{1}{t+3}]dt$

$=\dfrac{1}{15}[\log |t|-\log |t+3|]+C$

$=\dfrac{1}{15}\log |\dfrac{x^5}{x^5+3}|+C$

Find the following integrals:
i) $\displaystyle \int { \dfrac { { x }^{ 3 }-1 }{ { x }^{ 2 } } dx }$
ii) $\displaystyle \int { \left( { x }^{ \cfrac { 2 }{ 3 } }+1 \right) dx }$
iii) $\displaystyle \int { \left( { x }^{ \cfrac { 3 }{ 2 } }+2{ e }^{ x }-\dfrac { 1 }{ x } \right) dx }$

(i)

$\displaystyle \int {\cfrac{{{x^3} - 1}}{{{x^2}}}dx}$

$\displaystyle = \int {\left( {x - {x^{ - 2}}} \right)dx}$

$\displaystyle = \cfrac{{{x^2}}}{2} - \cfrac{{{x^{ - 1}}}}{{ - 1}} + c$

$\displaystyle = \cfrac{{{x^2}}}{2} + \cfrac{1}{x} + c$

(ii)

$\displaystyle \int_{}^{} {\left( {{x^{\frac{2}{3}}} + 1} \right)} dx$

$\displaystyle = \int_{}^{} {{x^{\frac{2}{3}}}dx + \int_{}^{} {dx} }$

$\displaystyle = \frac{3}{5}{x^{\frac{5}{3}}} + x + c$

(iii)

$\displaystyle \int_{}^{} {\left( {{x^{\frac{3}{2}}} + 2{e^x} - \frac{1}{x}} \right)} dx$

$\displaystyle = \int_{}^{} {{x^{\frac{3}{2}}}dx + \int_{}^{} {2{e^x}dx - \int_{}^{} {\frac{1}{x}dx} } }$

$\displaystyle = \frac{2}{5}{x^{\frac{5}{2}}} + 2{e^x} - \log \left| x \right| + c$

The acceleration of a motorcycle is given by $a_{x}(t) = At - Br^{2}$ , where $A = 1.50\ ms^{-2}$ and $B = 0.120\ ms^{-4}$ . The motorcycle is at rest at the origin at time $t = 0$ .
Find its position and velocity as functions of time.

When

$v_0=0$ and

$x_0=0$

$v_x=\int_0^t (At-Bt^2)dt=\dfrac{A}{2}t^2-\dfrac{B}{3}t^3$

$=(0.75 ms^{-3})t^3-(0.04 ms^{-4})t^3$

$x=\int_0^t\left(\dfrac{A}{2}t^2-\dfrac{B}{3}t^3\right)dt=\dfrac{A}{6}t^3-\dfrac{B}{12}t^4$

$=(0.25\, ms^{-3})t^3-(0.010 ms^{-4})t^4$

$\displaystyle _{0}^{5}\int \dfrac{4\sqrt{x+4}}{4\sqrt{x+4}+4\sqrt{9-x}}$

Consider the given integral.

$I=\int_{0}^{5}{\dfrac{4\sqrt{x+4}}{4\sqrt{x+4}+4\sqrt{9-x}}}dx$ ……. (1)

We know that

$\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$

Therefore,

$I=\int_{0}^{5}{\dfrac{4\sqrt{5-x+4}}{4\sqrt{5-x+4}+4\sqrt{9-\left( 5-x \right)}}}dx$

$I=\int_{0}^{5}{\dfrac{4\sqrt{9-x}}{4\sqrt{9-x}+4\sqrt{4+x}}}dx$

$I=\int_{0}^{5}{\dfrac{4\sqrt{9-x}}{4\sqrt{9-x}+4\sqrt{x+4}}}dx$ …… (2)

On adding equation (1) and (2), we get

$I=\int_{0}^{5}{\dfrac{4\sqrt{9-x}+4\sqrt{x+4}}{4\sqrt{9-x}+4\sqrt{x+4}}}dx$

$I=\int_{0}^{5}{1}dx$

$I=\left[ x \right]_{0}^{5}$

$I=5-0$

$I=5$

Hence, this is the answer.

Evaluate $\int {x\ln \sqrt {xdx} }$

$\displaystyle\int x\ln\sqrt{x}dx=\dfrac{1}{2}{\displaystyle\int x\ln xdx}$

$integrating\ by\ parts$

$=\dfrac{1}{2}[\ln x{\displaystyle\int x dx}-\displaystyle\int \dfrac{d(\ln x)}{dx}(\int xdx)\ dx]=\dfrac{1}{2}[\dfrac{x^{2} {\ln x}}{2}-\displaystyle\int \dfrac{1}{x}\times \dfrac{x^{2}}{2}dx]$

$=\dfrac{x^{2 }{\ln x}}{4}-\dfrac{x^{2}}{8}$

Solve: $\displaystyle\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin^5 x.\cos^{100} x \,dx$

We have

$\sin^5x.\cos^{100}x$ is an odd function.

$\sin^5(-x).\cos^{100}(-x)$

$=-\sin^5x.\cos^{100}x$

So,

$\displaystyle\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin^5 x.\cos^{100} x \,dx =0$ [ Using property of definite integral]

$\int_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}}$

Direct integral which is;

$\displaystyle \int \dfrac{1}{1+x^2}dx=\text{tan}^{-1}x$

$\displaystyle \int^{\sqrt{3}}_{1}\dfrac{1}{1+x^2}dx=[\text{tan}^{-1}x]^{\sqrt{3}}_{1}=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$

$\int \dfrac {1}{\sqrt {3x+5}-\sqrt {3x+2}}dx$

$\displaystyle \int \dfrac{1}{\sqrt{3{x}+5}-\sqrt{3{x}+2}}dx\\=\displaystyle\int \dfrac{(\sqrt{3{x}+5}+\sqrt{3{x}+2})}{(\sqrt{3{x}+5}-\sqrt{3{x}+2})(\sqrt{3{x}+5}+\sqrt{3{x}+2})}dx\\=\displaystyle\int \dfrac{\sqrt{3{x}+5}+\sqrt{3{x}+2}}{(3{x}+5)-(3{x}+2)}dx$

$=\dfrac{1}{3}\displaystyle\int (\sqrt{3{x}+5}+\sqrt{3{x}+2})dx$

$=\dfrac{1}{3}\bigg(\dfrac{2(3{x}+5)^{\dfrac{3}{2}}}{3}+\dfrac{2(3{x}+2)^{\dfrac{3}{2}}}{3}\bigg)\\=\dfrac{2}{9}((3{x}+5)^{\dfrac{3}{2}}+(3{x}+2)^{\dfrac{3}{2}})$

Solve $\displaystyle \int { \dfrac { 1 }{ \log _{ x }{ e } } dx }$

Let,

$=\int {1.\log _exdx}$

Integrating by parts taking

$log_ex$ as first function

$=\log _ex \times x-\int {\dfrac{1}{x} \times xdx}$

$=x\log _ex-x+c$

$\displaystyle \int_{0}^\cfrac{\pi}{4} \sin 3x \sin 2x\ dx$

$\int_0^{\cfrac{\pi }{4}} {\sin 3x\sin 2xdx}$

$= \cfrac{1}{2}\int_0^{\cfrac{\pi }{4}} {2\sin 3x\sin 2xdx}$

$= \cfrac{1}{2}\int_0^{\cfrac{\pi }{4}} {\left( {\cos x - \cos 5x} \right)} dx$

$= \cfrac{1}{2}\left[ {\sin x - \cfrac{{\sin 5x}}{5}} \right]_0^{\cfrac{\pi }{4}}$

$= \cfrac{1}{2}\left[ {\cfrac{1}{{\sqrt 2 }} - \cfrac{1}{5}\sin \cfrac{{5\pi }}{4}} \right]$

$= \cfrac{1}{2}\left[ {\cfrac{1}{{\sqrt 2 }} + \cfrac{1}{{5\sqrt 2 }}} \right]$

$= \cfrac{1}{2} \times \cfrac{6}{{5\sqrt 2 }}$

$= \cfrac{{3\sqrt 2 }}{{10}}$

Evaluate $\int {\dfrac {\sin x}{\sin (x+a)}}dx$

$\int_{}^{} {\cfrac{{\sin x}}{{\sin \left( {x + a} \right)}}dx}$

putting

$x+a=t$

$\Rightarrow dx=dt$

$= \int_{}^{} {\cfrac{{\sin \left( {t - a} \right)}}{{\sin t}}} dt$

$= \int_{}^{} {\cfrac{{\left( {\sin t\cos a - \cos t\sin a} \right)}}{{\sin t}}dt}$

$= \cos a\int_{}^{} {dt - \sin a\int_{}^{} {\cot tdt} }$

$= t\cos a - \sin a\log \left| {\sin t} \right| + c$

$= \left( {x + a} \right)\cos a - \sin a\log \left| {\sin \left( {x + a} \right)} \right| + c$

$\displaystyle\int e^{log x}dx$ .

1317499_1063464_ans_a00aa418c6ae411088bba15826b473e0.jpg

Integrate with respect to $x$ :
$\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx$

We have,

$I=\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx$

Let

$t=1+x^{-5}$

$\dfrac{dt}{dx}=0-\dfrac{5}{x^6}$

$\dfrac{dt}{5}=-\dfrac{dx}{x^6}$

Therefore,

$I=-\dfrac{1}{5}\displaystyle \int \dfrac{1}{t^{\frac{1}{5}}}dt$

$I=-\dfrac{1}{5}\dfrac{t^{-\frac{1}{5}+1}}{-\frac{1}{5}+1}+C$

$I=-\dfrac{1}{5}\dfrac{t^{\frac{4}{5}}}{\frac{4}{5}}+C$

$I=-\dfrac{t^{\frac{4}{5}}}{4}+C$

On putting the value of

$t$ , we get

$I=-\dfrac{(1+x^{-5})^{\frac{4}{5}}}{4}+C$

Hence, this is the answer.

Solve $\displaystyle\int\limits_0^x {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {8 - x} }}dx}$

$\int\limits_0^x {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {8 - x} }}dx}$

$\int_0^x {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {8 - x} }} \times \frac{{\sqrt x - \sqrt {8 - x} }}{{\sqrt x - \sqrt {8 - x} }}dx}$

$\int_0^x {{{x - \sqrt {8x - {x^2}} } \over {2x - 8}}dx}$

${1 \over 2}\int_0^x {{{x - \sqrt {8x - {x^2}} } \over {x - 4}}} dx$

${1 \over 2}\int {{{u + 4 - \sqrt {8u + 32 - {{\left( {u + 4} \right)}^2}} } \over u}} du$

${1 \over 2}\int {du + \int {{{du} \over u} - {1 \over 2}\int {{{\sqrt{ - {{\left( {u + 4} \right)}^2} + 8u + 32} } \over u}} } du}$

$v = {u^2} = du = {1 \over {2u}}du\int {{1 \over {{u^2}}} = {1 \over v}}$

${4 \over 2} + 2\log u - {1 \over 4}\int {{{\sqrt {16 - v} } \over v}} du$

${4 \over 2} + 2\log u - {1 \over 2}\int {{{{w^2}} \over {{w^2} - 16}}dw}$

${4 \over 2} + 2\log u - {1 \over 2}\int {{{{w^2} + 16 - 16} \over {{w^2} - 16}}dw}$

${4 \over 2} + 2\log u - {1 \over 2}w - 8\int {{{dw} \over {{w^2} - 16}}}$

${4 \over 2} + 2\log u - {1 \over 2}\sqrt {16 - {u^2}} - 8 \times {1 \over 8}\log \left| {{{w - 4} \over {w + 4}}} \right|$

${{x - 4} \over 2} + 2\log \left| {x - 4} \right| - {1 \over 2}\sqrt {16 - {{\left( {x - 4} \right)}^2}} - \log \left| {{{\sqrt {16 - {u^2}} - 4} \over {\sqrt {16 - {u^2}} + 4}}} \right|$

${{x - 4} \over 2} + 2\log \left| {x - 4} \right| - {1 \over 2}\sqrt {8x - {x^2}} - \log \left| {{{\sqrt {8x - {x^2}} - 4} \over {\sqrt {8x - {x^2}} + 4}}} \right|_0^x$

${{x - 4} \over 2} + 2\log \left| {x - 4} \right| - {1 \over 2}\sqrt {8x - {x^2}} - \log \left| {{{\sqrt {8x - {x^2}} - 4} \over {\sqrt {8x - {x^2}} + 4}}} \right|$

Solve : $\int \dfrac{dx}{x(n + 1) (n + 2)}$

$\int \dfrac{dx}{x(n+1)(n+2)}$

$=\dfrac{1}{(n+1)(n+2)}\int \dfrac{1}{x}dx$

$=\dfrac{1}{(n+1)(n+2)}\log x+C$

Solve:
$\int {\dfrac{1}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right|}$

We have,

$I=\int{\dfrac{dx}{{{a}^{2}}-{{x}^{2}}}}$

$I=-\int{\dfrac{dx}{\left( x-a \right)\left( x+a \right)}}$

$I=-\dfrac{1}{2a}\left[ \int{\dfrac{dx}{\left( x-a \right)}-\int{\dfrac{dx}{\left( x+a \right)}}} \right]$

$I=-\dfrac{1}{2a}\left[ \ln \left( x-a \right)-\ln \left( x+a \right) \right]+C$

$I=\dfrac{1}{2a}\left[ \ln \left| x+a \right|-\ln \left| x-a \right| \right]+C$

$I=\dfrac{1}{2a}\left[ \ln \left| a+x \right|-\ln \left| a-x \right| \right]+C$

$I=\dfrac{1}{2a}\left[ \ln \left| \dfrac{a+x}{a-x} \right| \right]+C$

Hence, proved.

Evaluate $\int \dfrac {1}{a^{x}b^{x}}dx$

$\int{\cfrac{1}{{a}^{x} {b}^{x}} dx}$

$= \int{\cfrac{1}{{\left( ab \right)}^{x}} dx}$

$= \int{{\left( ab \right)}^{-x} dx}$

$= - \cfrac{{\left( ab \right)}^{-x}}{\ln{\left| ab \right|}} + c \left[ \because \int{{a}^{-x} dx} = -\cfrac{{a}^{-x}}{\ln{\left| a \right|}} + c \right]$

Here, c is an arbitrary constant of integration.

Hence

$\int{\cfrac{1}{{a}^{x} {b}^{x}} dx} = - \cfrac{{\left( ab \right)}^{-x}}{\ln{\left| ab \right|}} + c$ .

Solve
$\displaystyle \int \dfrac {1-x^{2}}{x(1-2x)} =$

Consider the given integral.

$I=\int{\dfrac{1-{{x}^{2}}}{x\left( 1-2x \right)}}dx$

$I=\dfrac{1}{2}\int{\dfrac{x-2}{x\left( 2x-1 \right)}}dx+\dfrac{1}{2}\int{1}dx$

$I=\dfrac{1}{2}\int{\dfrac{2}{x}}dx-\dfrac{1}{2}\int{\dfrac{3}{\left( 2x-1 \right)}}dx+\dfrac{1}{2}\int{1}dx$

$I=\dfrac{1}{2}\int{\dfrac{2}{x}}dx-\dfrac{3}{4}\int{\dfrac{1}{\left( x-\dfrac{1}{2} \right)}}dx+\dfrac{1}{2}\int{1}dx$

$I=\ln x-\dfrac{3}{4}\ln \left( x-\dfrac{1}{2} \right)+\dfrac{x}{2}+C$

Solve $\displaystyle \overset{4}{\underset{2}{\int}} 2^x dx$ .

Now,

$\displaystyle \overset{4}{\underset{2}{\int}} 2^x dx$

$=\left[\dfrac{2^x}{\log 2}\right]_{x=2}^{x=4}$

$=\dfrac{1}{\log 2}.(2^4-2^2)$

$=\dfrac{12}{\log 2}$ .

The value of $\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { dx }{ 1+\tan ^{ 3 }{ x } } }$ is

Let

$I=\displaystyle\int ^{\pi/2}_0 \dfrac{dx}{1+\tan^{3}x}\cdots\cdots (1)$

as we know that

$\displaystyle\int ^{a}_b f(x)dx=\displaystyle\int ^{a}_b f(a+b-x)dx$

$\implies I=\displaystyle\int ^{\pi/2}_0 \dfrac{dx}{1+\tan^{3}(\dfrac{\pi}{2}-x)}=\displaystyle\int ^{\pi/2}_0 \dfrac{dx}{1+\cot^{3}x}=\displaystyle\int ^{\pi/2}_0 \dfrac{dx}{1+\dfrac{1}{\tan^{3}x}}=\displaystyle\int ^{\pi/2}_0 \dfrac{\tan^{3}x dx}{1+\tan^{3}x}...(2)$

adding

$(1)$ and

$(2)$

$2{I}=\displaystyle\int ^{\pi/2}_0 dx=[x]^{\pi/2}_0=\dfrac{\pi}{2}$

$\implies I=\dfrac{\pi}{4}$

$\int {x^2}{e^x} dx$ =

Integrating by parts,

$u = x^2$ ,

$v=e^x$

$u^{'} = 2x$ ,

$\int v dx = e^x$

$\implies x^2e^x - \int 2xe^xdx$

Again integrating by parts,we get

$u = x$ ,

$v = e^x$

$u^{'} = 1$ ,

$\int vdx = e^x$

$\implies x^2e^x - 2xe^x + 2\int e^xdx$

$\implies x^2e^x - 2xe^x + 2e^x + c$

Solve:
$\frac{1}{2}\int {\frac{{2x}}{{{{\left( {1 + x} \right)}^2}}}dx}$

Consider the given integral.

$I=\dfrac{1}{2}\int{\dfrac{2x}{{{\left( 1+x \right)}^{2}}}}dx$

$I=\int{\dfrac{x+1-1}{{{\left( 1+x \right)}^{2}}}}dx$

$I=\int{\dfrac{x+1}{{{\left( 1+x \right)}^{2}}}}dx-\int{\dfrac{1}{{{\left( 1+x \right)}^{2}}}}dx$

$I=\int{\dfrac{1}{\left( 1+x \right)}}dx-\int{\dfrac{1}{{{\left( 1+x \right)}^{2}}}}dx$

$I=\log \left( 1+x \right)-\left( -\dfrac{1}{1+x} \right)+C$

$I=\log \left( 1+x \right)+\dfrac{1}{1+x}+C$

Hence, the value is $\log \left( 1+x \right)+\dfrac{1}{1+x}+C$ .

$\int{\dfrac{{x}^{2}dx}{{(1+{x}^{2})}^{2}}}$ .

1317520_1064124_ans_f9b20db95dc64cdb83a6607aaeb9e667.jpg

Integerate:
$\int {x\,\cos \,x\,dx}$

Consider the given integral.

$I=\int{x\cos xdx}$

We know that

$\int{uvdx}=u\int{vdx-}\int{\left( \dfrac{d}{dx}\left( u \right)\int{vdx} \right)}dx$

Therefore,

$I=x\sin x-\int{1\times \left( \sin x \right)}dx$

$I=x\sin x-\int{\left( \sin x \right)}dx$

$I=x\sin x-\left( -\cos x \right)+C$

$I=x\sin x+\cos x+C$

Hence, this is the answer

$\int \dfrac {\cos 2x}{\cos^{2}x.\sin^{2}x}dx$

We have,

$\int{\dfrac{\cos 2x}{{{\cos }^{2}}x\ {{\sin }^{2}}x}dx}$

Divide and multiply by $4$ and we get,

$\dfrac{4}{4}\int{\dfrac{\cos 2x}{{{\cos }^{2}}x\ {{\sin }^{2}}x}dx}$

$=4\int{\dfrac{\cos 2x}{4{{\cos }^{2}}x\ {{\sin }^{2}}x}dx}$

$=4\int{\dfrac{\cos 2x}{{{\sin }^{2}}2x}dx}$

$=4\int{\csc 2xcot2xdx}$

$=-4\dfrac{\csc 2x}{2}+C\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \,\,\therefore \int{\csc x\cot x=-\csc x} \right)$

$=-2\csc 2x+C$

Hence, this is the answer.

Solve $\displaystyle\int_{}^{} {\dfrac{{\left( {3{{\tan }^2}x + 2} \right){{\sec }^2}x}}{{{{\left( {{{\tan }^3}x + 2\tan x + 5} \right)}^2}}}dx}$

Consider the given integral.

$I=\displaystyle\int{\dfrac{\left( 3{{\tan }^{2}}x+2 \right){{\sec }^{2}}x}{{{\left( {{\tan }^{3}}x+2\tan x+5 \right)}^{2}}}}dx$

Let $t={{\tan }^{3}}x+2\tan x+5$

$t={{\tan }^{3}}x+2\tan x+5$

$\dfrac{dt}{dx}=3{{\tan }^{2}}x{{\sec }^{2}}x+2{{\sec }^{2}}x+0$

$dt=\left( 3{{\tan }^{2}}x+2 \right){{\sec }^{2}}xdx$

Therefore,

$I=\displaystyle\int{\dfrac{dt}{{{t}^{2}}}}$

$I=-\dfrac{1}{t}+C$

On putting the value of $t$ , we get

$I=-\dfrac{1}{\left( {{\tan }^{3}}x+2\tan x+5 \right)}+C$

Hence, this is the answer.

Solve $\displaystyle\int\limits_b^a {\frac{x}{{\sqrt {{a^2} + {x^2}} }}dx}$

Consider the given integral.

$I=\displaystyle\int_{b}^{a}{\dfrac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx}$ …….. (1)

Let $t={{a}^{2}}+{{x}^{2}}$

$\dfrac{dt}{dx}=0+2x$

$\dfrac{dt}{2}=xdx$

Therefore,

$I=\dfrac{1}{2}\displaystyle\int_{{{a}^{2}}+{{b}^{2}}}^{2{{a}^{2}}}{\dfrac{1}{\sqrt{t}}dt}$

$I=\dfrac{1}{2}\left[ 2\sqrt{t} \right]_{{{a}^{2}}+{{b}^{2}}}^{2{{a}^{2}}}$

$I=\sqrt{2{{a}^{2}}}-\sqrt{{{a}^{2}}+{{b}^{2}}}$

Hence, this is the answer.

$\int x{\sin ^{ - 1}}x$

We have,

$I=\int{x{{\sin }^{-1}}xdx}$

We know that

$\int{uv}dx=u\int{vdx-\int{\left( \dfrac{d}{dx}\left( u \right)\int{vdx} \right)}}dx$

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}\times \left( \dfrac{{{x}^{2}}}{2} \right)}dx$

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx$

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}-1+1}{\sqrt{1-{{x}^{2}}}}}dx$

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\left[ \int{\dfrac{{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx+\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}dx \right]$

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\left[ -\int{\dfrac{1-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx+\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}dx \right]$

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\left[ -\int{\sqrt{1-{{x}^{2}}}}dx+\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}dx \right]$

We know that

$\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$

$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\dfrac{1}{2}x\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$

Therefore,

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\left[ -\left[ \dfrac{1}{2}x\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}x \right]+{{\sin }^{-1}}x \right]+C$

$I={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\dfrac{1}{2}\left[ -\dfrac{1}{2}x\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}x \right]+C$

$I=\dfrac{{{x}^{2}}}{2}{{\sin }^{-1}}x+\dfrac{1}{4}x\sqrt{1-{{x}^{2}}}-\dfrac{1}{4}{{\sin }^{-1}}x+C$

Hence, this is the answer.

Evalute $\int\limits_0^a x {\left( {{a^2} - {x^2}} \right)^{\frac{7}{2}}}dx$

We have,

$I={{\int_{0}^{a}{x\left( {{a}^{2}}-{{x}^{2}} \right)}}^{\frac{7}{2}}}dx$

Let

$t={{a}^{2}}-{{x}^{2}}$

$\dfrac{dt}{dx}=0-2x$

$-\dfrac{dt}{2}=xdx$

Therefore,

$I=-\dfrac{1}{2}{{\int_{{{a}^{2}}}^{0}{\left( t \right)}}^{\dfrac{7}{2}}}dt$

$I=-\dfrac{1}{2}\left[ \dfrac{{{t}^{\frac{7}{2}+1}}}{\dfrac{7}{2}+1} \right]_{{{a}^{2}}}^{0}$

$I=-\left[ \dfrac{{{t}^{\frac{9}{2}}}}{9} \right]_{{{a}^{2}}}^{0}$

$I=-\dfrac{1}{9}\left[ {{0}^{^{\frac{9}{2}}}}-{{\left( {{a}^{2}} \right)}^{^{\frac{9}{2}}}} \right]$

$I=\dfrac{{{a}^{9}}}{9}$

Hence, the value is $\dfrac{{{a}^{9}}}{9}$ .

Evaluate: $\int \,lnx \cdot \dfrac{1}{(x + 1)^2}dx$ equals

$\displaystyle \int x\dfrac{1}{(x+1)^2}dx$

Using Integration by parts,we get

$=\displaystyle ln x\int \dfrac{1}{(x+1)^2}dx-\int\left(\dfrac{d}{dx}lnx\int\dfrac{1}{(x+1)^2}dx\right)dx$

$\displaystyle =ln x\left(\dfrac{-1}{x+1}\right)-\int \dfrac{1}{x}\left(\dfrac{-1}{x+1}\right)dx$

$=\dfrac{-lnx}{x+1}+\int \dfrac{dx}{x(x+1)}$

$=\dfrac{1}{x(x+1)}=\dfrac{A}{x}+\dfrac{B}{x+1}$ (Partial fraction)

$1=A(x+1)+Bx$

put

$x=0$ we get

$A=1$

put

$x=-1$ ,we get

$B=-1$

$\displaystyle =\dfrac{-ln x}{x+1}+\int \dfrac{dx}{x}-\int \dfrac{dx}{x+1}$

$=\dfrac{-ln x}{x+1}+ln x -ln (x+1)$

$=\dfrac{-ln x}{x+1}+ln \left(\dfrac{x}{x+1}\right)+C$

$\boxed{\displaystyle \int \dfrac{ln x}{(x+1)^2}dx=\dfrac{-ln x}{x+1}+ln \left(\dfrac{x}{x+1}\right)+C}$

1078815_1094741_ans_5f86164af86b4f2983467d4e472f2430.png

The value of $\displaystyle\int_1^{37\pi } {\dfrac{{\pi \sin \left( {\pi \log x} \right)}}{x}} \,dx$

$=\displaystyle\int _{ 1 }^{ 37\pi }{ \frac { \pi \sin { (\pi \log { x } ) } }{ x } } dx$

put

$\pi log x =t$ and

$\frac {\pi}{x} dx =dt$

$=\int _{ 1 }^{ 37\pi }{ \sin { t } dt }$

now integrate

$={ \left[ -\cos { t } \right] }_{ 1 }^{ 37\pi }$

put

$\pi log x =t$

$={ \left[ -\cos ({ \pi \log { x } })\ \right] }_{ 1 }^{ 37\pi }$

$={ (-\cos { (\pi \log { (37\pi ))) } } }-(-\cos { (\pi \log { 1 } )) }$

$={ (-\cos { (\pi \log { (37\pi ))) } } }-(-\cos { (\pi (0))) }$

$={ (-\cos { (\pi \log { (37\pi ))) } } }-(-\cos {0) }$

$={ (-\cos { (\pi \log { (37\pi ))) } } }-(-1)$

$={ (-\cos { (\pi \log { (37\pi ))) } } }+1$

$={1 -\cos { (\pi \log { (37\pi )) } } }$

$\int \cot^{2}xdx =$

Consider the following integral.

$I=\int\limits_{{}}^{{}}{{{\cot }^{2}}xdx}$

$=\int\limits_{{}}^{{}}{\left( {{\csc }^{2}}x-1 \right)dx}$

$=\int\limits_{{}}^{{}}{\left( {{\csc }^{2}}x \right)-\int\limits_{{}}^{{}}{1}dx}$

$=-\tan x-x+C$

Hence, This is the required answer.

$\displaystyle\int 2x\sqrt{x} dx$ .

Given $$x^{\cfrac{1}{2}}x= x^{\cfrac{3}{2}}$$

$\therefore \int 2 x^{\cfrac{3}{2}}dx$

$=\cfrac{4x^{\cfrac{5}{2}}}{5} + C$

$\displaystyle\int \left(\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2dx$ .

$\displaystyle \int \Bigg(\cfrac{e^x + e^{-x}}{e^x - e^{-x}}\Bigg)^2 dx$

$\displaystyle \int \coth^2 x dx$

$\displaystyle \int (cosech^2 x + 1) dx$

$\displaystyle - \coth x + x + C$

$\displaystyle\int \dfrac{1}{1+\cot}dx$ .

$\displaystyle \int \cfrac{dx}{1+cotx}$

$\displaystyle \int \cfrac{cosec^2x \space dx}{(cot^2x+1)(1+cotx)}$

Putting

$u = cotx$

$du = cosec^2x dx$

$\int \cfrac{du}{(u^2+1)(u+1)}$

$\int \cfrac{1}{2} (\cfrac{1}{u+1} - \cfrac{u-1}{u^2+1}) du$

$\cfrac{ln|u+1|}{2} - \cfrac{1}{2}(\int \cfrac{u}{u^2+1} + \int \cfrac{1}{u^2+1})$

$\cfrac{ln|cot^2+1|}{4} -\cfrac{ln|cotx+1|}{2}-\cfrac{tan^{-1}cotx}{2}$

Solve : $\displaystyle \int \dfrac{x^2}{(4 + x)^{3/2}} dx$

$\displaystyle I=\int \dfrac{x^2}{(4+x)^\dfrac{3}{2}}dx$

$(4+x)=t \Rightarrow x=(t-4)$

$dx=dt$

$=\displaystyle I=\int \dfrac{(t-4)^2dt}{t^\dfrac{3}{2}}$

$=\displaystyle \int \dfrac{t^2-8^\dfrac{1}{2}+16}{t^\dfrac{3}{2}}dt$

$=\displaystyle \int [t^\dfrac{1}{2}-\dfrac{8}{t^\dfrac{1}{2}}+t^{\dfrac{-3}{2}}]dt$

$\Rightarrow (\dfrac{t^\dfrac{1}{2}+1}{\dfrac{1}{2}+1})-8(\dfrac{t^{\dfrac{-1}{2}+1}}{\dfrac{-1}{2}+1})+\dfrac{t^{\dfrac{-3}{2}+1}}{\dfrac{-3}{2}+1}+c$

$\Rightarrow \dfrac{2t^\dfrac{3}{2}}{3}-1b\,t^{\dfrac{1}{2}}-t\dfrac{t^{\dfrac{-1}{2}}}{\dfrac{-1}{2}}+c$

$\Rightarrow \dfrac{2t^{\dfrac{3}{2}}}{3}-16t^{\dfrac{1}{2}}-2t^{\dfrac{1}{2}}+c$

$\Rightarrow 2[\dfrac{t^{\dfrac{3}{2}}}{3}-8t^\dfrac{1}{2}-t^\dfrac{-1}{2}]+c$

$\Rightarrow 2[\frac{(4+x)^{\dfrac{3}{2}}}{3}-8(4+x)^{\dfrac{1}{2}}-\dfrac{1}{(4+x)^\dfrac{1}{2}}]+c$

$\Rightarrow 2[\dfrac{(4+x)^2-(8.3)(4+x)-1}{3(4+x)^\dfrac{1}{2}}]+c$

$\Rightarrow \dfrac{2}{3}[\dfrac{x^2+8x+16-96-24x-1}{(x+4)^\dfrac{1}{2}}]+c$

$\boxed{I\Rightarrow \dfrac{2}{3}[\dfrac{x^2-16x-81}{(x+4)^\dfrac{1}{2}}]}$

Evaluate :
$\displaystyle \int x.(\log x)^2 \ dx$

$\displaystyle I=\int x.(\log x)^2 \ dx$

Applying integration by parts, we get,

$\displaystyle I=(\log x)^2\int x \ dx-\int\left[\dfrac{d}{dx}(\log x)^2. \int x \ dx\right]dx$

$\displaystyle =(\log x)^2.\dfrac{x^2}{2}-\int \left[\dfrac{2\log x}{x} \times \dfrac{x^2}{2}\right]dx$

$\displaystyle =\dfrac{x^2}{2}(\log x)^2-\int x \log x \ dx$

Again, applying integration by parts, we get,

$\displaystyle I=\dfrac{x^2}{2}(\log x)^2-\left[\log x\int x \ dx-\int (\dfrac{d}{dx} \log x \int x \ dx)dx\right]$

$\displaystyle =\dfrac{x^2}{2}(\log x)^2-\left[\log x.\dfrac{x^2}{2}-\int \dfrac{1}{x} \times \dfrac{x^2}{2} dx\right]$

$\displaystyle =\dfrac{x^2}{2}(\log x)^2-\dfrac{x^2}{2}.\log x+\int \dfrac{x}{2} dx$

$=\dfrac{x^2}{2}(\log x)^2-\dfrac{x^2}{2}\log x+\dfrac{x^2}{4}+C$

Solve : $\displaystyle \int\limits_{-\pi/2}^{\pi/2} \sin^7x \ dx$

$I=\displaystyle \int\limits_{-\pi/2}^{\pi/2} \sin^7x \ dx$

$\sin^7(-x)=-\sin^7x$ , therefore,

$\sin^7x$ is an odd function.

Therefore, by property

$\displaystyle \int\limits_{-a}^{a} f(x) dx=0$ if

$f(x)=-f(-x)$

$I=\displaystyle \int\limits_{-\pi/2}^{\pi/2}\sin^7x \ dx=0$

Evaluate :
$\displaystyle \int\limits_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

$I=\int\limits_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ........(1)

$I=\int\limits_{0}^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ......(2)

$\left(\int\limits_{0}^{a} f(x) dx =\int\limits_{0}^{a} f(a-x) dx\right)$

Adding 1 and 2, we get,

$2I=\int\limits_{0}^{\pi/2} dx$

$2I=\dfrac{\pi}{2}$

$I=\dfrac{\pi}{4}$

Evaluate : $\displaystyle \int\limits_{0}^{\pi/4} \log(1+\tan x)dx$

$\displaystyle I=\int\limits_{0}^{\pi/4} \log(1+\tan x)dx$ .......(1)

$\displaystyle I=\int\limits_{0}^{\pi/4} \log\left(1+\tan\left(\dfrac{\pi}{4}-x\right)\right)dx$

$\left(\because \int\limits_{0}^{a} f(x)dx=\int\limits_{0}^{a} f(a-x)dx\right)$

$\displaystyle I=\int\limits_{0}^{\pi/4} \log\left[1+\dfrac{1-\tan x}{1+\tan x}\right]dx$

$\displaystyle I=\int\limits_{0}^{\pi/4} \log \dfrac{2}{1+\tan x} dx$

$\displaystyle I=\int\limits_{0}^{\pi/4}\log 2 dx-\int\limits_{0}^{\pi/4} \log(1+\tan x)dx$

$\displaystyle 2I=\int\limits_{0}^{\pi/4} \log 2 dx=\log 2\int\limits_{0}^{\pi/4} dx$

$2I=\dfrac{\pi}{4}\log 2$

$I=\dfrac{\pi}{8} \log 2$

Evaluate :
$\displaystyle \int x \ \sin x \ dx$

$I=\displaystyle \int x \ \sin x \ dx$

Taking x as first function and sin x as second function and integrating by parts, we get,

$\displaystyle I=x \int \ sin x \ dx-\int \left[\left(\dfrac{d}{dx}x\right)\int \sin x \ dx\right]dx$

$\displaystyle =x(-\cos x)-\int 1.(-\cos x)dx$

$\displaystyle =-x\cos x+\sin x +C$

Solve :-
$I = \displaystyle \int_0^{\infty}\dfrac{dx}{e^x+e^{-x}}$

$\displaystyle\int_{0}^{\infty}\dfrac{1}{e^x+e^{-x}}dx$

$=\displaystyle\int_{0}^{\infty}\dfrac{e^x}{e^{2x}+1}dx$

Let

$e^x=t$ , then

$e^xdx=dt$
When

$x=0, t=1$ ,

$x\to\infty, t\to\infty$

$=\displaystyle\int_{1}^{\infty}\dfrac{1}{t^{2}+1}dt$

$=\left[\tan^{-1}x\right]_{1}^{\infty}$

$=\dfrac{\pi}{2}-\dfrac{\pi}{4}$

$=\boxed{\dfrac{\pi}{4}}$

$\displaystyle\int x\left(\dfrac{\sec 2x-1}{\sec 2x+1}\right)dx$ .

$\displaystyle \int x \Bigg(\cfrac{1-\cos 2x}{1+\cos 2x}\Bigg) dx$

$\displaystyle \int x\tan^2x dx$

$\displaystyle x\sec x - \int \sec x dx$

$\displaystyle x\sec x - \ln|\sec x + \tan x| + C$

Evaluate:

$\int x.\sin{2x}dx$

$\displaystyle \int x sin2xdx$

$\displaystyle \dfrac{-x cos2x}{2}-\dfrac{1}{2}\int cos 2x dx$

$\Rightarrow -(\dfrac{xcos2x}{2}+\dfrac{sin2x}{4})$

1120322_1140093_ans_d8a53c5606ab45619aeff178602cd882.jpg

Evaluate :

$\displaystyle\int_{0}^{\pi} \dfrac{1}{1+\sin x} dx$

$\displaystyle \int_{0}^{\pi } \frac{1}{1+sinx}dx$

$\displaystyle = \int_{0}^{\pi } \frac{1-sinx}{1-sin^{2}x}dx$ Raticnaising the devotion

$\displaystyle = \int_{0}^{\pi } \frac{1-sinx}{cos^{2}x}dx$

$\displaystyle = \int_{0}^{\pi } (sec^{2}x-tan\,x sec\,x) dx$

$\displaystyle | tan x - sec\,x |_{0}^{\pi }$

$\displaystyle = tan \pi -tan\,0 - sec\pi +sec\,0$

$\displaystyle = 0-0-(-1)+1$

$\Rightarrow 1+1 = 2$

1120315_1140112_ans_fb3cb06a1d7b42a18edbdd37c91b4bac.jpg

$\displaystyle\int e^x\left(\dfrac{x-1}{x^2}\right)dx$ .

$\displaystyle \int e^x (\cfrac{1}{x} - \cfrac{1}{x^2})dx$

Using by parts,

$u = \cfrac{1}{x} , v = e^x$

$du = -\cfrac{1}{x^2} , \int v dv = e^x$

$\displaystyle \cfrac{e^x}{x} + \int \cfrac{e^x}{x^2}dx - \int \cfrac{e^x}{x^2}dx$

$\cfrac{e^x}{x} + C$

$\displaystyle\int (x+1)e^xlog (xe^x)dx$ .

Putting

$xe^x = t$

$(xe^x + e^x) dx = dt$

$\displaystyle \int \log t dt$

From by parts, we have

$t \log t - t + C$

$xe^x \log xe^x - xe^x + C$

$\displaystyle\int x \sin^3x dx$ .

$\displaystyle \int \sin^3x dx$

$\displaystyle \int \sin x(1-\cos^2x)dx$

$\displaystyle -\cos x - \int \cos^2 x \sin x dx$

$\displaystyle -\cos x + \int u^2 du$

$\displaystyle -\cos x + \cfrac{\cos^3x}{3} + C$

Integrating by parts, we get

$\displaystyle x(-\cos x +\cfrac{\cos^3x}{3}) - \int (-\cos x + \cfrac{\cos^3x}{3})dx$

$\displaystyle -x\cos x + x \cfrac{\cos^3x}{3} + \sin x -\int \cos^3x dx$

$\displaystyle -x\cos x + x \cfrac{\cos^3x}{3} + \sin x -\int \cos x(1-\sin^2x) dx$

$\displaystyle -x\cos x + x \cfrac{\cos^3x}{3} + \sin x -\int \cos x dx- \int \cos x \sin^2x dx$

$\displaystyle -x\cos x + x \cfrac{\cos^3x}{3} - \int u^2 du$

$\displaystyle -x\cos x + x \cfrac{\cos^3x}{3} - \cfrac{u^3}{3}$

$\displaystyle -x\cos x + x \cfrac{\cos^3x}{3} - \cfrac{\sin^3x}{3}$

Solve:
$\int_0^1 x^3 \sqrt{1 +3x^4 } dx$

$I= \displaystyle\int_{0}^{1}x^{3}\sqrt{1+3x^{4}}dx$

$x^{4}=t\Rightarrow 4x^{3}dx=dt$

$\Rightarrow x^{3}dx=\dfrac{dt}{4}$

$\therefore I= \displaystyle\int_{0}^{1}\sqrt{1+3t}\dfrac{dt}{4}$

$=\left[\dfrac{1}{\dfrac{3}{2}}(1+3t)^{\dfrac{3}{2}}.\dfrac{1}{4}\right]_{0}^{1}$

$=\dfrac{1}{6}\left[4^{\dfrac{3}{2}}-1^{\dfrac{3}{2}}\right]$

$=\dfrac{8-1}{6}=\dfrac{7}{6}$

Integrate:
$\int \sin^4xdx$

$\displaystyle I = \int sin^4 x dx = \int \frac{(1-cos 2x)^2}{4}dx$

$[\because \dfrac{1-cos2x}{2} = sinx]$

$\displaystyle = \frac{1}{4} \int (1-2cos2x+cos^2 2x )dx$

$\displaystyle =\frac{1}{4} \int 1-2cos2x+\frac{(1+cos4x)}{2}dx$

$\displaystyle = \frac{1}{4}\int (\frac{3}{2}-2cos2x+\frac{cos4x}{2})dx$

$\displaystyle = \frac{1}{4} (\frac{3}{2}x - \frac{2sin2x}{2}+\frac{sin{4}x}{2})+c$

$I = \dfrac{3}{8}x-\dfrac{1}{4}sin2x + \dfrac{1}{32}sin4x+c$

$\int \frac{2x^{2}}{3x^{4}2x} dx$

$\int \frac{2x^{2}}{3x^{4}2x}dx = \int \frac{x^{2}dx}{3x^{5}}$

$= \frac{1}{3}\int \frac{dx}{x^{3}}$

$= \frac{1}{3}\times (\frac{x^{-2}}{-2})$ [using

$\int x^{n}= \frac{x^{n+1}}{n+1}]$

$= \frac{-1}{6x^{2}}$

1114061_1200746_ans_b59211b246804e39853d3cdaccf5f220.jpg

$\int {\dfrac{{{e^x}dx}}{{\left( {1 + {e^{2x}}} \right)}}}$

We have,

$I=\int {\dfrac{{{e^x}dx}}{{\left( {1 + {e^{2x}}} \right)}}}$

Let

$t=e^x$

$\dfrac{dt}{dx}=e^x$

$dt=e^x\ dx$

Therefore,

$I=\int \dfrac{dt}{1+t^2}$

$I=\tan^{-1} t+C$

On putting the value of

$t$ , we get

$I=\tan^{-1} e^x+C$

Hence, this is the answer.

Evaluate :
$\displaystyle \int e^{ax} \cos(bx+c) \ dx$

1347407_1259038_ans_139ead3292cd4089a34b9d0495ff89ea.jpg

Evaluate; $\int {\left( {\frac{{3{x^4} - 5{x^3} + 4{x^2} - x + 2}}{{{x^3}}}} \right)dx}$

$\begin{array}{l} \int { \left( { \frac { { 3{ x^{ 4 } }-5{ x^{ 3 } }+4{ x^{ 2 } }-x+2 } }{ { { x^{ 3 } } } } } \right) dx } =\int { \left( { 3x-5+\frac { 4 }{ x } -\frac { 1 }{ { { x^{ 2 } } } } +\frac { 2 }{ { { x^{ 3 } } } } } \right) dx } \, \, \left[ { divide\, each\, term\, by\, { x^{ 3 } } } \right] \\ =3\int { xdx } -5\int { dx } +4\int { \frac { 1 }{ 4 } dx } -\int { { x^{ -2 } }dx } +2\int { { x^{ -3 } }dx } \\ =3.\frac { { { x^{ 2 } } } }{ 2 } -5x+4\log \left| x \right| -\left( { -\frac { 1 }{ x } } \right) +2\left( { \frac { { { x^{ -2 } } } }{ { -2 } } } \right) +C \\ =\frac { { 3{ x^{ 2 } } } }{ 2 } -5x+4\log \left| x \right| -\left( { -\frac { 1 }{ x } } \right) +2\left( { \frac { { { x^{ -2 } } } }{ { -2 } } } \right) +C \\ =\frac { { 3{ x^{ 2 } } } }{ 2 } -5x+4\log \left| x \right| +\frac { 1 }{ x } -\frac { 1 }{ { { x^{ 2 } } } } +C. \end{array}$

$I = \int\limits_0^1 {\frac{{\left( {1 - x} \right)dx}}{{\left( {1 + x} \right)}}}$

$\begin{array}{l} I=\int _{ 0 }^{ 1 }{ \frac { { \left( { 1-x } \right) dx } }{ { \left( { 1+x } \right) } } } \\ =\int _{ 0 }^{ 1 }{ \left( { -1+\frac { 2 }{ { x+1 } } } \right) dx } \, \, \, \, \, \, \, \, \left[ { on\, dividing\, \left( { -x+1 } \right) by\, \left( { x+1 } \right) } \right] \\ =\left[ { -x+2\log \left| { x+1 } \right| } \right] _{ 0 }^{ 1 } \\ =\left[ { \left( { 2\log 2 } \right) -1 } \right] . \end{array}$

Calculate $\int \frac{x}{x^{2}+1}dx$

$\int \dfrac{x}{x^2+1}=\dfrac{1}{2}\int \dfrac{2x}{x^2+1}dx$

Reading

$x^2+1$ as t

$I=\dfrac{1}{2}\int \dfrac{dt}{t}=\dfrac{1}{2}ln|t|=\dfrac{1}{2}ln|x^2+1|$

Solve: $\displaystyle \int \dfrac{x^3}{2x + 1}dx$ .

$\displaystyle\int \dfrac{x^3}{2x+1}dx$

$=\displaystyle\int \dfrac{(x^2/2-x/3+1/8)(2x+1)-1/8}{2x+1}dx$

$=\displaystyle\int \left(\dfrac{x^2}{2}-\dfrac{x}{4}+\dfrac{1}{8}\right)dx-\displaystyle\int \dfrac{dx}{8(2x+1)}$

$=\dfrac{x^3}{6}-\dfrac{x^2}{8}+\dfrac{x}{8}-\dfrac{1}{8}\dfrac{1}{2}ln(2x+1)+c$

$=\dfrac{x^3}{6}-\dfrac{x^2}{8}+\dfrac{x}{8}-\dfrac{1}{16}ln(2x+1)+c$

$\therefore \displaystyle\int \dfrac{x^3}{2x+1}dx=\dfrac{x^3}{6}-\dfrac{x^2}{8}+\dfrac{x}{8}-\dfrac{1}{16}ln(2x+1)+c$ .

1112229_1208590_ans_e8c48254147b4987842cd0e4ae93e333.jpg

Evaluate: $\displaystyle\int \dfrac { \dfrac{- 1} 6 } { ( x + 3 ) } d x + \int \dfrac { \dfrac1 6 } { ( x - 3 ) } d x$

$I= \dfrac{-1}{6} \int \dfrac{dx}{x+3} + \dfrac{1}{6}\int \dfrac{dx}{x-3}$

$I = \dfrac{-1}{6} log(x+3) + \dfrac{1}{6} log (x-3)+c$

$I = \dfrac{1}{6}\left [log(x-3)-log(x+3) \right ]+c$

$I = \dfrac{1}{6} log \left ( \dfrac{x-3}{x+3} \right )+c$

Find the following integrals:
$\int (ax^{2} + bx + c) dx$ .

$I=\displaystyle\int (ax^2+bx+c)dx$

$=\dfrac{ax^3}{3}+\dfrac{bx^2}{2}+cx+d$

$=\dfrac{1}{3}ax^3+\dfrac{1}{2}bx^2+cx+d$ (arbitrary constant).

1177541_1156558_ans_780fa56853fa493a9533fa0da3357a24.jpg

Evaluate
$\int \sqrt{1+t^{3}} dt$

We have,

$I=\int _{ }^{ }{ \sqrt { 1+{ t^{ 3 } } } } dt \\I ={ \int _{ }^{ }{ \left( { 1+{ t^{ 3 } } } \right) } ^{ 1/2 } }dt \\I =\int _{ }^{ }{ \left( { 1+\dfrac { 1 }{ 2 } { t^{ 3 } }-\dfrac { 1 }{ 8 } { t^{ 6 } }....... } \right) } dt \\I =\left( { t+\dfrac { 1 }{ 8 } { t^{ 4 } }-\dfrac { 1 }{ { 56 } } .{ t^{ 7 } }+...... } \right) +C$

Hence, this is the answer.

Solve:
$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)dx}$

$\int_{0}^{1}tan^{-1}(\frac{2x}{1-x^{2}})dx$

$\int_{0}^{1}tan^{-1}(tan x)dx$

$\int_{0}^{1} 2x dx = [2\frac{x^{2}}{2}]_{0}^{1}=1$

1210045_1306954_ans_a069a3e42a6248e1b3a9b3bfd746311a.jpg

Evaluate $\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$

Let

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$ . Then

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}dx$ . Then,

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sec^{2}x}{\sec^{2}x(\sec^{2}x+3\tan^{2}x)}\ dx$ [Diving

$N^{r}$ and

$D^{r}$ by

$\cos^{4}x$ ]

$\Rightarrow I=\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+t^{2})(1+4t^{2})}\ dt$ , where

$t=\tan x$

$\Rightarrow I=-\dfrac{1}{3}\displaystyle\int_{0}^{\infty}\left(\dfrac{1}{1+t^{2}}-\dfrac{4}{1+t^{2}}\right)dt=-\dfrac{1}{3}\left[\tan^{-1}t-2\tan^{-1}2t\right]_{0}^{\infty}$

$\Rightarrow =-\dfrac{1}{3}\left(\dfrac{\pi}{2}-\pi\right)=\dfrac{\pi}{6}$

Evaluate $\int_0^n { xdx}$ .

We have,

$I=\int_0^n { xdx}$

$I=\left[\dfrac{x^2}{2}\right]_0^n$

$I=\left[\dfrac{n^2}{2}-0\right]$

$I=\dfrac{n^2}{2}$

Hence, this is the answer.

$\displaystyle \int { \dfrac { 1 }{ \sqrt { 9-25{ x }^{ 2 } } } } dx$

$\int \frac{dx}{\sqrt{9-25x^{2}}}=\int \frac{dx}{5\sqrt{\frac{9}{25}-x^{2}}}$

$=\frac{1}{5}\int \frac{dx}{\sqrt{(\frac{3}{5})^{2}-x^{2}}}$

$(\because \int \frac{dx}{\sqrt{a^{2-x^{2}}}}=sin^{-1}(\frac{x}{a})+c)$

$=\frac{1}{5}sin^{-1}(\frac{5x}{3})+c$

1232309_1302604_ans_6be00fceb32f4a80a9b0a6d5062cac4e.jpg

Evaluate:
$\int \dfrac{\sqrt{x^{2}-a^{2}}}{x}dx$

We have,

$I=\int { \dfrac { { \sqrt { { x^{ 2 } }-{ a^{ 2 } } } } }{ x } dx }$

$\\ put\, \, x=a\sec \theta \, \, \,$

$dx=a\sec \theta .\tan \theta d\theta \\$

Therefore,

$I =\int { \dfrac { { \sqrt { { a^{ 2 } }{ { \sec }^{ 2 } }\theta -{ a^{ 2 } } } } }{ { a\sec \theta } } } .a\sec \theta \tan d\theta \\ =\int { a\sqrt { \left( { { { \sec }^{ 2 } }\theta -1 } \right) } } .\tan \theta d\theta \\ =a\int { { { \tan }^{ 2 } } } d\theta \\ =a\int { \left( { { { \sec }^{ 2 } }\theta- 1 } \right) } d\theta \\ =a\tan \theta -a.\theta +c$

On putting the value of

$\theta$ , we get

$\\ =a.\dfrac { { \sqrt { { x^{ 2 } }-{ a^{ 2 } } } } }{ a } .a{ \sec ^{ -1 } }\left( { \dfrac { x }{ a } } \right) +c \\ =\sqrt { { x^{ 2 } }-{ a^{ 2 } } } -a{ \sec ^{ -1 } }\left( { \dfrac { x }{ a } } \right) +c$

Hence, this is the answer.

Prove that:
$\displaystyle \int x^5(x+7)dx$

We have,

$I=\int { { x^{ 5 } }( x+7 ) } dx$

$I=\int x^6dx+\int 7x^5 dx$

$I=\dfrac{x^7}{7}+\dfrac{7x^6}{6}+C$

Hence, this is the answer.

Evaluate $\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

Let

$I=\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$ and let

$\sin 2t=u$ . Then,

$d(\sin 2t)=du\Rightarrow 2\cos 2t\ dt=du\Rightarrow \cos 2t dt=\dfrac{1}{2}du$

Also,

$t=0\Rightarrow u=\sin 0=0$ and

$t=\dfrac{\pi}{4}\Rightarrow u=\sin\dfrac{\pi}{2}=1$

$\therefore I=\dfrac{1}{2}\displaystyle\int_{0}^{1}u^{3}du=\dfrac{1}{8}\left[u^{4}\right]_{0}^{1}=\dfrac{1}{8}$

Evaluate : $\int _{ 0 }^{ \pi /2 }{ sinxdx }$ .

$I=\int_{0}^{\pi / 2} \sin x dx$

$I=[{-\cos x}]_{0}^{\pi /2}$

$I=-\cos \dfrac{\pi}{2} + \cos 0$

$I=1$

Solve $\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } x d x$

$\displaystyle\int ^{\pi/2}_{0}\cos^2 x d x=\dfrac{1}{2}\int ^{\pi/2}_{0} (1+\cos 2 x) d x=\dfrac{1}{2}[x+\frac{\sin 2 x }{2}]^{\pi/2}_{0}=\dfrac{\pi}{4}$

Solve it
$2I=\int _{ O }^{ Q }{ dx }$

$2I=\displaystyle\int_{0}^{Q}{dx}$

$\Rightarrow 2I=\left[x\right]_{0}^{Q}$

$\Rightarrow I=\dfrac{1}{2}\left[Q-0\right]$

$\therefore I=\dfrac{Q}{2}$

Evaluate:
$\displaystyle\int\limits_{0}^{1} \dfrac{dx}{1+x^2}$

Now,

$\displaystyle\int\limits_{0}^{1} \dfrac{dx}{1+x^2}$

$=\left[\tan^{-1}x\right]_0^{1}$

$=\dfrac{\pi}{4}-0$

$=\dfrac{\pi}{4}$ .

Prove that:
$\displaystyle \int \dfrac {1}{4x^{2}-1}dx$

$\int \frac{1}{4x^{2}-1}dx=\int \frac{1}{(2x)^{2}-1}dx$

$\Rightarrow$ we know

$\int \frac{1}{x^{2}-a^{2}}=\frac{1}{2a}In \left | \frac{x-a}{x+a} \right |$

then

$\Rightarrow \frac{1}{2} in \left | \frac{2x-1}{2x+1} \right |+c$

1209861_1362354_ans_8626bb0986314fe7aa66c50ec42b9c66.JPG

Evaluate $\displaystyle\int { \left( \dfrac { { x }^{ 2 }+3x+4 }{ \sqrt { x } } \right) } dx$

$\int (\frac{x^{2}+3x+4}{\sqrt{x}})dx$

$\Rightarrow \int (x)^{2-\frac{1}{2}}+3(x)^{1-\frac{1}{2}}+4(x)\frac{-1}{2}$

$\Rightarrow \int x^{3/2}+3x^{1/2}+4x^{-1/2}dx$

$\Rightarrow \frac{2}{5}x^{5/2}+2x^{3/2}+8x^{1/2}+c.$

1211603_1362776_ans_129be0256509482aa7c108c8fce44bff.jpg

Evaluate:
$\displaystyle \int _{0}^\dfrac {\pi}{2}{}(2\log \sin x-\log \sin 2x)dx$

$I=\displaystyle\int\dfrac{\pi}{2}(2log\sin x -log\sin 2x )dx$

$I=\displaystyle\dfrac{\pi}{2}\displaystyle\int_0^{\dfrac {\pi}{2}}[2log\sin x-log (2\sin x\cos x)]dx$

$I=\dfrac{\pi}{2}\displaystyle\int_0^{\dfrac{\pi}{2}}[2log\sin x-log\sin x - log\cos x-log 2]dx$

$I=\dfrac{\pi}{2}\displaystyle\int_0^{\dfrac{\pi}{2}}(2log\sin x-2log \sin x-log 2)dx$

$I=\dfrac{\pi^2}{2}\displaystyle\int^{\dfrac{\pi}{2}}_0(-log 2)dx$

$I=-\dfrac{\pi^2}{4}(log 2)$ .

1172994_1346826_ans_5f39fbd83fc444f6b50475bf57d10aee.jpg

Evaluate : $\displaystyle I = \int {\cfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {\,1 - {x^2}} }}} dx$

We have,

$\displaystyle I = \int {\cfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {\,1 - {x^2}} }}} dx$

Let

$t=\sin^{-1}x$

$\dfrac{dt}{dx}=\dfrac{1}{\sqrt {1-x^2}}$

$dt=\dfrac{dx}{\sqrt {1-x^2}}$

Therefore,

$I=\int t\ dt$

$I=\dfrac{t^2}{2}+C$

On putting the value of

$t$ , we get

$I=\dfrac{(\sin^{-1}x)^2}{2}+C$

Hence, this is the answer.

Evaluate the following definite integrals :
$\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$I=\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$=\displaystyle \int _{0}^{\pi /2} \dfrac {1+\cos 2x}{2} dx$ ........

$using \ (\cos2 x=2cos^2 x-1)$

$=\dfrac {1}{2} \left [x +\dfrac {\sin 2x}{2} \right]_0^{\pi /2}$

$=\dfrac {1}{2}\left [\left (\dfrac {\pi}{2}+\dfrac {\sin \pi}{2} \right) -\left (0+\dfrac {\sin 0}{2}\right) \right] =\dfrac {\pi}{4}$

Evaluate $\displaystyle\int_{0}^{\pi/3}\dfrac{\cos x}{3+4\sin x}dx$

$\displaystyle \int ^{\dfrac{\pi}{3}}_{0}\dfrac{cosx}{3+4sinx}dx$

$=\displaystyle \dfrac{1}{4}\times ln|3+4sinx|^{\dfrac{\pi}{3}}_{0}$

$\left[\displaystyle \because \int{\dfrac{f'(x)}{f(x)}}dx=\log|f(x)|\right]$

$=\dfrac{1}{4}\times \left(ln\left(3+4\left(\dfrac{\sqrt{3}}{2}\right)\right)-ln(3+4(0))\right)$

$=\dfrac{ln(3+2\sqrt{3})-ln(3)}{4}$

Evaluate the definite integral :
$\displaystyle \int_{0}^{5}x^4\ dx$

$\displaystyle \int_{0}^{5}x^4\ dx$

$=\left[\dfrac {x^5}5\right]_0^5$

$= \dfrac {5^5}{5}$

$=625$

Evaluate the definite integral:
$\displaystyle \int_{0}^{\pi /2} \cos x\ dx$

$\displaystyle \int_{0}^{\pi /2} \cos x\ dx$

$=\left[ -\sin x \right]_0^{\pi/2}$

$=-1-0$

$=-1$

Evaluate: $\displaystyle\int_{0}^{1}\dfrac{\sqrt{\tan^{-1}x}}{1+x^{2}}dx$

$I=\displaystyle\int_{0}^{1}\dfrac{\sqrt{\tan^{-1}x}}{1+x^{2}}dx$

Let

$\tan^{-1}x=t$

$\dfrac{1}{1+x^2}=\dfrac{dt}{dx}$ [ Differentiate both sides w.r.t

$x$ ]

$\dfrac{1}{1+x^2}dx=dt$

When,

$x=0$

$\tan^{-1}0=t$

$\therefore$

$t=0$

When,

$x=1$

$\tan^{-1}1=t$

$\therefore$

$t=\dfrac{\pi}{4}$

$\Rightarrow$

$I=\displaystyle\int_{0}^{1}\dfrac{\sqrt{\tan^{-1}x}}{1+x^{2}}dx$

$=\displaystyle\int_0^{\pi/4}\sqrt{t}dt$

$=\left[\dfrac{2t^{\frac{3}{2}}}{3}\right]_0^{\pi/4}$

$=\dfrac{2}{3}\left(\dfrac{\pi}{4}\right)^{\frac{3}{2}}$

$=\dfrac{1}{12}\pi^{\frac{3}{2}}$

Evaluate $\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}} \ dx$

Let

$I=\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}} \ dx$

and

$\tan^{-1}x=t$ .

Then,

$d(\tan^{-1}x)=dt\Rightarrow dx=(1+x^{2})dt$

Also,

$x=0\Rightarrow t=\tan^{-1}0=0$

$x=1\Rightarrow t=\tan^{-1}1=\dfrac{\pi}{4}$

$\therefore I=\displaystyle\int_{0}^{\pi/4}t\ dt=\left[\dfrac{t^{2}}{2}\right]_{0}^{\pi/4}=\dfrac{\pi^{2}}{32}$

Evaluate $\displaystyle\int_{4}^{12}x\ dx$

$\displaystyle\int_{4}^{12}x\ dx$

$=\left[\dfrac {x^2}2\right]_4^{12}$

$=\dfrac {144}2-\dfrac {16}2$

$=72-8=64$

Evaluate $\displaystyle \int_{2}^{3}x^{2}dx$

$\displaystyle^{3}_{2}\int x^2dx$

$=\left[\dfrac{x^3}{3}\right]^3_2$

$=\left[\dfrac{3^3}{3}-\dfrac{2^3}{3}\right]$

$=\left[\dfrac{3^3-2^3}{3}\right]$

$=\left[\dfrac{27-8}{3}\right]$

$=\dfrac{19}{3}$ .

1228259_1502039_ans_31200c74be714ec7a0a62fb0f09d5cb0.jpeg

$\int \dfrac{1}{1-cos\dfrac{x}{2}}$ dx

As we know that

$1-\cos 2\theta=2\sin^2 \theta$

$\displaystyle\int \dfrac{d x}{1-\cos \frac{x}{2}}=\int \dfrac{d x}{2\sin^2 \frac{x}{4}}$

$=\dfrac{1}{2}\displaystyle\int \text{cosec}^2 \frac{x}{4} d x$

$=\dfrac{1}{2}\times \dfrac{1}{\frac{1}{4}}(-\cot \frac{x}{4})+C$

$=-2\cot \dfrac{x}{4}+C$

Evaluate:
$\displaystyle\int \dfrac{1}{2x+3}dx$

$\displaystyle\int \dfrac {1}{2x+3}dx$

Let

$t=2x+3 \implies dt=2 dx$ ........(differentiating with respect to

$x$ )

Replacing we get,

$\displaystyle \int \dfrac 1{2t }dt$

$=\dfrac {1}{2} \log t$

$=\log t^\dfrac{1}{2}$

$=\log \sqrt {4x+6}+C$

Evaluate the following definite integrals :
$\displaystyle \int_{0}^{\frac {\pi} {2}}\sin^3 x\cos x\ dx$

$\displaystyle \int_{0}^{\pi /2}\sin^3 x\cos x\ dx$

Let

$\sin x =t\implies \cos x \ dx=dt$

$x\to 0\to \dfrac {\pi}2\\t \to 0\to 1$

$\displaystyle \int _0^1 t^3 dt$

$=\left[\dfrac {t^4}4\right]_0^1$

$=\dfrac 14$

Evaluate:
$\displaystyle\int_{2}^{1}|x-3|\ dx$

$\displaystyle\int_{1}^{2}|x-3|\ dx=\displaystyle\int_{1}^{2}-(x-3)\ dx$

$[\because x-3<0\ for\ 1\le x\le 2]$

$=-\left[\dfrac {x^2}2+3x\right]_2^1$

$=-\dfrac 12+3+\dfrac42 -6$

$=-\dfrac 32$

Evaluate the definite integral:
$\displaystyle\int_{-1}^{1}5x^{4}\sqrt{x^{5}+1}\ dx$

Let

$I=\displaystyle\int_{-1}^{1}5x^{4}\sqrt{x^{5}+1}\ dx$ and let

$x^{5}+1=t$ .

differentiating with respect to

$x$ , we get

$d(x^{5}+1)=dt\Rightarrow 5x^{4}dx=dt$

Also,

$x=-1\Rightarrow t=0$

$x=1\Rightarrow t=2$

$\therefore \displaystyle\int_{0}^{2}\sqrt{t}dt=\dfrac{2}{3}\left[t^{3/2}\right]_{0}^{2}=\dfrac{2}{3}\times 2^{3/2}=\dfrac{4\sqrt{2}}{3}$

Evaluate the definite integral :
$\displaystyle \int_{0}^{\pi /2} \sin x \ dx$

$\displaystyle \int_{0}^{\pi /2} \sin x \ dx$

$=\left[ \cos x\right]_0^{\pi/2}$

$= 0-1$

$=-1$

Evaluate $\displaystyle\int_{0}^{1}\sqrt x \ dx$

$\displaystyle\int_{0}^{1}\sqrt x \ dx$

$=\left [\dfrac {2x^{3/2}}{3} \right]_0^1$

$=\dfrac 23$

Evaluate the definite integral:
$\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$

Let

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$ . Then

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}dx$ . Then,

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sec^{2}x}{\sec^{2}x(\sec^{2}x+3\tan^{2}x)}\ dx$ [Diving

$Numerator$ and

$Denominator$ by

$\cos^{4}x$ ]

$\Rightarrow I=\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+t^{2})(1+4t^{2})}\ dt$ , where

$t=\tan x$

$\Rightarrow I=-\dfrac{1}{3}\displaystyle\int_{0}^{\infty}\left(\dfrac{1}{1+t^{2}}-\dfrac{4}{1+t^{2}}\right)dt=-\dfrac{1}{3}\left[\tan^{-1}t-2\tan^{-1}2t\right]_{0}^{\infty}$

$\Rightarrow =-\dfrac{1}{3}\left(\dfrac{\pi}{2}-\pi\right)=\dfrac{\pi}{6}$

Evaluate the definite integral:
$\displaystyle \int_{0}^{\pi /2} \sin 2x\ dx$

$\displaystyle \int_{0}^{\pi /2} \sin 2x\ dx\\$

$\displaystyle \int_{0}^{\pi /2} 2\sin x \cos x \ dx\\$

differentiating with respect to

$x$ , we get

$t=\sin x \implies dt=\cos x dx$

Replacing we get,

$\displaystyle \int 2tdt$

$=2 \cdot \dfrac {t^2}{2}+C$

$=t^2+c$

$=\sin ^2 x+c$

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$ is equal to:

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$

$\sin x=t\implies \cos x dx=dt$

As,

$x\to 0\to \dfrac \pi 2$

$t\to 0\to 1$

$\displaystyle \int _0^{1} t dt$

$=\left[\dfrac {t^2}2\right|^1_0$

$=\dfrac 12-0$

$=\dfrac 12$

Evaluate $\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cdot \cos 2t\ dt$

Let

$I=\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

and

let

$\sin 2t=u$ .

Then,

$d(\sin 2t)=du\Rightarrow 2\cos 2t\ dt=du\Rightarrow \cos 2t dt=\dfrac{1}{2}du$

Also,

$t=0\Rightarrow u=\sin 0=0$

$t=\dfrac{\pi}{4}\Rightarrow u=\sin\dfrac{\pi}{2}=1$

$\therefore I=\dfrac{1}{2}\displaystyle\int_{0}^{1}u^{3}du=\dfrac{1}{8}\left[u^{4}\right]_{0}^{1}=\dfrac{1}{8}$

Evaluate the following integral :
$\displaystyle\int_{0}^{\pi}x\sin x\cos^{4}x\ dx$

Let

$I=\displaystyle\int_{0}^{\pi}x\sin x\cos^{4}x\ dx$ ...........(i)

$\because \int_{a}^b f(x) \ dx= \int _{a}^{b} f(a+b-x) \ dx$

Then,

$I=\displaystyle\int_{0}^{\pi}(\pi-x)\sin (\pi-x)\cos^{4}(\pi-x)dx=\displaystyle\int_{0}^{\pi}(\pi-x)\sin x\cos^{4}x\ dx$ ..............(ii)

Adding

$(1)$ and

$(2)$ , we get

$\therefore 2I=-\pi\left[\dfrac{t^{5}}{5}\right]_{1}^{-1}$

$=-\pi\left(-\dfrac{1}{5}-\dfrac{1}{5}\right)$

$=\dfrac{2\pi}{5}$

$\Rightarrow I=\dfrac{\pi}{5}$

Evaluate the following definite integral:

$\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$

$I=\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$ .

Let

$x+2=t^{2}$ . Then ,

$dx=2t dt$

Also

$x=0\Rightarrow t^{2}=2\Rightarrow t=\sqrt{2}$ and,

$x=2\Rightarrow t^{2}=4\Rightarrow t=2$

$\therefore I=\displaystyle\int_{\sqrt{2}}^{2}(t^{2}-2)\sqrt{t^{2}}2t dt$

$=2\displaystyle\int_{\sqrt{2}}^{2}(t^{4}-2t^{2})dt$

$=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$

$=2\left[\left(\dfrac{2^{5}}{5}-\dfrac{2(2)^{3}}{3}\right)-\left(\dfrac{(\sqrt{2})^{5}}{5}-\dfrac{2(\sqrt{2})^{3}}{3}\right)\right]$

$\Rightarrow I=2\left[\left(\dfrac{32}{5}-\dfrac{16}{3}\right)-\left(\dfrac{4\sqrt{2}}{5}-\dfrac{4\sqrt{2}}{3}\right)\right]$

$=2\left(\dfrac{16}{15}+\dfrac{8\sqrt{2}}{15}\right)$

$=\dfrac{32+16\sqrt{2}}{15}$

Evaluate the following definite integral :
$\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$I=\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$=\displaystyle \int _{0}^{\pi /2} \dfrac {1+\cos 2x}{2} dx$ ...........................

$(Using \ \ 2cos^2{x}=1+sin 2x)$

$=\dfrac {1}{2} \left [x +\dfrac {\sin 2x}{2} \right]_0^{\pi /2}$

$=\dfrac {1}{2}\left [\left (\dfrac {\pi}{2}+\dfrac {\sin \pi}{2} \right) -\left (0+\dfrac {\sin 0}{2}\right) \right]$

$=\dfrac {\pi}{4}$

Evaluate $\displaystyle \int_{0}^{1} \dfrac {1}{\sqrt {1+x}-\sqrt {x}} \, dx$

$I=\displaystyle \int_{0}^1\dfrac {1}{\sqrt {1+x}-\sqrt x}dx$

Let's rationalize, Multiplying and Dividing by

${\sqrt {1+x}+\sqrt x}$

$I=\displaystyle \int_{0}^1\dfrac {1}{\sqrt {1+x}-\sqrt x} \dfrac{{\sqrt {1+x}+\sqrt x}}{{\sqrt {1+x}+\sqrt x}}dx$

$=\displaystyle \int_{0}^1\dfrac {\sqrt {1+x}+\sqrt x}{(1+x)-x}dx$

$=\displaystyle \int_{0}^1\left\{\sqrt {1+x}+\sqrt x \right\}dx$

$\Rightarrow \ I=\left [\dfrac {2}{3} (1+x)^\frac{3}{2} +\dfrac {2}{3} x^\frac{3}{2}\right]_0^1$

$=\dfrac {2}{3} (1+1)^\frac{3}{2}+\dfrac {2}{3} 1^\frac{3}{2}-\dfrac {2}{3} (1+0)^\frac{3}{2}-0$

$=\dfrac {4\sqrt 2}{3}+\dfrac {2}{3}-\dfrac {2}{3}$

$=\dfrac {4\sqrt 2}{3}$

$\displaystyle \int_{0}^{1} (1+x)^5 dx$

$\displaystyle \int_{0}^{1}{{\left( 1 + x \right)}^{5} \; dx}$

Let

$1 + x = t \\ \Rightarrow dx = dt$

Therefore,

$\displaystyle \int_{1}^{2}{{t}^{5} \; dt}$

$\displaystyle = \left[ \cfrac{{t}^{6}}{6} \right]_{1}^{2}$

$= \cfrac{{2}^{6}}{6} - \cfrac{{1}^{6}}{6}$

$= \cfrac{64 - 1}{6} = \cfrac{21}{2}$

Solve $\displaystyle \int_{0}^{\pi} {1+ \dfrac {x}{2}}dx$

$\displaystyle \int_{0}^{\pi} {1+ \dfrac {x}{2}}dx$

$=x+\dfrac{x^3}6\bigg|_0^\pi$

$=\left(\pi+\dfrac{\pi^3}6\right)-\left(0+\dfrac{0^3}6\right)$

$=\pi+\dfrac{{\pi}^3}6$

Evaluate the following integral :
$\displaystyle\int_{0}^{\pi} ({1+\sin x})\ dx$

$\displaystyle\int_{0}^{\pi} ({1+\sin x})\ dx$

Using

$\int \sin x \ dx=-\cos x+C$

$=\left [x-\cos x\right|_0^\pi$

$=\pi-(-1)-0+1$

$=\pi +2$

Evaluate the following definite integral:
$\displaystyle \int_{0}^{1} \dfrac {1}{\sqrt {1+x-\sqrt x}}dx$

$I=\displaystyle \int_{0}^1\dfrac {1}{\sqrt {1+x}-\sqrt x}dx$

$=\displaystyle \int_{0}^1\dfrac {\sqrt {1+x}+\sqrt x}{(\sqrt {1+x}-\sqrt x){(\sqrt {1+x}+\sqrt x})}dx$

$=\displaystyle \int_{0}^1\dfrac {\sqrt {1+x}+\sqrt x}{(1+x)-x}dx$ .......

${(a-b)(a+b)=a^2-b^2}$

$=\displaystyle \int_{0}^1\left\{\sqrt {1+x}+\sqrt x \right\}dx$

$\Rightarrow \ I=\left [\dfrac {2}{3} (1+x)^{3/2} +\dfrac {2}{3} x^{3/2} \right]_0^1$

$I=\left(\dfrac {2}{3} (1+1)^{3/2} +\dfrac {2}{3} 1^{3/2}\right)-\left(\dfrac {2}{3} (1+0)^{3/2} +\dfrac {2}{3} 0^{3/2}\right)$

$=\dfrac {4\sqrt 2}{3}+\dfrac {2}{3}-\dfrac {2}{3}$

$I=\dfrac {4\sqrt 2}{3}$

Evaluate the following integral:

$\displaystyle \int_{0}^{1} \ x+x^2 dx$

$\displaystyle \int_{0}^{1} \ x+x^2 dx$

$\implies\left.\dfrac {x^2}2+\dfrac {x^3}3\right|_0^1$

$\implies \left(\dfrac {1^2}{2}+\dfrac {1^3}{3} \right) - \left(\dfrac {0^2}{2}+\dfrac {0^3}{3} \right)$

$\implies \dfrac 12+\dfrac 13=\dfrac 56$

Evaluate $\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$

Let

$I=\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$

and

$\tan^{-1}x=t$ .

Then,

$d(\tan^{-1}x)=dt\Rightarrow dx=(1+x^{2})dt$

Also,

$x=0\Rightarrow t=\tan^{-1}0=0$

$x=1\Rightarrow t=\tan^{-1}1=\dfrac{\pi}{4}$

$\therefore I=\displaystyle\int_{0}^{\pi/4}t\ dt=\left[\dfrac{t^{2}}{2}\right]_{0}^{\pi/4}=\dfrac{\pi^{2}}{32}$

Evaluate the following definite integral:
$\displaystyle \int_{1}^{2}\dfrac {1}{\sqrt {(x-1) (2-x)}}dx$

$I=\displaystyle \int_{1}^{2} \dfrac {1}{\sqrt {-x^2 +3x-2}}dx =\displaystyle \int_{1}^{2} \dfrac {1}{\sqrt {-\left\{ \left(x-\dfrac {3}{2}\right)^2-\left(\dfrac {1}{2}\right)^2 \right\}}}dx =\displaystyle \int_{1}^{2} \dfrac {1}{\sqrt {\left (\dfrac {1}{2}\right)^2-\left (x-\dfrac {3}{2}\right)^2}}dx$

$\displaystyle \int \dfrac 1{\sqrt {a^2-x^2}}=\sin ^{-1}\dfrac xa \\\displaystyle \int _1^2\dfrac 1{\sqrt {\left(\dfrac 12\right)^2-\left(x-\dfrac 32\right)^2}}=\left.\sin ^{-1}\left(\dfrac {x-\dfrac {3}{2}}{\dfrac 12}\right)\right|_1^2\\\left.\sin^{-1} (2x-3)\right|_1^2=\sin ^{-1}(1)-\sin ^{-1}(-1)=\dfrac {\pi}2+\dfrac {\pi}2=\pi$

Evaluate the following definite integral:
$\displaystyle\int_{0}^{1}\sqrt x \ dx$

Consider,

$I=\displaystyle\int_{0}^{1}\sqrt x \ dx$

$I=\left [\dfrac {2x^{3/2}}{3} \right]_0^1$

$I=\dfrac 23$

Evaluate $\displaystyle \int_{1}^{2} \dfrac {x}{(x+1)(x+2)}dx$

$I=\displaystyle \int_{1}^2\dfrac {x}{(x+1)(x+2)}dx$

$=\displaystyle \int_{1}^2\dfrac {2x+2-x-2}{(x+1)(x+2)}dx$

$=\displaystyle \int_{1}^2\dfrac {2(x+1)-(x+2)}{(x+1)(x+2)}dx$

$\displaystyle \int_{1}^2 \left (-\dfrac {1}{x+1} +\dfrac {2}{x+2} \right)dx$

$\Rightarrow \ I=\left [-\log (x+1)+2 \log (x+2)\right]_1^2$

$=\left [\log \dfrac {(x+2)^2}{x+1}\right]_1^2$

$=\left [\log \dfrac {(2+2)^2}{2+1}\right]-\left [\log \dfrac {(1+2)^2}{1+1}\right]$

$=\log \dfrac {16}{3}-\log \dfrac {9}{2}$

$=\log \dfrac {32}{27}$

$\displaystyle \int_{0}^{\pi /2}\frac{\cos ^{2}x\sin x}{\sqrt{(1+\cos ^{2}x)}}dx=\frac{1}{n}[\sqrt{k}-log(\sqrt{m}+1)]$ .Find $k+m+n$ ?

Let

$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { \cos ^{ 2 }{ x } \sin { x } dx }{ \sqrt { \left( 1+\cos ^{ 2 }{ x } \right) } } }$
Substitute

$\cos { x } =\tan { t } \Rightarrow -\sin { x } dx=\sec ^{ 2 }{ t } dt$

$\displaystyle \therefore I=-\int _{ -\dfrac { \pi }{ 4 } }^{ 0 }{ \dfrac { \tan ^{ 2 }{ t } .\sec ^{ 2 }{ t } }{ \sec { t } } } dt\Rightarrow I=\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \tan { t } . } \sec { t } .\tan { t } dt$

$\displaystyle =\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \sqrt { \sec ^{ 2 }{ t-1 } } } \sec { t } \tan { t } dt$
Substitute

$\sec { t } =z$

$\displaystyle \therefore I=\int _{ 1 }^{ \sqrt { 2 } }{ \sqrt { { z }^{ 2 }-1 } } dx=\left[ \dfrac { z }{ 2 } \sqrt { { z }^{ 2 }-1 } -\dfrac { 1 }{ 2 } \log { z } +\sqrt { { z }^{ 2 }+1 } \right] _{ 1 }^{ \sqrt { 2 } }$

$\displaystyle =\dfrac { 1 }{ 2 } \sqrt { 2 } -\dfrac { 1 }{ 2 } \left[ \log { \left( \sqrt { 2 } +1 \right) -0 } \right] =\dfrac { 1 }{ 2 } \left[ \sqrt { 2 } -\log { \left( \sqrt { 2 } +1 \right) } \right]$

$\displaystyle \int_{-1}^{1}\frac{x\sin ^{-1}x}{\sqrt{\left ( 1-x^{2} \right )}}dx= ??$

Let

$\displaystyle I=\int _{ -1 }^{ 1 } \frac { x\sin ^{ -1 } x }{ \sqrt { \left( 1-x^{ 2 } \right) } } dx$
And

$\displaystyle f\left( x \right) =\frac { x\sin ^{ -1 } x }{ \sqrt { \left( 1-x^{ 2 } \right) } }$

$\displaystyle f\left( -x \right) =\frac { \left( -x \right) \sin ^{ -1 } \left( -x \right) }{ \sqrt { \left[ 1-\left( -x \right) ^{ 2 } \right] } } =\frac { x\sin ^{ -1 } x }{ \sqrt { \left( 1-x^{ 2 } \right) } } =f\left( x \right)$

$\displaystyle I=2\int _{ 0 }^{ 1 } \frac { x\sin ^{ -1 } x }{ \sqrt { \left( 1-x^{ 2 } \right) } } dx$
Substitute

$x=\sin \theta$

$\displaystyle \therefore I=2\int _{ 0 }^{ \pi /2 } \theta \frac { \sin \theta \cos \theta d\theta }{ \cos \theta } =2\int _{ 0 }^{ \pi /2 } \theta \sin \theta d\theta \\ =2\left\{ \theta \left( -\cos \theta \right) +\int _{ 0 }^{ 1 } \cos \theta .1d\theta \right\} \\ =2\left\{ -\theta \cos \theta +\sin \theta \right\} _{ 0 }^{ \pi /2 }=2\left[ \left( 0+1 \right) -\left( 0 \right) \right] =2$

The value of $\displaystyle 3\int _{ 0 }^{ \pi/2 }{ \sqrt { \cos x-\cos ^{ 3 } x } dx }$ is

Let

$I=\int { \sqrt { \cos x-\cos ^{ 3 } x } } \: dx=\int { \sqrt { \cos x\sin ^{ 2 }{ x } } } \: dx=\int { \sin { x } \sqrt { \cos x } } \: dx$
Substituting

$u=\cos { x } \Rightarrow du=-\sin { x } dx$

$I=-\int { \sqrt { u } du } =-\cfrac { 2{ u }^{ 2/3 } }{ 3 } +c=-\cfrac { 2{ \cos { x } }^{ 2/3 } }{ 3 } +c$
Therefore,

$3\int _{ 0 }^{ \pi/2 }{ \sqrt { \cos x-\cos ^{ 3 } x } dx } =2$

The value of $\displaystyle \int_{0}^{1/2}\frac{dx}{\sqrt{\left ( x-x^{2} \right )}}$ is $\displaystyle \frac{\pi }{k}.$ Find the value of $k$ .

$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ \sqrt { x-{ x }^{ 2 } } } dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ \frac { 1 }{ \sqrt { \frac { 1 }{ 4 } -{ \left( x-\frac { 1 }{ 2 } \right) }^{ 2 } } } dx }$

Put

$\displaystyle t=x-\frac { 1 }{ 2 } \Rightarrow dt=dx$

$\displaystyle I=\int _{ -\frac { 1 }{ 2 } }^{ 0 }{ \frac { 1 }{ \sqrt { \frac { 1 }{ 4 } -{ t }^{ 2 } } } } dt=2\int _{ -\frac { 1 }{ 2 } }^{ 0 }{ \frac { 1 }{ \sqrt { 1-{ 4t }^{ 2 } } } } dt$

$\displaystyle =\left[ sin^{ -1 }{ 2t } \right] _{ -\frac { 1 }{ 2 } }^{ 0 }=0-sin^{ -1 }{ \left( \frac { -2 }{ 2 } \right) }=\frac { \pi }{ 2 }$

$\displaystyle \int_{0}^{\pi. }\frac{dx}{1+2\sin ^{2}x}= \frac{\pi }{\sqrt{\left ( k \right )}}$
what is k?

Let

$\displaystyle I=\int _{ 0 }^{ \pi } \frac { dx }{ 1+2\sin ^{ 2 } x } =2\int _{ 0 }^{ \pi /2 } \frac { dx }{ 1+2\sin ^{ 2 } x }$

$\displaystyle =2\int _{ 0 }^{ \pi /2 } \frac { \sec ^{ 2 } xdx }{ 1+\tan ^{ 2 } x+2\tan ^{ 2 } x }$
Put

$\tan x=t$

$\displaystyle \therefore I=2\int _{ 0 }^{ \infty } \frac { dt }{ 1+3t^{ 2 } } =\frac { 2 }{ \sqrt { \left( 3 \right) } } \left[ \tan ^{ -1 } t\sqrt { \left( 3 \right) } \right] _{ 0 }^{ \infty }$

$\displaystyle =\frac { 2 }{ \sqrt { \left( 3 \right) } } \left( \frac { \pi }{ 2 } -0 \right) =\frac { \pi }{ \sqrt { \left( 3 \right) } }$

Solve $\displaystyle \int_{0}^{\infty }\log \left ( x+\frac{1}{x} \right )\frac{dx}{1+x^{2}}$

$\displaystyle \int_{0}^{\infty }\log \left ( x+\frac{1}{x} \right )\frac{dx}{1+x^{2}}$

Put

$\displaystyle x= \tan \theta$

$\displaystyle \therefore dx= \sec ^{2}\theta d\theta$

When

$x=0 \Rightarrow \theta=0$

When

$x=\infty \Rightarrow \theta=\cfrac{\pi}{2}$

$\displaystyle \therefore I= \int_{0}^{\pi /2}\log \left ( \tan \theta +\cot \theta \right )d\theta$

$\displaystyle = \int_{0}^{\pi /2}\log \frac{\sin ^{2}\theta +\cos ^{2}\theta }{\sin \theta \cos \theta }d\theta$

$\displaystyle = -\int_{0}^{\pi /2}\log \sin \theta \cos \theta d\theta$

We know that

$\displaystyle \int_{0}^{\pi/2} \log\sin\theta =-\cfrac{\pi}{2}\log 2$

$\displaystyle \int_{0}^{\pi/2} \log\cos\theta =-\cfrac{\pi}{2}\log 2$

$\displaystyle = -\left [ \int_{0}^{\pi /2}\log \sin \theta d\theta +\int_{0}^{\pi /2}\log \cos \theta d\theta \right ]$

$\displaystyle = -\left [ -\frac{\pi }{2}\log 2-\frac{\pi }{2}\log 2\right ]$

$\displaystyle I = \pi \log 2.$

Prove that $\displaystyle \int_{0}^{\infty }\frac{\log \left ( 1+x^{2} \right )}{1+x^{2}}dx= {\pi} \log 2$

Substitute

$\displaystyle x= \tan \theta \therefore dx= \sec ^{2}\theta d\theta .$
The limits are adjusted as 0 to

$\displaystyle \pi /2$

$\displaystyle \therefore I= \int_{0}^{\pi /2}\frac{\log \sec ^{2}\theta }{\sec ^{2}\theta }\sec ^{2}\theta d\theta = \int_{0}^{\pi /2}\log \sec ^{2}\theta d\theta$
or

$\displaystyle I= 2\int_{0}^{\pi /2}\log \sec \theta d\theta = -2\int_{0}^{\pi /2}\log \cos \theta d\theta$

$\displaystyle = -2.\frac{\pi }{2}\log \frac{1}{2},$

$\displaystyle = -\pi \log \frac{1}{2}= \pi \log 2.$

$\displaystyle I= \int_{0}^{\pi }\frac{x\tan x}{\sec x+\tan x}dx= \frac{\pi }{C}\left ( \pi -2 \right ).$
What is C?

$\displaystyle I=\int _{ 0 }^{ \pi } \frac { x\tan x }{ \sec x+\tan x } dx$ ...(1)
Using

$\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx }$

$\displaystyle \therefore I=\int _{ 0 }^{ \pi } \frac { \left( \pi -x \right) \tan \left( \pi -x \right) }{ \sec \left( \pi -x \right) +\tan \left( \pi -x \right) } dx=\int _{ 0 }^{ \pi } \frac { \left( \pi -x \right) \tan x }{ \sec x+\tan x }$ ...(2)
Adding (1) and (2)

$\displaystyle 2I=\pi \int _{ 0 }^{ \pi } \frac { \tan xdx }{ \sec x+\tan x } =\pi \int _{ 0 }^{ \pi } \frac { \tan x\left( \sec x-\tan x \right) }{ 1 } dx$

$\displaystyle \therefore I=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi } \left[ \sec x\tan x-\left( \sec ^{ 2 } x-1 \right) \right] dx=\frac { \pi }{ 2 } \left[ x-\tan x+\sec x \right] _{ 0 }^{ \pi }$

$\displaystyle =\frac { \pi }{ 2 } \left[ \left( \pi -0-1 \right) -\left( 0-0+1 \right) \right] =\frac { \pi }{ 2 } \left( \pi -2 \right)$

Evaluate: $\int_{2}^{4} \dfrac {x}{x^{2} + 1}dx$

$\int _{ 2 }^{ 4 }{ \dfrac { x }{ { x }^{ 2 }+1 } } dx$

let

${x}^{2}+1$ =

$t$ , we get

$2xdx = dt$ ( the limits of

$t$ are

$5$ to

$17$ )

Therefore the given expression reduces to

$\int _{ 5 }^{ 17 }{ \dfrac { 1 } { 2t} }dt = \dfrac { 1 }{ 2 } (\ln { 17 } -\ln { 5 } ) .$

Evaluate $\displaystyle\int _{ 0 }^{ \pi }{ \log { \left( 1-\cos { x } \right) } dx }$ .

$\displaystyle \int _{ 0 }^{ \pi }{ \log(1+\cos x)dx } =\int _{ 0 }^{ \pi /2 }{ \log(1+\cos x)dx } +\int _{ 0 }^{ \pi /2 }{ \log(1+\cos(\pi -x))dx }$

$\displaystyle=\int _{ 0 }^{ \pi /2 }{ \log({\sin }^{ 2 }x)dx } =2\int _{ 0 }^{ \pi/2 }{\log({\sin} x)dx }$

$I=\displaystyle\int _{ 0 }^{ \pi }{\log({\sin }x)dx } =2\int _{ 0 }^{ \pi /2 }{\log({\sin }2t)dt } =2\int _{ 0 }^{ \pi /2 }{\log(2\sin t\cos t)dt }$

$=\pi \log2+2\int _{ 0 }^{ \pi /2 }{\log(\sin t)dt } +2\int _{ 0 }^{ \pi /2 }{\log(\cos t)dt } \\ =\pi \log 2+2I$

$\Rightarrow I=-\pi \log2$

$\therefore \int _{ 0 }^{ \pi }{ log(1+cosx)dx } =-\pi \log2$

If $\int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx=\dfrac{(\alpha x^6+\beta x^4+\gamma x^2)^{3/2}}{18}+C$ where C is constant then find the value of $(\beta + \gamma-\alpha).$

$\int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx = \int (x^5+x^3+x)\sqrt{2x^6+3x^4+6x^2}dx$

Let

$2x^6+3x^4+6x^2=t$

$\therefore dt=12(x^5+x^3+x)$

$\implies \int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx = \int \dfrac{1}{12}\sqrt{t}dt$

$=\dfrac{1}{12}\times\dfrac{2}{3}t^{\dfrac{3}{2}}+c$

$=\dfrac{1}{18}(2x^6+3x^4+6x^2)^{\dfrac{3}{2}}+c$

$\implies \beta+\gamma-\alpha = 7$

Evaluate:
$\displaystyle \int _{ 1 }^{ e }{ \left( \frac{1}{\sqrt{x \ln x}} + \sqrt{\frac{\ln x}{x}} \right) dx }$

Let

$I=\displaystyle \int _{ 1 }^{ e }{ \left( \frac{1}{\sqrt{x ln x}} + \sqrt{\frac{ln x}{x}} \right) dx }$

Take

$x={ e }^{ y }$ ,

$\Rightarrow dx={ e }^{ y }dy$

So,

$x\rightarrow1 \Rightarrow e^y \rightarrow \ln 1$ and

$x\rightarrow e \Rightarrow e^y \rightarrow \ln e$

Substituting in the integral and changing it's limits,

$\Rightarrow \displaystyle \int _{ \ln { 1 } }^{ \ln { e } }{ \left(\dfrac { { e }^{ y } }{ \sqrt { y{ e }^{ y } } } +\sqrt { \dfrac { y }{ { e }^{ y } } } { e }^{ y }\right) } dy$

$\displaystyle =2\int _{ 0 }^{ 1 }{ { e }^{ y/2 } } \left(\dfrac { 1 }{ 2 } \dfrac { 1 }{ \sqrt { y } } +\dfrac { 1 }{ 2 } \sqrt { y } \right)dy$

$\displaystyle =2\sqrt { y{ e }^{ y } } { ] }_{ 0 }^{ 1 }=2\sqrt { e }$

Solve $\int {r^5} {e^x} dx$ :
(Given $r$ is constant)

$\int{r^5 e^x}dx = r^5 \int {e^x} dx$ since,

$r$ is a constant.

Therefore

$\int {e^x} dx = e^x$

$\int{r^5 e^x} dx = r^5 e^x$

Evaluate: $\displaystyle \int_{2}^{3} \dfrac {x\ dx}{x^{2} + 1}$ .

$\displaystyle \int_{2}^{2} \dfrac{x\ dx}{x^{2} + 1} = \int_{5}{10} \dfrac {\dfrac {dt}{2}}{t}$ Put

$x^{2} + 1 = t$

$= \dfrac {1}{2}\displaystyle \int_{5}^{5} \dfrac {dt}{t} = \dfrac {1}{2}\log |t||_{5}^{10}\ 2x\ dx = dt\Rightarrow x\ dx = \dfrac {dt}{2}$

$= \dfrac {1}{2}[\log 10 - \log 5]$ when

$x = 2\Rightarrow t = 5$
when

$x =3\Rightarrow t = 10$

$= \dfrac {1}{2}\left (\log \dfrac {10}{5}\right ) = \dfrac {1}{2}\log 2$ .

Prove that $\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a} f(a - x)dx$ and hence evaluate $\displaystyle \int_{0}^{a} \dfrac {\sqrt {x}}{\sqrt {x} + \sqrt {a - x}} dx$ .

Let

$\displaystyle I = \int_{0}^{a} f(x) dx$ Put

$x = a - t\ \Rightarrow dx = -dt$
when

$x = 0\Rightarrow t = a$ ; when

$x = a\Rightarrow t = 0$

$\displaystyle I = \int_{a}^{0}f(a - t)(-dt)$

$\displaystyle = \int_{0}^{a}f(a - t)dt$ (using property of definite integrals)

$\displaystyle = \int_{0}^{a} f(a - x)dx$ (changing variable

$t$ to

$x$ )

Consider

$\displaystyle I = \int_{0}^{a} \dfrac {\sqrt {x}dx}{\sqrt {x} + \sqrt {a - x}} .... (1)$

$\displaystyle = \int_{0}^{a} \dfrac {\sqrt {a - x}dx}{\sqrt {a - x} + \sqrt {a - (a - x)}} = \int_{0}^{a} \dfrac {\sqrt {a - x}dx}{\sqrt {a - x} + \sqrt {a - a + x}}$

$\displaystyle I = \int_{0}^{a} \dfrac {\sqrt {a - x}dx}{\sqrt {a - x} + \sqrt {x}} ..... (2)$

Adding

$(1)$ and

$(2)$ we get

$\displaystyle 2I = \int_{0}^{a} \dfrac {\sqrt {x} + \sqrt {a - x}}{\sqrt {a - x} + \sqrt {x}} dx = \int_{0}^{a}1\ dx = x|_{0}^{a}$

$\therefore I = \dfrac {a}{2}$ .

Solve:
$\displaystyle \int { \dfrac { \sqrt { 1+{ x }^{ 2 } } }{ { x }^4 } } dx$

We have,

$I=\int{\dfrac{\sqrt{1+{{x}^{2}}}}{{{x}^{4}}}dx}$

Let put

$x=\tan \theta \,\,.......\,\,\left( 1 \right)$

Differentiation this and we get,

$dx={{\sec }^{2}}\theta d\theta$

Then,

$I=\int{\dfrac{\sqrt{1+{{\tan }^{2}}\theta }}{{{\tan }^{4}}\theta }{{\sec }^{2}}\theta d\theta }$

$=\int{\dfrac{\sqrt{{{\sec }^{2}}\theta }}{{{\tan }^{4}}\theta }}{{\sec }^{2}}\theta d\theta$

$=\int{\dfrac{{{\sec }^{3}}\theta }{{{\tan }^{4}}\theta }d\theta }$

$=\int{\dfrac{{{\sec }^{3}}\theta }{{{\tan }^{3}}\theta \tan \theta }d\theta }\,\,\,\,\,\,\,\,\,\,\,\,\because \dfrac{\sec \theta }{\tan \theta }=\csc \theta$

$=\int{{{\csc }^{3}}\theta }\cot \theta d\theta$

$=\int{cs{{c}^{2}}\theta (\csc \theta }\cot \theta )d\theta$

Again let,

$\csc \theta =t$ $...... (2)$

Again differentiating and we get

$-\csc \theta \cot \theta d\theta =dt$

$\csc \theta \cot \theta d\theta =-dt$

Then,

$=\int{{{t}^{2}}}(-dt)$

$=-\int{{{t}^{2}}}dt$

On integrating and we get,

$=-\dfrac{{{t}^{3}}}{3}+C\,\,\,\,\,\,\,\,\,\,\because \left( \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1} \right)$

Put $t=\csc \theta$ by equation (2) to, and we get,

$=-\dfrac{({{\csc }}\theta )^3}{3}+C$

By equation (1) to,

$x=\tan \theta$

$\tan \theta =x$

$\theta ={{\tan }^{-1}}x$

Then,

$=-\dfrac{{{\left[ \csc \left( {{\tan }^{-1}}x \right) \right]}^{3}}}{3}+C$

$=-\dfrac{{{\left[ \csc \left( {{\csc }^{-1}}\dfrac{\sqrt{{{x}^{2}}+1}}{x} \right) \right]}^{3}}}{3}+C\,\,\,\,\,\,\,\,\because {{\tan }^{-1}}x=\dfrac{\sqrt{{{x}^{2}}+1}}{x}$

$=-\dfrac{{{\left( \dfrac{\sqrt{{{x}^{2}}+1}}{x} \right)}^{3}}}{3}+C$

$=-\dfrac{{{\left( \sqrt{{{x}^{2}}+1} \right)}^{3}}}{3{{x}^{3}}}+C$

Hence, this is the answer.

For what $a < 0$ does the inequality $\displaystyle\, \int_{a}^{0}(3^{-2x} \, \, 2.3^{-x})dx \geqslant 0$ hold true?

$\int_{a}^{0} 3^{-2x} 2 . 3^{-x} dx = 2\int_{a}^{0} 3^{-3x} dx$

Apply substituion:

$-3x = u$

$-3dx = du$

$dx= \dfrac {-1}{3} du$

$= 2 \int_{-3a}^{0} 3^{u} \left (\dfrac {-1}{3}\right )du$

$= \dfrac {-2}{3} \int_{-3a}^{0} 3^{u} du = \dfrac {-2}{3} \left [\dfrac {3^{u}}{\ln 3}|_{-3a}^{0} \right ]$

$= \dfrac {-2}{3} \left [\dfrac {1}{\ln 3} - \dfrac {3^{-3a}}{\ln 3}\right ] \geq 0$

$\Rightarrow [1 - 3^{-3a} ] \leq 0$

For this term to be

$'\leq 0'$

$1 - 3^{-3a} \leq 0$

$1\leq 3^{-3a}\Rightarrow 3^{3a} \leq 1$

$-1 < a < 0$ then

$3a \geq 1$ and then

$3^{3a} \not {\leq} 1$

Therefore,

$a\leq -1$ .

Calculate the following integrals.
$\displaystyle \int_{0}^{2e} \, \frac{dx}{0.5x \, + \, 1}.$

$\displaystyle\int_{0}^{2e} \dfrac {dx}{0.5x + 1}$

$\displaystyle\Rightarrow \int_{0}^{2e} \dfrac {dx}{(0.5)\left [x + \dfrac {1}{0.5}\right ]} = \int_{0}^{2e} \dfrac {dx}{(0.5)[x + 2]}$

$= \dfrac {1}{0.5} \left (\log (x + 2)|_{0}^{2e}\right )$

$= \dfrac {1}{0.5} (\log (2e) - \log 2)$

$= \dfrac {1}{0.5} (\log 2 + \log e - \log 2)$

$= \dfrac {1}{0.5}(\log e)$

$= \dfrac {1}{0.5}$ .

Evaluate $\int_{0}^{1} x \tan^{-1} x\ dx$ .

$\displaystyle \int f(x) g(x) dx= f(x) \displaystyle \int g(x) dx= \displaystyle \int (f(x) \displaystyle \int g(x)dx)dx$

$\displaystyle \int_0^1 x\tan^{-1}x\ dx=\displaystyle \int_0^1 (\tan^{-1}x)x\ dx$

$=\tan^{-1}x \displaystyle \int_0^1x\ dx-\displaystyle \int_0^1\left (\dfrac {d(\tan^{-1}x)}{dx} \displaystyle \int x. dx\right) dx$

$=\left [\tan^{-1}x. \dfrac {x^2}{2}\right]_0^1 -\displaystyle \int_0^1 \dfrac {1}{1+x^2}dx$

$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]-\dfrac {1}{2} \displaystyle \int_0^1 \dfrac {x^2 +1-1}{x^2 +1}dx$

$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]_0^1 -\dfrac {1}{2} \left [\displaystyle \int_0^1 \dfrac {x^2 +1}{x^2 +1}dx -\displaystyle \int_0^1 \dfrac {dx}{x^2 +1}\right]$

$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]-\dfrac {1}{2} \left [\displaystyle \int_0^1 dx-\displaystyle \int_0^1 \dfrac {dx}{x^2 +1}\right]$

$\left [Using\ \displaystyle \int \dfrac {dx}{a^2 +x^2}=\dfrac {1}{a} \tan^{-1} \dfrac {x}{a}+C\right]$

$=\left [\dfrac {x^2}{2}\tan^{-1}x-\dfrac {1}{2}x +\dfrac {1}{2}+\dfrac {1}{1}\tan^{-1}\dfrac {x}{1}\right]_0^1 C$

$=\left [\dfrac {x^2}{2}\tan^{-1}x -\dfrac {x}{2}+\dfrac {1}{2}\tan^{-1}x\right]_0^1$

$=\left[ \dfrac { { \left( 1 \right) }^{ 2 } }{ 2 } \tan ^{ 1 }{ \left( 1 \right) } -\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \tan ^{ -1 }{ 1 } \right] -\left [\dfrac {0^2}{2}\tan^{-1}0-\dfrac {0}{2}+\dfrac {1}{2}\tan^{-1} (0)\right]$

$=\left [\dfrac {\pi}{8}-\dfrac {1}{2}+\dfrac {\pi}{8}\right]-[0-0+0]$

$= \pi /4 -\dfrac {1}{2}$

Evaluate: $\int_{0}^{\pi/2} \dfrac {x}{a^{2}\cos^{2} x + b^{2}\sin^{2}x} dx$ .

$\int_{0}^{\pi/2} \dfrac {x}{a^{2}\cos^{2} x + b^{2}\sin^{2}x} dx$

Divide numinator and denominator by

$cos^2x$

$\int_{0}^{\pi/2} \dfrac {x\ \sec^2x}{a^{2}+ b^{2}\tan^{2}x} dx$ .

$I = \dfrac {\pi}{2ab} \left [\tan^{-1} \dfrac {b}{a} \tan x\right ]_{0}^{\pi}$

$= \dfrac {\pi}{2ab} \left [\tan^{-1} \dfrac {b}{a} \tan^{-1}\pi - \tan^{-1} \dfrac {b}{a}\tan 0\right )$

$= \dfrac {\pi}{2ab} \left [\tan^{-1} 0 - \tan^{-1} 0\right ]$

$= 0$ .

Evaluate: $\int { \tan ^{ -1 }{ x } } dx$

$\int1\times\tan^{-1}x$

$=x\tan^{-1}x-\displaystyle\int x\times\dfrac{1}{1+x^2}$

$=x\tan^{-1}x-\dfrac{1}{2}\displaystyle\int \dfrac{2x}{1+x^2}$

$=x\tan^{-1}x-\dfrac{1}{2}\ln|1+x^2|+K$

Sita is driving along a straight highway in her car. At time $t = 0$ , when Sita is moving at $10\ ms^{-1}$ in the positive x-direction, she passes a signpost at $x = 50\ m$ . Here acceleration is a function of time:
$a = 2.0\ ms^{-2} - \left (\dfrac {1}{10} ms^{-3}\right )t$
Derive expressions for her velocity and position as functions of time.

At time

$t = 0$ , Sita's position is

$x_{0} = 50$ , and her velocity is

$v_{0} = 10\ ms^{-1}$ . Since we know that acceleration

$a$ can be given as

$a = dv/ dt$ . Hence,

$dv = a\ dt$ .
Integrating both sides, we get

$\int_{v_{0}}^{v}dv \int_{0}^{t} a\ dt$ .

$[v - v_{0}] = \int_{0}^{t}\left [2 - \dfrac {1}{10} t\right ] dt$ .

$v - 10 = \left [2t - \dfrac {t^{2}}{20}\right ]_{0}^{t}$

$v = 2t - \dfrac {t^{2}}{20} + 10 (ms^{-1}) = 10 + 2t - \dfrac {t^{2}}{20}(ms^{-1}) .... (i)$
We know that velocity in relation with position is given by

$v = \dfrac {dx}{dt}$ .

$dx = dt$
Integrating both sides, we get

$\int_{x_{0}}^{x} dx = \int_{0}^{t} \left (2t - \dfrac {t^{2}}{20} + 10\right )dt$

$[x]_{50}^x=\left[2\dfrac{t^2}{2}-\dfrac{t^3}{60}+10t\right]_0^t$

$x = 50 + 10t + t^{2} - \dfrac {t^{2}}{60} (m) ..... (ii)$ .

$\int\limits_0^{\pi /2} {{1 \over {{a^2}{{\sin }^2}x + {b^2}{{\cos }^2}x}}dx}$

Given,

$\int _0^{\frac{\pi }{2}}\dfrac{1}{a^2\sin ^2\left(x\right)+b^2\cos ^2\left(x\right)}dx$

substitute

$x=\tan ^{-1}u$

$=\int \dfrac{1}{a^2u^2+b^2}du$

substitute

$u=\dfrac{b}{a}v$

$=\int \dfrac{1}{ab\left(v^2+1\right)}dv$

$=\dfrac{1}{ab}\tan ^{-1}v$

$=\dfrac{1}{ab}\tan ^{-1}\left ( \dfrac{a}{b}\tan (x) \right )+c$

$=\left [ \dfrac{1}{ab}\tan ^{-1}\left ( \dfrac{a}{b}\tan (x) \right ) \right ]_0 ^{\frac{\pi}{2}}$

$=\dfrac {\pi |a|}{2a^2|b|}-0$

$=\dfrac {\pi |a|}{2a^2|b|}$

Evaluate the integral:
$\displaystyle \int e^x\left(\frac{1+sinx}{1+cosx}\right) dx$

$\displaystyle \int e^x\left(\frac{1+\sin x}{1+\cos x}\right) dx = \int e^x \left(\frac{\sec^2\frac{x}{2}}{2}+ \tan \frac{x}{2}\right) dx$

It is of the form

$\displaystyle \int e^x(f(x) + f'(x) dx)= e^xf(x) +c$

Putting,

$f(x) = \tan \dfrac{x}{2}$

$\therefore f'(x) = \sec^2\dfrac{x}{2}\left(\dfrac{1}{2}\right)= \dfrac{\sec^2\frac{x}{2}}{2}$
Thus,

$\displaystyle \int e^x \left(\frac{\sec^2\frac{x}{2}}{2}+ \tan \frac{x}{2}\right) dx= \int e^x\left(\frac{1+\sin x}{1+\cos x}\right) dx =e^x \tan\dfrac{x}{2}$

Find $\int { \dfrac { ({ x }^{ 2 }+1)({ { x }^{ 2 } }+2) }{ ({ { x }^{ 2 } }+3)({ { x }^{ 2 } }+4) } dx }$

1205475_1037347_ans_84f901e7ee8947709cafaebd7a5020a1.JPG

Integrate:
$\int { \dfrac { 2+\sin { 2x } }{ 1+\cos { 2x } } } { e }^{ 2 }dx$

$\displaystyle \int\left [\frac{2+\sin 2 x}{1 + (2\cos^2 2x -1)}\right]e^2 = \left[\frac{2 + \sin 2x}{2\cos^2 x}\right] \quad \because (\cos 2x = 2\cos^2 x-1)$

$\displaystyle \int \left[\frac{2(1+\cos x\sin x)}{2 \cos^2 x}\right] \quad \because \sin 2x = \sin x \cos x$

$\displaystyle \int \left[\frac{1}{\cos^2 x}+\frac{\cos x\sin x}{ \cos^2 x}\right] =\int [\sec^2 x + \tan x] dx$

$\displaystyle = \int \sec^2 x \,dx + \int \tan x \, dx$

$\displaystyle = e^2 [\tan x - ln |\cos x | + c]$

1044010_1038046_ans_708acbd9fe534ce4a04877911073b60c.jpg

Integrate:
$\int { \dfrac { \left( x-3 \right) { e }^{ x } }{ { \left( x-1 \right) }^{ 3 } } dx }$

$\displaystyle \int \dfrac{(x-3)e^{x}}{(x-1)^{3}}dx$ (By ILATE)

let

$x-1 = m$ or

$m+1 = x$

$dx =dm$

$\Rightarrow \displaystyle \int \dfrac{(m-2)e^{m+1}}{m^{3}}dm$

Here (1)

$u = (m-2)e^{m+1}, V = \dfrac{1}{m^{3}}$

$(\because \displaystyle \int uv \, dx = u \int v dx - \int \left [ \dfrac{dv}{dx} \int v \,dx \right ])$

$\Rightarrow \dfrac{-e^{m+1}(m-2)}{2m^{2}} - \displaystyle \int -\frac{e^{m+1}+e^{m+1}m}{2m^{2}}dm$

$\displaystyle -\dfrac{1}{2} \int \dfrac{-e^{m+1}+e^{m+1}m}{m^{2}}dm$

Here (2)

$m = e^{m+1}+e^{m+1} m , v = \dfrac{1}{m^{2}}$

$\Rightarrow \frac{-1}{2} \left [ -\left ( \dfrac{-e^{m+1}+e^{m+1}m}{m} \right )-e^{m+1} dn \right ]$

$(-e^{m+1})$

$\Rightarrow \dfrac{-1}{2} \left ( -\dfrac{e^{m+1}+e^{m+1}m}{m} -(-e^{1+m})\right )$

$\dfrac{-e^{m+1}}{2m}$

$\Rightarrow \dfrac{-e^{m+1}(m-2)}{2m^{2}} + \dfrac{e^{m+1}}{2m}+c$

$\Rightarrow \boxed{ \dfrac{e^{x}(x-3)}{2(x-1)^{2}}+ \dfrac{e^{x}}{2(x-1)}+c}$

1104542_1038050_ans_829232f0f5fd42b792be389a2ab3a34c.jpeg

Evaluate :
$e^{2x}\left(\dfrac{1-\sin 2x}{1-\cos 2x}\right)dx$

$I=\displaystyle\int e^{2x}\dfrac{(1-\sin 2x)}{(1-\cos 2x)}dx$

$=\displaystyle\int e^{2x}\dfrac{(1-2\sin x\cos x)}{2\sin^{2}x}dx$

$[\because \cos 2x=1-2\sin^{2}x\sin 2x=2\sin x\cos x]$

$=\dfrac{1}{2}\left[\displaystyle\int e^{2}x.\csc^{2}x dx-2\int e^{2x}\cot x dx\right]$

Integrating by parts

$=\dfrac{1}{2}\left[e^{2x}\displaystyle\int \csc^{2} x\ dx-\int e^{2x}2\int \csc^{2}x\ dx\right]-\int e^{2x}\cot x\ dx$

$=\dfrac{1}{2}e^{x}(-\cot x)+\displaystyle\int e^{2x}\cot x\ dx-\int e^{2x}\cot x\ dx$

$\left[\because \dfrac{d}{dx}\cot x=-\csc^{2}x\right]$

$=\dfrac{-e^{x}\cot x}{2}+C$

Evaluate :
$\int\limits_0^{\pi /2} {\dfrac{{\cos x\,dx}}{{\left( {\,\cos \,x\, + \,\sin x} \right)}}}$

We have,

$I=\int\limits_0^{\pi /2} {\dfrac{{\cos x\,dx}}{{\left( {\,\cos \,x\, + \,\sin x} \right)}}}$

$.........(1)$

We know that

$\int\limits_b^{a}f(x)dx=\int \limits_b^a\ f(a+b-x)dx$

Therefore,

$I=\int\limits_0^{\pi /2} {\dfrac{{\sin x\,dx}}{{\left( {\,\sin \,x\, + \,\cos x} \right)}}}$

$............(2)$

On adding equations

$(1)$ and

$(2)$ , we get

$2I=\int\limits_0^{\pi /2} {\dfrac{{\sin x+\cos x}}{{\left( {\,\sin \,x\, + \,\cos x} \right)}}} dx$

$2I=\int\limits_0^{\pi /2} 1 dx$

$2I=[x]_0^\frac{\pi}{2}$

$2I=\dfrac{\pi}{2}-0$

$2I=\dfrac{\pi}{2}$

$I=\dfrac{\pi}{4}$

Hence, this is the answer.

$\displaystyle\int\limits_0^{\frac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)dx}$

$I=\displaystyle\int_{0}^{\dfrac{\pi }{2}}(2 \log \sin x- \log \sin 2x)dx$

we have

$f(x)=2 \log \sin x- \log \sin 2x$

$f(x)=\log \dfrac{\sin^{2}x}{\sin 2x}$

$=\log \tan x-\log 2$

$I_{1}=\displaystyle\int_{0}^{\dfrac{\pi }{2}}\log \tan x\ dx$

Let

$y=\dfrac{\pi }{2}-x, dy=-dx, \tan x= \cot y$

$I_{1}= -\displaystyle\int_{0}^{\dfrac{\pi }{2}}\log \cot y \ dy=-\int_{0}^{\dfrac{\pi }{2}}\log \tan x\ dx=-I$

$\Rightarrow 2I_{1}=0$

$\Rightarrow I_{1}=0$

Now

$I= \displaystyle\int_{0}^{\dfrac{\pi }{2}}\log \tan x dx-\int_{0}^{\dfrac{\pi }{2}}\log 2 \ dx$

$I=\dfrac{\pi }{2} \log 2$

$\int \sin^{-1} \ (\dfrac{2x +2}{\sqrt{4x^{2} +8x +13}}\ )dx$ =

$\dfrac{2x+2}{\sqrt{4x^2+8x+3}}=\dfrac{2(x+1)}{\sqrt{4(x^2(4+1)+3^2}}$

$=\dfrac{2(x+1)}{\sqrt{4(x+1)^2+3^2}}$

set

$x+1=\dfrac{3}{2}\tan\theta$

$=\dfrac{2\cdot \dfrac{3}{2}\tan \theta}{\sqrt{4\times \dfrac{9}{4}\tan^2\theta +3^2}}=\dfrac{3\tan\theta}{\sqrt{3^2(1+\tan^2\theta)}}$

$=\dfrac{3\tan\theta}{3\sin\theta}=\dfrac{\sin\theta}{\cos\theta}\cdot\cos\theta =\sin \theta$

$=\sin^{-1}\left(\dfrac{2x+2}{\sqrt{4x^2+\theta x+13}}\right)=\sin^{-1}(\sin \theta)$

$=\theta$

$x+1=\dfrac{3}{2}\tan\theta$

$dx=\dfrac{3}{2}\sin^2\theta d\theta$

$\tan\theta =\dfrac{2}{3}(x+1)$

$\sin\theta =\sqrt{1+\tan^2\theta}$

$=\sqrt{1+\dfrac{4}{9}(x+1)^2}$

$=\dfrac{1}{3}\sqrt{4x^2+8x+13}$

$I=\displaystyle\int\theta \cdot\dfrac{3}{2}\sin^2\theta\cdot d\theta$

$=\dfrac{3}{2}\displaystyle\int \theta\cdot \sec^2\theta\cdot d\theta$

$=\dfrac{3}{2}[\theta \cdot \tan \theta-\displaystyle\int \tan\theta d\theta]$

$=\dfrac{3}{2}[\theta \tan \theta -ln \sec\theta]+c$

$=\dfrac{3}{2}[\dfrac{2}{3}(x+1)\tan \{\dfrac{2}{3}(x+1)\}$

$-ln\left(\dfrac{\sqrt{4n^2+dn+13}}{3}\right)+c$

$=\dfrac{3}{2}\cdot\dfrac{2}{3}(n+1)\tan^{-1}\left\{\dfrac{2}{3}(n+1)\right\}-\dfrac{3}{2}[ln \sqrt{4n^2+dn+13}-ln 3]+c$

$=(x+1)\tan^{-1}\left\{\dfrac{2}{3}(n+1)\right\}-\dfrac{3}{2}\cdot \dfrac{1}{2}ln(4n^2+dn+13)+\dfrac{3}{2}ln 3+c$ New laws

$=(x+1)\tan^{-1}\left\{\dfrac{2}{3}(x+1)\right\}-\dfrac{3}{4}ln(4n^2+dn+13)+c^2$ .

Integrate $\int x\log 2x \, dx$

$I=\displaystyle\int{x\log{2x}dx}$

Let

$u=\log{2x}\Rightarrow\,du=\dfrac{1}{2x}\times 2\,dx=\dfrac{dx}{x}$

$dv=x\,dx\Rightarrow\,v=\dfrac{{x}^{2}}{2}$

$I=\dfrac{{x}^{2}}{2}\log{2x}-\displaystyle\int{\dfrac{{x}^{2}}{2}\dfrac{dx}{x}}$

$=\dfrac{{x}^{2}}{2}\log{2x}-\dfrac{1}{2}\displaystyle\int{x\,dx}$

$=\dfrac{{x}^{2}}{2}\log{2x}-\dfrac{{x}^{2}}{4}+c$

$\displaystyle\int^{1/2}_0\dfrac{x\ dx}{(y^2+x^2)^{3/2}}$

We have,

$I=\displaystyle\int^{1/2}_0\dfrac{x\ dx}{(y^2+x^2)^{3/2}}$

Let

$t=y^2+x^2$

$\dfrac{dt}{dx}=0+2x$

$\dfrac{dt}{2}=x\ dx$

Therefore,

$I=\dfrac{1}{2}\displaystyle\int^{(y^2+1/4)}_{y^2}\dfrac{dt}{(t)^{3/2}}$

$I=\dfrac{1}{2}\displaystyle\int^{(y^2+1/4)}_{y^2}(t)^{-3/2} \ dt$

$I=\dfrac{1}{2}\displaystyle \left[\dfrac{t^{-1/2}}{-1/2}\right]^{(y^2+1/4)}_{y^2}$

$I=\displaystyle \left[t^{-1/2}\right]^{(y^2+1/4)}_{y^2}$

$I=\displaystyle \left[\dfrac{1}{\sqrt t }\right]^{(y^2+1/4)}_{y^2}$

$I=\displaystyle \left[\dfrac{1}{\sqrt {y^2+1/4} }-\dfrac{1}{\sqrt {y^2}}\right]$

$I=\displaystyle \left[\dfrac{2}{\sqrt {4y^2+1} }-\dfrac{1}{y}\right]$

Hence, this is the answer.

$\displaystyle \int_{0}^{2}2x dx=$

$\displaystyle \int_{0}^{2}2x dx\\\left.x^2\right|_0^2\\2^2-0^2=4$

$\int (\log x)^{2}dx$

$\displaystyle\int{{\left(\log{x}\right)}^{2}dx}$

Let

$dv=dx\Rightarrow v=x$

$u={\left(\log{x}\right)}^{2}\Rightarrow du=2\log{x}\times\dfrac{dx}{x}$

$=x{\left(\log{x}\right)}^{2}-\displaystyle\int{x\times 2\log{x}\times\dfrac{dx}{x}}$

$=x{\left(\log{x}\right)}^{2}-2\displaystyle\int{\log{x}dx}$

Let

$u=\log{x}\Rightarrow du=\dfrac{1}{x}dx$ and

$dv=dx\Rightarrow v=x$

$=x{\left(\log{x}\right)}^{2}-2\left[x\log{x}-\displaystyle\int{x\times \dfrac{dx}{x}}\right]$

$=x{\left(\log{x}\right)}^{2}-2x\log{x}+2\displaystyle\int{dx}$

$=x{\left(\log{x}\right)}^{2}-2x\log{x}+2x+c$ where

$c$ is the constant of integration.

Solve $\displaystyle\int { \left[ \dfrac { \sqrt { { x }^{ 2 }+1 } \left[ \ln { \left( { x }^{ 2 }+1 \right) -2\ln { x } } \right] }{ { x }^{ 4 } } \right] } dx$

$\displaystyle\int \dfrac{\sqrt{x^{2}+1}[\ln (x^{2}+1)-2\ln x]}{x^4}dx=\displaystyle\int \sqrt{\dfrac{x^{2}+1}{x^{2}}}\times \ln \dfrac{x^{2}+1}{x^{2}}\times \dfrac{dx}{x^{3}}\ ...(1)$

Put

$t=\dfrac{x^{2}+1}{x^{2}}\implies dt=-\dfrac{2}{x^{3}}dx$

$(1)$ becomes

$\displaystyle\int \sqrt{t}\ln t(-\dfrac{dt}{2})=-\dfrac{1}{2}\bigg[\ln t{\displaystyle \int\sqrt{t}dt}-\displaystyle\int \dfrac{d(\ln t)}{dt}\times \displaystyle\int \sqrt{t}dt\ \times dt\bigg]$

$=-\dfrac{1}{2}\bigg[\dfrac{2}{3}t^{\dfrac{3}{2}}\ln t-\displaystyle\int \dfrac{1}{t}\times \dfrac{2}{3}\times t^{\dfrac{3}{2}}dt\bigg]$

$=-\dfrac{1}{2}\bigg[\dfrac{2}{3}t^{\dfrac{3}{2}}\ln t-\dfrac{4}{9}t^{\dfrac{3}{2}}\bigg]=-\dfrac{1}{9}\bigg[3{t^{\dfrac{3}{2}}}\ln t-2{t^{\dfrac{3}{2}}}\bigg]$

$\int { \sqrt { \left( \dfrac { x-1 }{ x+1 } \right) } } dx$

$\int{(\sqrt{\dfrac{x-1}{x+1}})}dx$

This can be written as

$\Rightarrow \int{\sqrt{\dfrac{(x-1)(x-1)}{(x+1)(x-1)}}}dx$

$\rightarrow \int{\sqrt{\dfrac{(x-1)^2}{x^2-1}}}dx$

$\Rightarrow \int{{\dfrac{x-1}{\sqrt{x^2-1}}}}dx$

$\Rightarrow \int{\dfrac{x}{\sqrt{x^2-1}}}dx-\int{\dfrac{1}{\sqrt{x^2-1}}}dx$

In the first interval take

$x^2-1=t\space\Rightarrow 2xdx=dt$

Second interval is in the form of

$\int\dfrac{1}{\sqrt{x^2-a^2}}dx=\log|\sqrt{x^2-a^2}+x|+C$

Using these results we get

$\Rightarrow \dfrac{1}{2}\int{\dfrac{dt}{\sqrt t}}-\log|\sqrt{x^2-1}+x|+C$

$\Rightarrow (\sqrt t)-\log|\sqrt{x^2-1}+x|+C$

$\Rightarrow (\sqrt {x^2-1})-\log|\sqrt{x^2-1}+x|+C$

Find the value of $\displaystyle\int { \dfrac { d\left( { x }^{ 2 }+1 \right) }{ \sqrt { \left( { x }^{ 2 }+2 \right) } } }$

$\int\dfrac{d(x^2+1)}{\sqrt{(x^2+2)}}$

$\Rightarrow \int\dfrac{d(x^2+1)}{\sqrt{(x^2+1+1)}}$

Take

$x^2+1=t$ ,

So the integral changes to

$\Rightarrow \int\dfrac{dt}{\sqrt{t+1}}$

Take

$t+1=y$

$\Rightarrow dt=dy$

So the integral changes to

$\Rightarrow \int\dfrac{dy}{\sqrt{y}}$

$\Rightarrow \int\dfrac{dy}{\sqrt{y}}=\int{(y)^{\dfrac{-1}{2}}}dy$

$\Rightarrow \int{\dfrac{dy}{\sqrt y}}=\dfrac{y^{\dfrac{-1}{2}+1}}{\dfrac{-1}{2}+1}=2y^\dfrac{1}{2}=2\sqrt y$

But

$y=t+1$

$\Rightarrow 2\sqrt y=2\sqrt{t+1}$

But

$t=x^2+1$

$\Rightarrow 2\sqrt y=2\sqrt{t+1}=2\sqrt{x^2+1+1}=2\sqrt{x^2+2}$

Therefore,

$\int\dfrac{d(x^2+1)}{\sqrt{(x^2+2)}}=2\sqrt{x^2+1}$

$\int_{0}^{\frac{\pi}{2}}\dfrac{\sin\ x-\cos\ x}{1+\sin x\ \cos\ x}dx$

As,

$\begin{array}{l}\int_0^a {f\left( x \right)dx = \int_x^a {f\left( {x - a} \right)dx} } \\So,\\I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} \\ = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {\frac{\pi }{2} - x} \right) - \cos \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}dx} \\ = \int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}dx} \end{array}$

Adding two equations,

$\begin{array}{c}I + I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} + \int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}dx} \\2I = \int\limits_0^{\frac{\pi }{2}} {\frac{{sinx - cosx + \cos x - \sin x}}{{1 + \sin x\cos x}}dx} \\ = \int_0^{\frac{\pi }{2}} {\frac{0}{{1 + \sin x\cos x}}dx} \\2I = 0\\I = 0\end{array}$

Thus value of integral is 0.

Solve:
$\displaystyle \int x^{2} \sin^{2}x dx$

1144602_1060490_ans_3ea7bc9283eb47d39e197687fe946dbd.png

Integrate with respect to $x$ .:
$e^{x}(\text{sec}^2 {x}+\tan x)$

$\int e^{x} (f(x)+ f (x))dx = e^{ x} f (x)$

$\Rightarrow \int e^{x} (\tan + \sec^{2} x)= e^{x} \tan x +c$

$\int { x\left( \dfrac { \sec { 2x } -1 }{ \sec { 2x } +1 } \right) dx }$

1141018_1063148_ans_ca0946ed801844f1833442ee56a6df38.JPG

Evaluate:
$\displaystyle\int{{(\ln x)}^{4}dx}$

$\displaystyle \int (ln \, x)dx$

Let

$ln\, x = t$

$x = e^t$

$dx =e^t dt$

$\displaystyle \int t^4e^tdt$

Using integration by parts,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$\displaystyle = t^4 \int e^tdt - \int 4t^3e^tdt$

$\displaystyle = t^4 e^t - 4 \left[t^3 \int e^t dt - \int 3t^2 e^t dt\right]$ ..... again by parts

$\displaystyle = t^4 e^t - 4\left[t^3e^t - 3 \int t^2 e^t dt\right]$

$\displaystyle = t^4e^t -4 t^3e^t + 12\left[t^2\int e^t dt - \int 2t e^t dt\right]$ .... again by parts

$\displaystyle = t^4 e^t - 4t^3e^t+ 12 \left[ t^2e^t - 2 \int te^tdt]\right]$

$\displaystyle =t^4e^t - 4t63e^t + 12 \left[t^2e^t - 2 (te^t - \int e^tdt)\right]$ .... again by parts

$\displaystyle= t^4e^t - 4t^3e^t + 12t^2 e^t - 24te^t + 24e^t + C$

Put the value of

$t$

$= (ln\, x)^4 x- 4 (ln\, x)^3 x+ 12 (ln \,x)^2 x - 24(ln\, x)x + 24x + C$

$\int { \dfrac { arc\sin { \sqrt { x } } }{ \sqrt { 1-x } } dx }$

1323242_1063202_ans_0ffccd73d9564be5b851ac3e91255d15.jpg

$\displaystyle\int x\cos^3x\sin x dx$ .

We are given,

$I= \int { x\quad cos^{ 3 }.sinxdx. }$

as we know,

$\int { uvdx\quad =\quad u\int { vdx\quad -\quad \int { \left( \frac { dy }{ dx } -\int { vdx } \right) } } } dx$

let

$u=x$ and

$v =cos^{ 3 }x.sinx$

so,

$I=x\int { cos^{ 3 }x.sindx-\int { \left( 1.\int { cos^{ ^{ 3 } }sinx.dx } \right) } dx }$

let

$cosx=t\Rightarrow -sin dx = dt$

so, differentiating wrt

$x$ , we get,

$I=x\int { t^{ 3 } } (-dt)-\int { \left[ \int { t^{ 3 } } (-dt) \right] } dx$

$=x\left[ \frac { -t^{ 4 } }{ 4 } \right] -\int { \frac { t^{ 4 } }{ 4 } } dx$

$\therefore I=-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 4 } \int { cos^{ 4 } } xdx$

$\therefore I=-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 4 } \int { \left( \frac { cos2x+1 }{ 2 } \right) ^{ 2 }dx }$

$\displaystyle =-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 4\times 4 } \left[ \int { \left( cos^{ 2 }2x+2cos2x+1 \right) } dx \right]$

$\displaystyle =-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 16 } \left[ \int { \left( \frac { cos4x+1 }{ 2 } +2cos2x+1 \right) } dx \right]$

$\displaystyle =-\frac { x }{ 4 } cos^{ 4 }x-\frac { 1 }{ 32 } \int { cos4xdx\quad -\frac { 1 }{ 8 } \int { cos2xdx } } -\int { \frac { 3 }{ 32 } dx }$

$\displaystyle I=\frac { -x }{ 4 } .cos^{ 4 }x-\frac { 1 }{ 32 } .\frac { sin^{ 4 }x }{ 4 } -\frac { 1 }{ 8 } .\frac { sin^{ 2 }x }{ 2 } -\frac { 3 }{ 32 } x+c$

$\displaystyle I=\quad \frac { -sin^4x }{ 128 } -\frac { sin^2x }{ 16 } -\frac { x.cos^{ 2 }x }{ 4 } -\frac { 3x }{ 32 } +c$

$\displaystyle \int { xcos^{ 3 }x \, sinx \, dx=\frac { -sin^{ 4 }x }{ 128 } } -\frac { sin^{ 2 }x }{ 16 } -\frac { x.cos^{ 4 }x }{ 4 } -\frac { 3x }{ 32 } +c$

$\int {x{{\sin }^2} x\times dx}$

$\int xsin^{2}xdx=\int x(\frac{1-cosx}{2})dx$

$=\int\frac{x}{2}dx -\int \frac{x cosx}{2}dx$

$=\frac {x^{2}}{4}+\frac{xsinx+cosx}{2}+c$

$\int {x\sin 3xdx}$

$\int xsin3x dx=x(-cos3x)\frac{1}{3}-\int1\times(-cos3x)\frac{1}{3}dx$

$=\dfrac {-xcos3x}{3}+\dfrac {sin3x}{9} +c$

Solve $\displaystyle\int x\sin^{-1} xdx$ .

Integrating by parts, we get

$u = sin^{-1}x,v=x$

$du = \cfrac{1}{\sqrt{1-x^2}}, \displaystyle\int v dv = \cfrac{x^2}{2}$

$\cfrac{x^2 \space sin^{-1}x}{2}- \cfrac{1}{2}\displaystyle\int \cfrac{x^2dx}{\sqrt{1-x^2}}$

Putting

$x=sin\theta$

$dx = cos \theta d\theta$

$=\cfrac{x^2 \space sin^{-1}x}{2}- \cfrac{1}{2}\displaystyle\int \cfrac{sin^2\theta \space cos\theta \space d\theta}{\sqrt{1-sin^2\theta}}$

$=\cfrac{x^2 \space sin^{-1}x}{2}- \cfrac{1}{2}\displaystyle\int sin^2\theta d\theta$

$=\cfrac{x^2 \space sin^{-1}x}{2}- \cfrac{1}{4}\displaystyle\int (1-cos2\theta) d\theta$

$=\cfrac{x^2 \space sin^{-1}x}{2}- \cfrac{1}{4} (\theta -\cfrac{1}{2} sin2\theta)$

$=\cfrac{x^2 \space sin^{-1}x}{2}- \cfrac{1}{4} (sin^{-1}x -\cfrac{1}{2} sin(2sin^{-1}\theta))$

Let $f:\left[ -2,3 \right] \rightarrow \left[ 0,\infty \right)$ be a continuous function such that $f\left( 1-x \right) -f\left( x \right)$ for all $x\epsilon \left[ -2,3 \right]$ .
If ${ R }_{ 1 }$ is the numerical value of the area of the region bounded by $y=f\left( x \right) ,x=-2,x=3$ and the axis of x and ${ R }_{ 2 }=\int _{ 2 }^{ 3 }{ xf\left( x \right) dx }$ , then:-

${ R }_{ 1 }=\int _{ -2 }^{ 3 }{ f\left( x \right) dx }$

${ R }_{ 2 }=\int _{ -2 }^{ 3 }{ xf\left( x \right) dx }$

$=\int _{ -2 }^{ 3 }{ \cfrac { xf\left( x \right) +\left( 1-x \right) f\left( 1-x \right) }{ 2 } dx }$

$=\int _{ -2 }^{ 3 }{ \cfrac { xf\left( x \right) +(1-x)f\left( x \right) }{ 2 } dx }$

$=\cfrac { 1 }{ 3 } \int _{ -2 }^{ 3 }{ f\left( x \right) dx } =\cfrac { { R }_{ 1 } }{ 2 }$

Evaluate $\displaystyle\int\limits_4^9 {\dfrac{{dx}}{{\sqrt {\left( {9 - x} \right)\left( {x - 4} \right)} }}}$

Consider the given integral.

$I=\int_{4}^{9}{\dfrac{dx}{\sqrt{\left( 9-x \right)\left( x-4 \right)}}}$

$I=\int_{4}^{9}{\dfrac{dx}{\sqrt{9x-36-{{x}^{2}}+4x}}}$

$I=\int_{4}^{9}{\dfrac{dx}{\sqrt{13x-36-{{x}^{2}}}}}$

$I=\int_{4}^{9}{\dfrac{dx}{\sqrt{13x+\dfrac{169}{4}-\dfrac{169}{4}-36-{{x}^{2}}}}}$

$I=\int_{4}^{9}{\dfrac{dx}{\sqrt{\dfrac{25}{4}-\left( -13x+\dfrac{169}{4}+{{x}^{2}} \right)}}}$

$I=\int_{4}^{9}{\dfrac{dx}{\sqrt{{{\left( \dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{13}{2}-x \right)}^{2}}}}}$

Let

$t=\dfrac{13}{2}-x$

$dt=-dx$

Therefore,

$I=-\int_{\dfrac{5}{2}}^{-\dfrac{5}{2}}{\dfrac{dt}{\sqrt{{{\left( \dfrac{5}{2} \right)}^{2}}-{{t}^{2}}}}}$

We know that

$\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$

Therefore,

$I=-\left[ {{\sin }^{-1}}\left( \dfrac{t}{\dfrac{5}{2}} \right) \right]_{\dfrac{5}{2}}^{-\dfrac{5}{2}}$

$I=-\left[ {{\sin }^{-1}}\left( \dfrac{2\left( -\dfrac{5}{2} \right)}{5} \right)-{{\sin }^{-1}}\left( \dfrac{2\left( \dfrac{5}{2} \right)}{5} \right) \right]$

$I={{\sin }^{-1}}\left( 1 \right)-{{\sin }^{-1}}\left( 1 \right)$

$I=0$

Hence, the value is $0$ .

Integrate:
$\displaystyle \int \dfrac{2}{(x^{2}+1)^{2}}dx$

Let,

$I=\displaystyle\int{\dfrac{2}{{{\left( {{x}^{2}}+1 \right)}^{2}}}dx}$

$I=2 \displaystyle\int{\dfrac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}dx}$

Applying reduction formula, we have

$\displaystyle\int{\dfrac{1}{{{\left( a{{x}^{2}}+b \right)}^{n}}}dx=\dfrac{2n-3}{2b\left( n-1 \right)}\int{\dfrac{1}{{{\left( a{{x}^{2}}+b \right)}^{n-1}}}dx+\dfrac{x}{2b\left( n-1 \right){{\left( a{{x}^{2}}+b \right)}^{n-1}}}}}$

Therefore,

$I=\dfrac{x}{\left( {{x}^{2}}+1 \right)}+\displaystyle\int{\dfrac{1}{{{x}^{2}}+1}dx}+C$

$I=\dfrac{x}{\left( {{x}^{2}}+1 \right)}+{{\tan }^{-1}}x+C$

Hence, this is the required result.

$\displaystyle\int \dfrac{dx}{3x^2+13x -10}$

Let's first factorize quadratic

$3x^{2}+ 13x- 10 = 3 \left( x^{2} +\dfrac{13x}{3}- \dfrac{10}{3} \right)$

$= 3 \left( \left( x+ \dfrac{13}{6} \right)^{2}- \left( \dfrac{13}{6} \right)^{2} -\dfrac{10}{3} \right)$

$= 3 \left( \left( x+\dfrac{19}{6} \right)^{2} - \left( \dfrac{17}{6} \right)^{2} \right)$

$= 3 \left( x-\dfrac{2}{3} \right) (x+5)$

$\Rightarrow \int \dfrac{dx}{3x^{2}+ 13x -10} = \dfrac{1}{3} \int \dfrac{dx}{\left( x- \dfrac{2}{3} \right)} (x+5)$

$=\dfrac{1}{3} \times \dfrac{1}{5+ \dfrac{2}{3} } \left[ \int \dfrac{dx}{x-\dfrac{2}{3}} - \int \dfrac{dx}{x+5} \right]$

$=\dfrac{1}{17} \left[ \ln \left| \dfrac{3x-2}{3x+15} \right| \right]+c$

Integrate
$\int_0^1 {\frac{{{x^3}}}{{{x^2} + 1}}dx}$

Consider the given integral.

$I=\int_{0}^{1}{\dfrac{{{x}^{3}}}{{{x}^{2}}+1}}dx$

$I=\int_{0}^{1}{\left( x-\dfrac{x}{{{x}^{2}}+1} \right)}dx$

$I=\int_{0}^{1}{\left( x-\dfrac{2x}{2\left( {{x}^{2}}+1 \right)} \right)}dx$

$I=\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1}-\dfrac{1}{2}\left[ {{\log }_{e}}\left( {{x}^{2}}+1 \right) \right]_{0}^{1}$

$I=\left[ \dfrac{{{1}^{2}}}{2}-0 \right]-\dfrac{1}{2}\left[ {{\log }_{e}}\left( {{1}^{2}}+1 \right)-{{\log }_{e}}\left( 0+1 \right) \right]$

$I=\dfrac{1}{2}-\dfrac{1}{2}\left[ {{\log }_{e}}\left( 2 \right)-{{\log }_{e}}\left( 1 \right) \right]$

$I=\dfrac{1}{2}-\dfrac{1}{2}\left[ {{\log }_{e}}\left( 2 \right)-0 \right]$

$I=\dfrac{1}{2}-\dfrac{1}{2}{{\log }_{e}}2$

Hence, this is the answer.

$\int_{0}^{\pi/2}\dfrac {\sin x \cos x}{1+\sin^{4}x}dx$

$\begin{array}{l} \int _{ }^{ }{ \frac { { \cos \left( x \right) \sin \left( x \right) } }{ { { { \sin }^{ 4 } }\left( x \right) +1 } } dx } \\ Substitute\, u={ \sin ^{ 2 } }x,dx=\frac { 1 }{ { 2\cos x\sin x } }du \\ =\frac { 1 }{ 2 } \int _{ }^{ }{ \frac { 1 }{ { 1+{ u^{ 2 } } } } du } \\ =\frac { 1 }{ 2 } { \tan ^{ -1 } }u+C \end{array}$

Apply the limit, we get

$\int_0^{\frac{\pi }{2}} {\frac{{\cos \left( x \right)\sin \left( x \right)}}{{{{\sin }^4}\left( x \right) + 1}}dx = \frac{\pi }{8}}$

Solve $\int\sin 4x\sin 8xdx$

Consider the given integral.

$I=\int{\sin 4x\sin 8xdx}$

$I=\int{\sin 8x\sin 4xdx}$

We know that

$2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$

Therefore,

$I=\dfrac{1}{2}\int{2\sin 8x\sin 4xdx}$

$I=\dfrac{1}{2}\int{\left( \cos \left( 8x-4x \right)-\cos \left( 8x+4x \right) \right)dx}$

$I=\dfrac{1}{2}\int{\left( \cos 4x-\cos 12x \right)dx}$

$I=\dfrac{1}{2}\left[ \dfrac{\sin 4x}{4}-\dfrac{\sin 12x}{12} \right]+C$

Hence, this is the answer.

Solve : $\displaystyle \underset{0}{\overset{\infty}{\int}} \dfrac{x \, \ln \, x}{(1 + x^2)^2}$

$\int \dfrac{x\ln x}{(1+x^2)^2}dx$

Integration by parts

$=-\dfrac{\ln x}{2(1+x^2)}-\int -\dfrac{1}{2x(1+x^2)}dx$

$=-\dfrac{\ln x}{2(1+x^2)}-\left ( -\dfrac{1}{2}\left ( \left ( \ln |x| \right )-\dfrac{1}{2}\left ( \ln |x^2+1| \right ) \right ) \right )$

$\therefore \int \dfrac{x\ln x}{(1+x^2)^2}dx=-\dfrac{\ln x}{2(1+x^2)}+\dfrac{1}{2}\left ( \left ( \ln |x| \right )-\dfrac{1}{2}\left ( \ln |x^2+1| \right ) \right)+C$

$\therefore \int_{0}^{\infty }\dfrac{x\ln x}{(1+x^2)^2}dx=0-0=0$

Solve :
$\displaystyle \int_0^{\pi}\dfrac{1}{a^2-2a \cos x+1}dx$

$I=\int _{ 0 }^{ \pi }{ \dfrac { 1 }{ { a }^{ 2 }-2a\cos x+1 } dx } \\ putting\quad the\quad value\quad of\quad \cos x\quad as\\ \cos x=\dfrac { 1-{ \tan }^{ 2 }\dfrac { x }{ 2 } }{ 1+{ \tan }^{ 2 }\dfrac { x }{ 2 } } \\ \therefore I=\int _{ 0 }^{ \pi }{ \dfrac { 1 }{ { a }^{ 2 }-2a.\dfrac { 1-{ \tan }^{ 2 }\dfrac { x }{ 2 } }{ 1+{ \tan }^{ 2 }\dfrac { x }{ 2 } } +1 } dx } \\ =\int _{ 0 }^{ \pi }{ \dfrac { 1+{ \tan }^{ 2 }\dfrac { x }{ 2 } }{ ({ a }^{ 2 }+1)\left( 1+{ \tan }^{ 2 }\dfrac { x }{ 2 } \right) -2a\left( 1{ -\tan }^{ 2 }\dfrac { x }{ 2 } \right) } dx } \\ =\int _{ 0 }^{ \pi }{ \dfrac { { \sec }^{ 2 }\dfrac { x }{ 2 } }{ ({ a }^{ 2 }+1)\left( 1+{ \tan }^{ 2 }\dfrac { x }{ 2 } \right) -2a\left( 1{ -\tan }^{ 2 }\dfrac { x }{ 2 } \right) } dx } \\$

$Let\quad \tan\dfrac { x }{ 2 } =t\\ \Rightarrow { \sec }^{ 2 }\dfrac { x }{ 2 } dx=2dt\\ also,\quad x=0\Rightarrow t=\tan0=0\\ and,\quad x=\pi \Rightarrow t=\tan\dfrac { \pi }{ 2 } =\infty \\ \therefore I=\int _{ 0 }^{ \infty }{ \dfrac { 2dt }{ \left( { a }^{ 2 }+1 \right) \left( 1+{ t }^{ 2 } \right) -2a\left( 1-{ t }^{ 2 } \right) } } \\ =2\int _{ 0 }^{ \infty }{ \dfrac { dt }{ { \left( 1-a \right) }^{ 2 }+{ \left( (1+a)t \right) }^{ 2 } } } \\ =\dfrac { 2 }{ { \left( 1-a \right) }^{ 2 } } \int _{ 0 }^{ \infty } \dfrac { dt }{ 1+{ \left[ \dfrac { 1+a }{ 1-a } t \right] }^{ 2 } } \\ again\quad Let\quad \dfrac { 1+a }{ 1-a } t=u\\ \Rightarrow dt=du\left( \dfrac { 1-a }{ 1+a } \right) \\ \therefore I=\dfrac { 2 }{ (1-a)(1+a) } \int _{ 0 }^{ \infty }{ \dfrac { du }{ 1+{ u }^{ 2 } } } \\ =\dfrac { 2 }{ (1-a)(1+a) } { \left[ { \tan }^{ -1 }u \right] }_{ 0 }^{ \infty }\\ =\dfrac { 2 }{ 1-{ a }^{ 2 } } \left( \dfrac { \pi }{ 2 } -0 \right) \\ =\dfrac { \pi }{ 1-{ a }^{ 2 } }$

Evaluate:
$\displaystyle \int _{ 0 }^{ 1 } x^2+ 2x$ dx

$\displaystyle \int_0^1 x^2+2x dx$

$=\displaystyle \int _0^1 x^2dx +\int _0^1 2x dx$

By using

$\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$

$=\left.\dfrac {x^3}{3}+x^2\right|_0^1$

$=\dfrac 13+1=\dfrac 43$

Solve $\displaystyle\int\dfrac{{\cos x}}{{1 + \cos x}}dx$

Consider the given integral.

$I=\displaystyle\int{\dfrac{\cos x}{1+\cos x}}dx$

$I=\displaystyle\int{\dfrac{1+\cos x-1}{1+\cos x}}dx$

$I=\displaystyle\int{\dfrac{1+\cos x}{1+\cos x}}dx-\displaystyle\int{\dfrac{1}{1+\cos x}}dx$

We know that

$\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$

Therefore,

$I=\displaystyle\int{1}dx-\displaystyle\int{\dfrac{1}{1+2{{\cos }^{2}}\dfrac{x}{2}-1}}dx$

$I=\displaystyle\int{1}dx-\dfrac{1}{2}\displaystyle\int{\dfrac{1}{{{\cos }^{2}}\dfrac{x}{2}}}dx$

$I=\displaystyle\int{1}dx-\dfrac{1}{2}\displaystyle\int{{{\sec }^{2}}\dfrac{x}{2}}dx$

$I=x-\dfrac{1}{2}\dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}}+C$

$I=x-\tan \dfrac{x}{2}+C$

Hence, this is the answer.

Solve : $\displaystyle \int \, \dfrac{dx}{\sqrt{2x - x^2}}$

$I=\displaystyle\int \dfrac{dx}{\sqrt{2x-x^2}}=\displaystyle\int \dfrac{dx}{\sqrt{1-1+2x-x^2}}\cdot \displaystyle\int \dfrac{dx}{\sqrt{1-(x^2-2x+1)}}$

$I=\displaystyle\int\dfrac{dx}{\sqrt{1^2-(x-1)^2}}$

Put

$(x-1)=t$

$dx=dt$ .

$I=\displaystyle\int\dfrac{dt}{\sqrt{1-t^2}}$

Put

$t=\sin\theta$

$dt=\cos\theta d\theta$

$I=\displaystyle\int \dfrac{\cos \theta d\theta}{\sqrt{1-\sin^2\theta}}=\displaystyle\int \dfrac{\cos\theta d\theta}{\cos\theta}=\displaystyle\int d\theta$

$I=\theta +c=\sin^{-1}(t)+c=\sin^{-1}(x-1)+c$

$\therefore \displaystyle\int \dfrac{dx}{\sqrt{2x-x^2}}=\sin^{-1}(x-1)+c$ .

Evaluate the integrals using substitution.
$\displaystyle\int^1_0\dfrac{x^2}{x^2+1}dx$ .

$\int _{ 0 }^{ 12 }{ \cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } } dx$

$x=\tan { \theta } \Rightarrow dx=\sec ^{ 2 }{ \theta } d\theta \quad$

$x=0\Rightarrow \theta ={ 0 }^{ o },x=1\Rightarrow \theta =\cfrac { \pi }{ 4 }$

$\quad =\int _{ 0 }^{ \pi /4 }{ \cfrac { \tan ^{ 2 }{ \theta } }{ \sec ^{ 2 }{ \theta } } .\sec ^{ 2 }{ \theta } .d\theta } =\int _{ 0 }^{ \pi /4 }{ \tan ^{ 2 }{ \theta } } .d\theta$

$=\int _{ 0 }^{ \pi /4 }{ \left( \sec ^{ 2 }{ \theta } -1 \right) } .d\theta ={ \left[ \tan { \theta } -\theta \right] }_{ 0 }^{ \pi /4 }=1-\cfrac { \pi }{ 4 }$

$\int x\log (1-x^{2})dx =$

Let

$1-x^{2}=t$

$-2xdx=dt$

$xdx=\dfrac{-dt}{2}$

$\int x\log (1-x^{2})dx= \dfrac{-1}{2}\int\log tdt$

$=\dfrac{-1}{2}(t\log t-\int1dt)$

$=\dfrac{-1}{2}t(\log t-1) +c$

$=\dfrac{-1}{2}(1-x^{2})(\log (1-x^{2})-1)+c$

Solve:
$\int {\sin x\left( {2 + 5x} \right)} dx$

$\displaystyle\int \sin x(2+5x)dx=\displaystyle\int (2\sin x+5x\sin x)dx$

$=\displaystyle\int 2\sin xdx+\displaystyle\int 5x\sin xdx$

$=-2\cos x+5\left[x\displaystyle\int \sin x dx-\displaystyle\int\left\{\dfrac{d}{dx}(x)\displaystyle\int \sin x dx\right\}dx\right]$

$=-2\cos x+5\left[x(-\cos x)-\displaystyle\int (-\cos x)dx\right]$

$=-2\cos x+5\left[-x\cos x+\displaystyle\int\cos xdx\right]$

$=-2\cos x-5x\cos x+5\sin x+c$

$\displaystyle\int \sin x(2+5x)dx=5\sin x-2\cos x-5x\cos x+c$ .

Solve:
$\int {{e^x}} \left( {\frac{{x - 1}}{{{x^2}}}} \right)dx$

$\int e^x(\dfrac{x-1}{x^2})dx$

$=\int e^x (\dfrac{1}{x}-\dfrac{1}{x^2})dx$

$=\int e^x(\dfrac{1}{x})dx$ $-$ $\int e^x\dfrac{1}{x^2}dx$

$=\dfrac{1}{x}\int e^x dx$ $-$ $\int \dfrac{d}{dx} \dfrac{1}{x}$ $\int e^x dx)dx$ $-$ $\int e^x \dfrac{1}{x^2}dx$

$=\dfrac{1}{x} e^x$ $-$ $\int \dfrac{-1}{x^2} e ^xdx$ $-$ $\int e^x \dfrac{1}{x^2} dx$

$=\dfrac{e^x}{x}$ $+$ $c$

$\displaystyle\int^1_0\tan^{-1}(1-x+x^2)dx$ .

$\displaystyle \int_0^1 \cfrac{\pi}{2} -\tan^{-1} \cfrac{1}{1-x+x^2} dx$

$I = \displaystyle \cfrac{\pi}{2} - \int_0^1 (\tan^{-1} x - \tan^{-1} (x-1))dx$

$I = \displaystyle \cfrac{\pi}{2} - \int_0^1 (\tan^{-1}(1-x) - \tan^{-1} (-x))dx$

$2I = \displaystyle \pi - 2\int_0^1 \tan^{-1} x dx$

$2I = \displaystyle \pi - \Bigg[2x \tan^{-1} x + \ln(1+x^2)\Bigg]_0^1$

$I = \displaystyle \cfrac{\pi}{2} - \cfrac{\pi}{4} + \cfrac{\ln 2}{2} = \cfrac{\pi}{4} + \cfrac{\ln 2}{2}$

Solve : $\int \dfrac{t^4 - 3 t^2 + 2}{t^2 (1 + t^2)} , dt$

$I=\displaystyle \int \dfrac {t^4 -3t^2 +2}{t^2 (1+t^2)}dt$

$=\displaystyle \int \dfrac {t^4 +t^2 4t^2+2}{t^2 (1+t^2)}dt$

$\Rightarrow \ \displaystyle \int dt-\displaystyle \int \dfrac {4t^2}{t^2 (1+t^2)}dt +\displaystyle \int \dfrac {2}{t^2 (1+t^2)}dt$

$I\Rightarrow \ t-\displaystyle \int \dfrac {4}{(1+t^2)}dt+\displaystyle \int 2 \left[\dfrac {1}{t^2}-\dfrac {1}{1+t^2}\right] dt$

$I\Rightarrow \ t-4\tan^{-1} (t)+2 \displaystyle \int \dfrac {1}{t^2}dt+\displaystyle \int \dfrac {2}{(1+t^2)}dt$

$I=t-4\tan^{-1}t+2\left(\dfrac {t^{-2+1}}{-2+1}\right)-2\tan^{-1} (t)+C$

$\boxed {I\Rightarrow \ t-\dfrac {2}{t}-6\tan^{-1} (t)+C}$

$\therefore \ \boxed {\displaystyle \int \dfrac {t^4-3t^2+2}{t^2 (1+t^2)}dt=t-\dfrac {2}{t}-6\tan^{-1} (t)+C}$

Solve : $\displaystyle2 \underset{0}{\overset{\pi / 2}{\int}} x \, sin \, x \, dx$

$I=2\displaystyle\int^{\dfrac{\pi}{2}}_0x\sin xdx$

Integration by parts

$\displaystyle\int^b_af(x)g(x)dx=f(x)\displaystyle\int g(x)dx\displaystyle\int^b_a-\displaystyle\int^b_af'(x)g(x)dx$

$I=2\left[x[-\cos x]^{\dfrac{\pi}{2}}_0-\displaystyle\int^{\dfrac{\pi}{2}}_0(1)(-\cos x)dx\right]$

$=2\left[-\left(\dfrac{\pi}{2}\cos \dfrac{\pi}{2}-0\right)+\displaystyle\int^{\dfrac{\pi}{2}}_0\cos xdx\right]$

$=2\left[\left[(0-0)+\sin x\right]^{\dfrac{\pi}{2}}_0\right]$

$=2\left[\sin\dfrac{\pi}{2}-\sin (0)\right]\Rightarrow 2\left[1-0\right]\Rightarrow 2$

$\therefore 2\displaystyle\int^{\dfrac{\pi}{2}}_0x\sin xdx=2$ .

Solve: $\displaystyle \int_0^{\pi}(\sqrt{1+\sin t}).dt$

$\displaystyle\int_{0}^{\pi}\sqrt{1+\sin t}\ dt$

$=\displaystyle\int_{0}^{\pi}\sqrt{\sin^{2}\left(\dfrac{t}{2}\right)+\cos^{2}\left(\dfrac{t}{2}\right)+2\sin \left(\dfrac{t}{2}\right)\cos \left(\dfrac{t}{2}\right)} dt$

$\left\{\because 1=\sin^{2}\theta+\cos^{2}\theta\sin (\theta)=\sin \left(\dfrac{\theta}{2}\right)\omega \left(\dfrac{\theta}{2}\right)\right\}$

$=\displaystyle\int_{0}^{\pi}\sqrt{\left[\sin\left(\dfrac{t}{2}\right)+\cos \left(\dfrac{t}{2}\right)\right]^{2}}\ dt$

$=\displaystyle\int_{0}^{\pi}\left[\sin\left(\dfrac{t}{2}\right)+\cos\left(\dfrac{t}{2}\right)\right]\ dt$

$=\displaystyle\int_{0}^{\pi}\left(\sin \dfrac{t}{2}+\cos\dfrac{t}{2}\right)\ dt$

$=2\left[-\cos\left(\dfrac{t}{2}\right)+\sin \left(\dfrac{t}{2}\right)\right]_{0}^{\pi}$

$=2[1+1]=\boxed{4}$

$\displaystyle\int^2_1\left(\dfrac{1}{x}-\dfrac{1}{2x^2}\right)e^{2x}dx$ .

As we can see that

$d\Bigg(\cfrac{e^{2x}}{2x}\Bigg) = e^{2x} \Bigg(\cfrac{1}{x} - \cfrac{1}{2x^2}\Bigg) dx$

$\displaystyle \int_1^2 d\Bigg(\cfrac{e^{2x}}{2x}\Bigg)$

$\Bigg[\cfrac{e^{2x}}{2x}\Bigg]^2_1$

$\cfrac{e^4}{4} - \cfrac{e^2}{2}$

$\displaystyle\int x^2\sin^{-1}x dx$ .

$\displaystyle \int x^2 \sin^{-1} x dx$

Using byparts,we get

$\displaystyle \cfrac{1}{3} x^3 \sin^{-1} x - \cfrac{1}{3} \int \cfrac{x^3}{\sqrt{1-x^2}} dx$

Putting

$x^2 = m$

$2x dx = dm$

$\displaystyle \cfrac{1}{3} x^3 \sin^{-1} x -\cfrac{1}{6} \int \cfrac{m}{\sqrt{1-m }} dm$

$\displaystyle \cfrac{1}{3} x^3 \sin^{-1} x - \cfrac{1}{6} \int \cfrac{1-(1-m)}{\sqrt{1-m}}dm$

$\cfrac{1}{3} x^3 \sin^{-1} x - \cfrac{1}{3} \sqrt{1-x^2} + \cfrac{1}{9} \sqrt{(1-x^2)^3}$

$\displaystyle \int { \dfrac { 1 }{ 2 } \cos ^{ -1 }{ x } dx }$

Let,

$\dfrac{1}{2}cos^{-1}x=t$

$\implies x=cos2t$

$\implies dx=-2sin2tdt$

$\implies \int -2t.sin2t.dt$

$\implies -2\int t.sin2t.dt$

Solving this, By parts, we get

$\implies -2[\dfrac{-t.cos2t}{2}+\dfrac{sin2t}{4}]+C$

$\implies t.cos2t+\dfrac{sin2t}{2}+C$

$\implies \dfrac{1}{2}cos^{-1}x.cos2(\dfrac{1}{2}cos^{-1}x)+\dfrac{sin2(\dfrac{1}{2}cos^{-1}x)}{2}+C$

$\implies \dfrac{1}{2}(cos^{-1}x)^2+\dfrac{sin(cos^{-1}x)}{2}+C$

Evaluate:
$\displaystyle\int\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2dx$ .

$\begin{array}{l} \int \begin{array}{l} \left( { \frac { { x-1 } }{ { \sqrt { x } } } } \right)^2 dx \\ \end{array} \\ \int { { { \left( { \sqrt { x } } \right) }^{ 2 } }-2\sqrt { x } .\frac { 1 }{ { \sqrt { x } } } } +{ \left( { \frac { 1 }{ { \sqrt { x } } } } \right) ^{ 2 } }dx \\ \int { xdx-2\int { dx+\int { \frac { 1 }{ x } dx } } } \\ \frac { { { x^{ 2 } } } }{ 2 } -2x+\log x+c \end{array}$

Evaluate: $\displaystyle\int \dfrac {dx}{x(1 + x^{2})}$ .

1360225_1133813_ans_2705b0ba6c634985b9dca7d32effe710.jpg

Evaluate :

$\displaystyle\int _{0}^{\pi/2} x\cos{2x}dx$

$\displaystyle\int _{ 0 }^{ \pi / 2 }{ x\cos { 2xdx } }$

$I=\displaystyle\int _{ 0 }^{ { \pi / }{ 2 } } \cos\ 2x\ dx-\int _{ 0 }^{ { \pi }{ /2 } }{ { \left[ \dfrac { d }{ dx } x\left( \int { \cos { 2x\ dx } } \right) \right] }dx }$

$={ { \left[ x\dfrac { \sin { 2x } }{ 2 } \right] }_{ 0 }^{ { \pi / 2 } } }-{ \left[ \dfrac { 1 }{ 2 } \left( \dfrac { -\cos\ { 2x } }{ 2 } \right) \right] }_{ 0 }^{ { \pi / 2 } }+C$

$I=\left[ \dfrac { \pi }{ 2 } \left( 0 \right) -0 \right] +\dfrac { 1 }{ 4 } \left[ \cos { \pi } -\cos { 0 } \right]+ C$

$=\dfrac { 1 }{ 4 } \left( -1-1 \right) =\dfrac { 1 }{ 4 } \left( -2 \right) =\dfrac { -1 }{ 2 }$

Solve: $\displaystyle \int_0^1 \dfrac{x\tan^{-1}x}{(1+x^2)^{3/2}}dx$

1968236_1140428_ans_1b2a46b05d5c4b72aa596fb3c58a1bb6.jpg

$I=\int_{0}^{1}{\tan^{-1}(\frac{2x-1}{1+x-x^{2}})}$

$I=\displaystyle \int _0^1 \tan^{-1} \left (\dfrac {2x-1}{1+x-x^2} \right)dx$

$Hx-x^2=t$

$(1-2n)dx=dt$

$=\dfrac {1}{1+\left (\dfrac {2n-1}{1-x-x^2}\right)^2}\displaystyle \int _0^1 +C$

$\Rightarrow \ \left [\dfrac {1}{1+\left (\dfrac {2(1)-1}{1-1-1}\right)^2} -\dfrac {1}{1+\left (\dfrac {2(0)-1}{1-0-0}\right)^2} \right]+C$

$\Rightarrow \ \left [\dfrac {1}{1+\left (\dfrac {2-1}{-1}\right)^2} -\dfrac {1}{1-\left (\dfrac {0-1}{1}\right)^2} \right]+C$

$=\left [\dfrac {1}{1+1} -\dfrac {1}{1-1}\right]$

$\Rightarrow \ \dfrac {1}{2}+C$

Evaluate $\int x ^ { 3 } \log 2 x d x$

1521178_1141378_ans_bb69af045101400cb815dbce15b507a1.jpg

Solve: $\displaystyle \overset{\frac{\pi}{2}}{\underset{0}{\int}} \sqrt{\cos \theta}\cdot \sin^3 \theta d\theta$

$\displaystyle\int_{0}^{\frac{\pi}{2}}{\sqrt{\cos{\theta}}{\sin}^{3}{\theta}d\theta}$

$=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sqrt{\cos{\theta}}{\sin}^{2}{\theta}\sin{\theta}d\theta}$

$=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sqrt{\cos{\theta}}\left(1-{\cos}^{2}{\theta}\right)\sin{\theta}d\theta}$

Let

$t=\cos{\theta}\Rightarrow\,dt=-\sin{\theta}d\theta$

When

$\theta=0\Rightarrow\,t=1$

When

$\theta=\dfrac{\pi}{2}\Rightarrow\,t=0$

$=-\displaystyle\int_{1}^{0}{\sqrt{t}\left(1-{t}^{2}\right)dt}$

$=-\displaystyle\int_{1}^{0}{\left({t}^{\frac{1}{2}}-{t}^{\frac{1}{2}+2}\right)dt}$

$=-\displaystyle\int_{1}^{0}{\left({t}^{\frac{1}{2}}-{t}^{\frac{5}{2}}\right)dt}$

$=-\left[\dfrac{{t}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}-\dfrac{{t}^{\frac{5}{2}+1}}{\dfrac{5}{2}+1}\right]_{1}^{0}$

$=-\left[\dfrac{{t}^{\frac{3}{2}}}{\dfrac{3}{2}}-\dfrac{{t}^{\frac{7}{2}}}{\dfrac{7}{2}}\right]_{1}^{0}$

$=-\dfrac{2}{3}\left[{t}^{\frac{3}{2}}\right]_{1}^{0}+\dfrac{2}{7}\left[{t}^{\frac{7}{2}}\right]_{1}^{0}$

$=-\dfrac{2}{3}\left[0-1\right]+\dfrac{2}{7}\left[0-1\right]$

$=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14-6}{21}=\dfrac{8}{21}$

$\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } }$

$\displaystyle \int _{-1}^{1}\dfrac {dx}{x^2+zx+5}$

$\displaystyle \int _{-1}^{1}\dfrac {1}{(x+1)^2+4}dx$

$\dfrac {1}{2}\tan^{-1}\dfrac {(x+1)}{2}|_{-1}^{1}$

$\dfrac {1}{2}\tan^{-1} \left (\dfrac {2}{2}\right)-\dfrac {1}{2}\tan^{-1}\left (\dfrac {-1+1}{2}\right)$

$\dfrac {1}{2}\tan^{-1} 1-\dfrac {1}{2}\tan^{-1}0$

$\dfrac {1}{2} \left (\dfrac {\pi}{4}\right)-\dfrac {1}{2}(0)\ \Rightarrow \dfrac {\pi}{8}-0=\boxed {\dfrac {\pi}{8}}$

$\int x\sqrt{3x-2}dx$

$\displaystyle \int x \sqrt {3x-2}dx$

$\dfrac {1}{3} \left (\dfrac {1}{3}\right) \displaystyle \int (t+2)\sqrt t dt$

$\dfrac {1}{9} \displaystyle \int (t^{3/2 +2t^{1/2}})dt$

$\dfrac {1}{9} \displaystyle \int \left [2 \dfrac {t ^{5/2}}{5} +\dfrac {2t^{3/2}}{3} \right]$

$\dfrac {1}{9} \left [\dfrac {2(3x-2)^{5/2}}{5}+\dfrac {4}{3} (3x-2)^{3/2} \right]$

Find the following integral.
$\displaystyle\int (x^{\frac{2}{3}}+1)dx$ .

$\int { \left( { { x^{ \frac { 2 }{ 3 } } }+1 } \right) dx } \\ \int { { x^{ \frac { 2 }{ 3 } } }dx+\int { dx } } \\ \Rightarrow \frac { { { x^{ \frac { 5 }{ 3 } } } } }{ { \frac { 5 }{ 3 } } } +x+c,\, \, \, \frac { 3 }{ 5 } { x^{ \frac { 5 }{ { 13 } } } }+x+0\, \, \,$

Evaluate $\displaystyle \int { \frac { 1 }{(1-2x) } }dx$

$\displaystyle \int \dfrac{1}{1-2x} dx\\t=1-2x\implies dt=-2dx\\\displaystyle \int \dfrac 1{-2}\dfrac 1t dt\\\dfrac 1{-2}\log t+c$

If $g(x) =\displaystyle \int _{ 0 }^{ x }{ \cos { 4t\ dt, } }$ then $g(x+\pi)$ equals

$g(x) =\displaystyle \int _{ 0 }^{ x }{ \cos { 4t\ dt, } }$

$=\left[ \dfrac{\sin 4 t}{4}\right]_0^x$

$g(x)=\dfrac{\sin 4 x}{4}-0$

$g(x)=\dfrac{\sin 4 x}{4}$

$g (x+ \pi)= \dfrac{\sin 4 (x+ \pi)}{4}$

$= \dfrac{\sin (4 \pi+4x)}{4}$

$=\dfrac{\sin 4x}{4}$

$\int\limits_0^{\pi /2} {\dfrac{{\cos x}}{{1 + \cos x + \sin x}}dx}$

$I= \int_{0}^{\pi/2} \dfrac{\cos x}{1+\cos x+ \sin x}$

$=\int_{0}^{\pi/2} \dfrac{\cos^{2} x/2 -\sin^{2} x/2 }{2 \cos^{2} x/2 +2 \sin x/2 \cos x/2}dx$

$=\int_{0}^{\pi/2} \dfrac{(\cos^{2} x/2 - \sin^{2} x/2)}{2 \cos x/2 (\cos x/2 + \sin x/2)} dx$

$=\int_{0}^{\pi/2} \dfrac{(\cos x/2- \sin x/2)(\cos x/2 + \sin x/2)}{2 \cos x/2 (\cos x/2 + \sin x/2)}dx$

$=\int_{0}^{\pi/2} \dfrac{(\cos x/2- \sin x/2)}{2 \cos x/2} dx$

$=\dfrac{1}{2 }\int_{0}^{\pi/2} dx - \dfrac{1}{2} \int_{0}^{\pi/2} \tan \dfrac{x}{2} dx$

$=\left[ \dfrac{1}{2} x -\dfrac{1}{2} \log \sec \dfrac{x}{2} . 2 \right]^{\pi/2}_{0}$

$=\left[ \dfrac{x}{2}- \log \sec \dfrac{x}{2} \right]^{\pi/2}_{0}$

$= \dfrac{1}{2} \left[ \dfrac{\pi}{2}- \log 2 \right]$

Evaluate
$\int\limits_0^\infty {\frac{{\log \left( {1 + {x^2}} \right)dx}}{{1 + {x^2}}}} =$

1311124_1170295_ans_faedc3e7446648edae36bc4887578e3c.jpg

$\displaystyle\int_{\frac{1}{9}}^{\frac{1}{4}} {\dfrac{{dx}}{{(1 - x)\sqrt x }}} = \,\log \dfrac{3}{2}$

$I=\displaystyle \int_{1/9}^{1/4}\dfrac{dx}{(1-x)\sqrt{x}}$

Let

$\sqrt{x}=t$ limit , at

$\sqrt{1/9}=t=\dfrac{1}{3}$

$\dfrac{1}{2\sqrt{x}}dx=dt$ at

$x=\dfrac{1}{4},\sqrt{\dfrac{1}{4}}=t=\dfrac{1}{2}$

$\dfrac{dx}{\sqrt{x}}=2dt$

$\therefore I =\displaystyle \int_{1/3}^{1/2} \dfrac{2dt}{(1-t^2)}$

$=\left[\dfrac{2}{2}\log\left|\dfrac{1+t}{1-t}\right|\right]_{1/3}^{1/2}$

$\dfrac{2}{2}\left[\log\left|\dfrac{1+1/2}{1-1/2}\right|-\log\left|\dfrac{1+1/3}{1-1/3}\right|\right]$

$=\log \dfrac{3}{2}$

Evaluate:
$\displaystyle \int { { \left( a\tan { x+b\cot { x } } \right) }^{ 2 }dx }$

$I=\displaystyle\int (a\tan x+b\cot x)^2dx$

$=\displaystyle\int (a^2\tan^2x+b^2\cot^2x+2ab)dx$

$I=\displaystyle\int a^2(\sec^2x-1)+b^2(cosec^2x-1)+2ab)dx$

$I=\displaystyle\int a^2\sec^2xdx+\displaystyle\int b^2 cosec^2dx-\displaystyle\int (a-b)^2dx$

$I=a^2\tan x+\displaystyle\int b^2 cosec^2xdx-\displaystyle\int (a-b)^2dx$

$I=a^2\tan x-b^2\cot x-\displaystyle\int (a-b)^2dx$

$I=a^2\tan x-b^2\cot x-(a-b)^2x+c$ .

Evaluate : $\int { { 2x }^{ 3 }{ e }^{ { x }^{ 2 } } } dx$

Let,

$x^2=t$

So,

$2x.dx=dt$

$\implies \int t.e^tdt$

Solving By parts, we get

$\implies t.e^t-e^t+C$

$\implies x^2e^{x^2}-e^{x^2}+C$

$\implies e^{x^2}(x^2-1)+C$

Integrate $\int x.sin2xdx$

$\int{x . \sin{2x} \; dx} = ?$

Using ILATE function, we have

$\int{ \underset{I}{x} . \underset{II}{\sin{2x}} \; dx}$

$= x \left( -\cfrac{1}{2} \cos{2x} \right) - \int{\left( 1 . \left( -\cfrac{1}{2} \cos{2x} \right) \right) \; dx}$

$= - \cfrac{x \cos{2x}}{2} + \cfrac{1}{2} \int{\cos{2x} \; dx}$

$= - \cfrac{x \cos{2x}}{2} + \cfrac{1}{2} \left( \cfrac{\sin{2x}}{2} \right) + C$

$= \cfrac{\sin{2x}}{4} - \cfrac{x \cos{2x}}{2} + C$

Hence

$\int{x . \sin{2x} \; dx} = \cfrac{\sin{2x}}{4} - \cfrac{x \cos{2x}}{2} + C$ .

$\int_{\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{sin x +cos x}{\sqrt{sin 2x}}dx$

We have,

$\int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$

$=\int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\sin x+\cos x}{\sqrt{1-1+\sin 2x}}dx}$

$=\int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}dx}$

$=\int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\dfrac{d}{dx}\left( \sin x-\cos x \right)}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}dx}\,\,\,\,\,\,\therefore \dfrac{1}{\sqrt{1-{{x}^{2}}}}dx={{\sin }^{-1}}x$

$={{\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin x-\cos x \right) \right]}_{-\dfrac{\pi }{6}}}^{\dfrac{\pi }{6}}$

$=\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin \left( -\dfrac{\pi }{6} \right)-\cos \left( -\dfrac{\pi }{6} \right) \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right) \right) \right]$

$=\left[ \sqrt{2}{{\sin }^{-1}}\left( -\sin \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right) \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right) \right) \right]$

$=\left[ \sqrt{2}{{\sin }^{-1}}\left( -\dfrac{1}{2}-\dfrac{\sqrt{3}}{2} \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{2} \right) \right]$

$=\left[ \sqrt{2}{{\sin }^{-1}}\left( \dfrac{-1-\sqrt{3}}{2} \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \dfrac{1-\sqrt{3}}{2} \right) \right]$

$=\sqrt{2}\left[ {{\sin }^{-1}}\left( \dfrac{-1-\sqrt{3}}{2} \right) \right]-\left[ {{\sin }^{-1}}\left( \dfrac{1-\sqrt{3}}{2} \right) \right]$

Hence, this is the answer.

Integrate $\int_{-\dfrac {\pi}{4}}^{\dfrac {\pi}{4}} \dfrac {1}{1 - \sin x} dx$ .

$I=\int _{ -\pi /4 }^{ \pi /4 }{ \cfrac { 1 }{ 1-\sin { x } } dx }$

$I=\int _{ -\pi /4 }^{ \pi /4 }{ \cfrac { 1 }{ 1-\sin { \left( \cfrac { \pi }{ 4 } -\left( -\cfrac { \pi }{ 4 } \right) -x \right) } } dx } =\int _{ -\pi /4 }^{ \pi /4 }{ \cfrac { 1 }{ 1+\sin { x } } dx }$

$2I=\int _{ -\pi /4 }^{ \pi /4 }{ \left( \cfrac { 1 }{ 1-\sin { x } } +\cfrac { 1 }{ 1+\sin { x } } \right) dx } =\int _{ -\pi /4 }^{ \pi /4 }{ \cfrac { 2 }{ \cos { x } } dx }$

$\Rightarrow I=\int _{ -\pi /4 }^{ \pi /4 }{ \sec { x } dx } =\log { \left| \sec { x } +\tan { x } \right| { ] }_{ -\pi /4 }^{ \pi /4 } }$

$=\log { \left| \sqrt { 2 } +1 \right| } -\log { \left| \sqrt { 2 } -1 \right| } =\log { \left| \cfrac { \sqrt { 2 } +1 }{ \sqrt { 2 } -1 } \right| }$

$=\log { \left| 3+2\sqrt { 2 } \right| }$

Solve: $\displaystyle \int_0^{2\pi}\dfrac{\cos x }{1+\sin^2 x} dx$

$\int _{ 0 }^{ 2\pi }{ \dfrac { \cos x }{ 1+{ \sin }^{ 2 }x } dx } \\ Let\quad \sin x=t\\ \Rightarrow \cos xdx=dt\\ \because \int { \dfrac { \cos x }{ 1+{ \sin }^{ 2 }x } dx=\int { \dfrac { dt }{ 1+{ t }^{ 2 } } } } \\ ={ \tan }^{ -1 }t\\ ={ \tan }^{ -1 }(\sin x)\\ \therefore \int _{ 0 }^{ 2\pi }{ \dfrac { \cos x }{ 1+{ \sin }^{ 2 }x } dx } ={ \left[ { \tan }^{ -1 }(\sin x) \right] }_{ 0 }^{ 2\pi }\\ ={ \tan }^{ -1 }(\sin2\pi )-{ \tan }^{ -1 }(\sin0)\\ =0-0\\ =0$

Evaluate
$\int {x\,{e^{ - x}}} \,\,dx$

$\displaystyle\int{x{e}^{-x}dx}$

Integrating by parts,

Let

$u=x\Rightarrow\,du=dx$

$dv={e}^{-x}dx\Rightarrow\,v=-{e}^{-x}$

$=-x{e}^{-x}-\displaystyle\int{-{e}^{-x}dx}$

$=-x{e}^{-x}+\displaystyle\int{{e}^{-x}dx}$

$=-x{e}^{-x}-{e}^{-x}+c$

Solve:
$\int \dfrac{1}{\sqrt{1-e^{2x}}}dx$

$\int \dfrac{1}{\sqrt{1 - e^{2x}}} dx$

$= \int \dfrac{1}{\sqrt{1 - (e^x)^2}} dx$

$= \sin^{-1} (e^x) + c$

If $f(x)=\displaystyle \int_{0}^{x}{t(\sin x-\sin t)dt}$ then $f'(x)$ ?

Given,

$f(x)= \int_{0}^{x} t (\sin x- \sin t) dt;$

Now applying Libnitz's rule fon differentiating under integration we get,

$f'(x)= x (\sin x- \sin x ).1-0+ \int_{0}^{x} t \cos x dt$

on,

$f'(x)= (\cos x)(x^{2}/2)$

$\int\limits_0^{\pi /2} {\dfrac{{a\sin x + b\cos x}}{{\sin x + \cos x}}dx}$

$I=\displaystyle \displaystyle \int_{0}^{\pi /2}\dfrac {a\sin x+b \cos x}{\sin x+ \cos x}dx-------(1)$

$I=\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac{ a\sin { \left( \dfrac{ \pi }{ 2 } +o-x \right) +b\cos { \left( \dfrac{ \pi }{ 2 } +o-x \right) } } }{ \sin { \left( \dfrac{ \pi }{ 2 } +o-x \right) +\cos { \left( \dfrac{ \pi }{ 2 } +o-x \right) } } } dx }$

$I=\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac{ a\sin { \left( \dfrac{ \pi }{ 2 } -x \right) + } b\cos { \left( \dfrac{ \pi }{ 2 } -x \right) } }{ \sin { \left( \dfrac{ \pi }{ 2 } -x \right) + } \cos { \left( \dfrac{ \pi }{ 2 } -x \right) } } }$

$I=\displaystyle \displaystyle \int_{0}^{\pi /2}\dfrac {a\cos x+b\sin x}{\cos x+\sin x}dx-----(2)$

Adding

$(1)$ &

$(2)$

$2I=\displaystyle \displaystyle \int_{0}^{\pi /2}\dfrac {a(\sin x+\cos x)+b(\sin x+\cos x)}{(\cos x+\sin x)}dx$

$2I=\displaystyle \displaystyle \int_{0}^{\pi /2} (a+b)dx$

$I=(a+b)\dfrac {\pi}{4}$

Solve: $\displaystyle \int_1^2 \dfrac{1}{x(1+x^2)}dx$

Given,

$\int _{ 1 }^{ 2 }{ \dfrac { 1 }{ x(1+{ x }^{ 2 }) } } dx$

We can write

$\dfrac { 1 }{ x(1+{ x }^{ 2 }) }$ as

$\dfrac { 1 }{ x } -\dfrac { x }{ 1+{ x }^{ 2 } }$

$\therefore \int _{ 1 }^{ 2 }{ \dfrac { 1 }{ x(1+{ x }^{ 2 }) } } dx=\int _{ 1 }^{ 2 }{ (\dfrac { 1 }{ x } -\dfrac { x }{ 1+{ x }^{ 2 } } ) } dx$

Now,

$\int _{ 1 }^{ 2 }{ (\dfrac { 1 }{ x } -\dfrac { x }{ 1+{ x }^{ 2 } } ) } dx\\ =\int _{ 1 }^{ 2 }{ \dfrac { 1 }{ x } } dx-\int _{ 1 }^{ 2 }{ \dfrac { x }{ 1+{ x }^{ 2 } } } dx$

So,

$\int _{ 1 }^{ 2 }{ \dfrac { 1 }{ x } } dx={ \left[ \ln { \left| x \right| } \right] }_{ 1 }^{ 2 }+{ C }_{ 1 }$

for,

$\int _{ 1 }^{ 2 }{ \dfrac { x }{ 1+{ x }^{ 2 } } } dx$ ,

Let,

$u=1+{ x }^{ 2 }\\ \Rightarrow \dfrac { du }{ dx } =2x\\ \Rightarrow du=2xdx\\ \Rightarrow \dfrac { 1 }{ 2 } du=xdx$

Substituting these values we get,

$\int _{ 1 }^{ 2 }{ \dfrac { x }{ 1+{ x }^{ 2 } } } dx\\ =\int _{ 1 }^{ 2 }{ \dfrac { 1 }{ 2u } } du\\ =\dfrac { 1 }{ 2 } \int _{ 1 }^{ 2 }{ \dfrac { 1 }{ u } } du\\ =\dfrac { 1 }{ 2 } { \left[ \ln { \left| u \right| } \right] }_{ 1 }^{ 2 }+{ C }_{ 2 }$

$\therefore \int _{ 1 }^{ 2 }{ \dfrac { 1 }{ x } } dx-\int _{ 1 }^{ 2 }{ \dfrac { x }{ 1+{ x }^{ 2 } } } dx\\ ={ \left[ \ln { \left| x \right| } \right] }_{ 1 }^{ 2 }-\dfrac { 1 }{ 2 } { \left[ \ln { \left| u \right| } \right] }_{ 1 }^{ 2 }+{ C }\\ ={ \left[ \ln { \left| x \right| } \right] }_{ 1 }^{ 2 }-\dfrac { 1 }{ 2 } { \left[ \ln { \left| 1+{ x }^{ 2 } \right| } \right] }_{ 1 }^{ 2 }+C\\ =\left[ \ln { 2 } -\ln { 1 } \right] -\dfrac { 1 }{ 2 } \left[ \ln { (1+{ 2 }^{ 2 })- } \ln { (1+{ 1 }^{ 2 }) } \right] +C\\ =\left[ \ln { 2 } -\ln { 1 } \right] -\dfrac { 1 }{ 2 } \left[ \ln { 5 } -\ln { 2 } \right] +C\\ =\ln { (\dfrac { 2 }{ 1 } ) } -\dfrac { 1 }{ 2 } \ln { (\dfrac { 5 }{ 2 } ) } +C\\ \therefore \int _{ 1 }^{ 2 }{ \dfrac { 1 }{ x(1+{ x }^{ 2 }) } } dx=\ln { (\dfrac { 2 }{ 1 } ) } -\dfrac { 1 }{ 2 } \ln { (\dfrac { 5 }{ 2 } ) } +C.$

$\displaystyle \int \left(\dfrac{1-\tan{x}}{1-\tan{x}}\right)^{2}dx$

$\int { { \left( \dfrac { 1-tanx }{ 1-tanx } \right) }^{ 2 } } dx$

$=\int { { 1 }^{ 2 } } dx$

$=\int { 1 } dx$

$=\int { dx }$

$=x+c$

$\int\limits_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx}$

$\begin{array} { l l } { \text { Let } x = \tan \theta } & { \Longrightarrow \theta = \tan ^ { - 1 } x } \\ { } & { \Longrightarrow \sin ^ { - 1 } \frac { 2 x } { 1 + x ^ { 2 } } = \sin ^ { - 1 } \frac { 2 \tan \theta } { 1 + \tan ^ { 2 } \theta } } \\ { } & { = \sin ^ { - 1 } ( \sin 2 \theta ) \quad \text { using } ( 1 ) } \\ { } & { = 2 \theta = 2 \tan ^ { - 1 } x } \end{array}$

$\begin{array} { l } { \text { Now the integral becomes fairly simple } } \\ { I = 2 \int _ { 1 } ^ { 0 } \tan ^ { - 1 } x \cdot 1 d x } \\ { \text { Integrating by parts } } \\ { \int u \cdot v d x = u \int v d x - \int \left( \frac { d u } { d x } \cdot \int v d x \right) d x } \end{array}$

$\begin{array} { l } { \text { Taking } v = 1 \quad \& \quad u = \tan ^ { - 1 } x } \\ { \Longrightarrow I = x \cdot \tan ^ { - 1 } \left. x \right| _ { 0 } ^ { 1 } - \left. \frac { \ln \left| 1 + x ^ { 2 } \right| } { 2 } \right| _ { 0 } ^ { 1 } } \end{array}$

$\Longrightarrow I = \frac { \pi } { 4 } - \frac { \ln 2 } { 2 }$

Evaluate; $\displaystyle\int _ { 0 } ^ { \pi/2 } \log \sin { 2 } x d x$

$\\I=\int_{0}^{(\frac{\pi}{2})}log sin2x dx\\Let 2x=t\\2dx=dt\\\therefore dx=(\frac{1}{2})dt\\and x\rightarrow(\frac{\pi}{2}), t\rightarrow\pi\\x\rightarrow 0, t\rightarrow 0\\I=\int_{0}^{\pi}logsin(t)(\frac{1}{2})dt\\=(\frac{1}{2})\int_{0}^{\pi}logsin(\pi-t)dt\\=(\frac{1}{2})\int_{0}^{\pi}logsin(\pi-t)dt\\=(\frac{1}{2})\int_{0}^{\pi}logsin\>t\>dt\\=(\frac{1}{2})\times2\int_{0}^{(\frac{\pi}{2})}log\>sint\>dt\\=-(\frac{\pi}{2})log2(standard\>result)$

Integrate : $\displaystyle\int _ { 1 } ^ { 2 } \frac { \ln x } { x ^ { 2 } } d x$

$\Rightarrow \ \displaystyle \int_{1}^{2}\dfrac {lnx}{x^2}dx$

let

$\ln x=t$

$\dfrac {1}{x}=\dfrac {dt}{dx}$

$\dfrac {1}{x}dx=dt$

$I=\displaystyle \int_{1}^{2}\dfrac {lnx}{x^2}dx=\displaystyle \int_{|n|=0}^{ln2}\dfrac {te^t}{e^{2t}}dt=\displaystyle \int_{0}^{ln2} t^{e^{-t}}dt$

Applying By parts gives

$I=(-te^{-t})_{0}^{ln2}+\displaystyle \int _{0}^{ln2}e^{-t}dt$

$I=-\ln2 \left (\dfrac {1}{2}\right)+(-e^{-t})_0^{ln2}$

$I=\dfrac { -\ln2 }{ 2 } +\left[ \dfrac { -1 }{ 2 } +1 \right] =\dfrac { 1 }{ 2 } -\dfrac { \ln2 }{ 2 }$

$\therefore \ \displaystyle \int_1^2 \dfrac {\ln2}{x^2}dx=\dfrac {1-\ln2}{2}$

Prove:
$\int _ { 0 } ^ { \pi } \dfrac { x \sin x } { 1 + \cos ^ { 2 } x } d x = \dfrac { \pi ^ { 2 } } { 4 }$

$I = \int_{0}^{\pi} \dfrac {x \, sin \,x}{1+ cos^2x}dx$ ...(1)

using property

$\infty$ definite integral

$\int_{a}^{b} f(x)dx =\int_{a}^{b} f(a+b-x)dx$

$I = \int_{a}^{b}\dfrac {(\pi-x)sin (\pi-x)}{1+cos^2(\pi-x)}dx$

$I = \int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}dx$ ....(2)

adding

$(1)$ &

$(2)$ , we get

$I+I=\int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}dx + \int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}$

$2I = \int_{a}^{b}\dfrac {(\pi- Sin x)sinx}{1+cos^2x}dx$

$I =\dfrac {\pi}{2}\int_{0}^{\pi} \dfrac {Sin x}{1+Cos^2x}dx$

$=\dfrac {\pi}{2}\cdot 2 \int_{0}^{\pi/2} \dfrac {Sin x}{1+Cos^2x}dx$

$= -\pi \int_{0}^{\pi/2} \frac {d(cos\,x)}{1+Cos^2x}dx$

$= - \pi [tan^{-1}(cos\,x)]_0^{\pi/2}$

$-\pi [ tan^{-1} cos \dfrac{\pi}2-tan^{-1}(cos \,0)]$

$-\pi [ 0- \dfrac{\pi} 4]$

$=\dfrac{\pi^2}4$

Evaluate: $\displaystyle \int^{\frac{\pi}{2}}_{0} log \,\sin x \,dx$

$I=\displaystyle\int^{\pi/2}_0log \sin xdx$ ………

$(1)$

$I=\displaystyle\int^{\pi/2}_0log\sin(\pi/2-x)dx$

$I=\displaystyle\int^{\pi/2}_0log\cos xdx$ …….

$(2)$

Add

$(1)$ &

$(2)$

$(\displaystyle\int^a_0f(x)dx=\displaystyle\int^a_0f(a-x)dx)$

$2I=\displaystyle\int^{\pi/2}_0(log\sin x+log\cos x)dx$

$2I=\displaystyle\int^{\pi/2}_0log\sin x\cos xdx(log a+log b=log ab)$

$2I=\displaystyle\int^{\pi/2}log\dfrac{2\sin x\cos x}{2}dx$

$2I=\displaystyle\int^{\pi/2}_0log\dfrac{\sin 2x}{2}dx$

$2I=\displaystyle\int^{\pi/2}_0log\sin 2xdx-\displaystyle\int^{\pi/2}_0log 2dx$

$(\because log\dfrac{a}{b}=log a-log b)$

$2I=\displaystyle\int^{\pi/2}_0log\sin 2xdx-log 2x\displaystyle\int^{\pi/2}_0$

Let

$2x=t$

$2dx=dt$

$dx=\dfrac{dt}{2}$

$2I=\displaystyle\int^{\pi}_0\dfrac{1}{2}log \sin tdt-\dfrac{\pi}{2}log 2$

$\left(\displaystyle\int^{2a}_0f(x)dx=2\displaystyle\int^a_0f(x)dx\right.$ if

$\left.f(x)=f(2a-x)\right)$

Using above property we get

$2I=\dfrac{2}{2}\displaystyle\int^{\pi/2}_0log\sin tdt+\dfrac{\pi}{2}log 2$

$2I=I-\dfrac{\pi}{2}log 2\left(\displaystyle\int^a_0f(x)dx=\displaystyle\int^a_0f(t)dt\right)$

$I=-\dfrac{\pi}{2}log 2$

$\therefore \displaystyle\int^{\pi/2}_0log \sin xdx=-\pi/2log 2$ .

Evaluate $\int _ { 0 } ^ { 1 } \frac { 2 x + 3 } { 5 x ^ { 2 } + 1 } d x$

$\\I=\int_{0}^{1}(\frac{2x}{5x^2+1})dx+\int_{0}^{1}(\frac{3}{5x^2+1})dx\\=(\frac{1}{5})\int_{0}^{1}(\frac{10x}{5x^2+1})dx+(\frac{3}{5})\int_{0}^{1}(\frac{1}{x^2+(\frac{1}{\sqrt{5}})^2})dx\\=(\frac{1}{5})\left[log(5x^2+1)\right]_{0}^{1}+(\frac{3}{5})(\frac{1}{(\frac{1}{\sqrt{5}})})\left[tan^{-1}(\frac{x}{(\frac{1}{\sqrt{5}})})\right]_{0}^{1}\\=(\frac{1}{5})log6+(\frac{3}{\sqrt{5}})tan^{-1}\sqrt{5}$

Evaluate : $\displaystyle \int _{0}^{\pi/2}2\sin{x}\cos{x}\tan^{-1}{(\sin{x})}dx$

1994100_1224598_ans_3f444cf4c4cb45eb9500bb5ea5b8cb74.png

Solve
$\int\dfrac{\cos 5x+\cos 4x}{1-2 \cos 3x} dx$

$\int \dfrac{\cos 5x+ \cos 4x}{1-2 \cos 3x} dx$

$\cos 5x+ \cos 4x=2 \cos \dfrac{9x}{2} \cos \dfrac{x}{2}$

$1-2 \cos 3x= 1-2 \left( 2\cos^{2} \dfrac{3x}{2}-1 \right)$

$=3-4 \cos^{2} \dfrac{3x}{2}$

$\dfrac{2\cos \dfrac{9x}{2}. \cos \dfrac{x}{2} }{3-4 \cos^{2} \dfrac{3x}{2}} \times \dfrac{\cos 3 \dfrac{x}{2} }{\cos 3 \dfrac{x}{2} }$

$=\dfrac{2 \cos \dfrac{x}{2}. \cos \dfrac{3x}{2}. \cos \dfrac{9x}{2}}{-\left( 4 \cos \dfrac{3x}{2}-3 \cos \dfrac{3x}{2} \right)}$

By using

$\cos 3 \theta=4 \cos^{3} \theta-3 \cos \theta$

$=\dfrac{2 \cos \dfrac{x}{2} \cos \dfrac{3x}{2}. \cos \dfrac{9x}{2} }{-\cos \dfrac{9x}{2}}$

$=-2 \cos\dfrac{x}{2} \cos \dfrac{3x}{2}$

$=- (\cos 2x+ \cos x)$

Now

$I=\int - \cos 2x- \cos x. dx.$

$=\dfrac{-\sin 2x}{2}-\sin x+c$ .

If $\int \frac { 1 } { 5 + 4 \cos 2 \theta } d \theta = A \tan ^ { - 1 } ( B \tan \theta ) + c$ then $( A , B ) =$

$\begin{array}{l} \int { \frac { { d\theta } }{ { 5+4\cos 2\theta } } =\int { \frac { { d\theta } }{ { 5+\frac { { 4\left( { 1-\tan \theta } \right) } }{ { 1+{ { \tan }^{ 2 } }\theta } } } } } } \\ =\int { \frac { { { { \sec }^{ 2 } }\theta d\theta } }{ { 9+{ { \tan }^{ 2 } }\theta } } } \\ Let\, \tan \theta =u \\ \frac { { du } }{ { d\theta } } ={ \sec ^{ 2 } }\theta \\ Now, \\ =\int { \frac { { du } }{ { 9+{ u^{ 2 } } } } =\frac { 1 }{ 3 } { { \tan }^{ -1 } }\left( { \frac { { \tan \theta } }{ 3 } } \right) +C } \\ \left( { A\, ,B } \right) =\left( { \frac { 1 }{ 3 } ,\, \frac { 1 }{ 3 } } \right) \end{array}$

Evaluate:-
$\int\limits_0^{\pi /2} {\operatorname{\sin x} .\sin 2xdx}$

1215630_1235623_ans_7bcee091cff0446ea94bb05d5f000094.jpg

$\displaystyle \int _{0}^{1}\cot^{-1}{(1-x+x^{2})}dx$

1970896_1224458_ans_e69439d67143401aac52068ec3424f02.PNG

Integrate: $\dfrac { 3 x ^ { 2 } } { x ^ { 6 } + 1 }$

$I=\displaystyle \int \dfrac{3x^2}{x^6 + 1} dx$

$x^3 = t$

$3x^2 dx = dt$

$I=\displaystyle \int \dfrac{3x^2}{t^2 + 1} \times \dfrac{dt}{3x^2} = \int \dfrac{dt}{t^2 + 1}$

$= \tan^{-1} (t) + c$

$= \tan^{-1} (x^3) + C$

Solve:
$\displaystyle \int \dfrac { \sin ^ { 3 } x + \cos ^ { 3 } x } { \sin ^ { 2 } x \cos ^ { 2 } x } d x$

$\int \dfrac{(sin^{3}+cos^{3}x)}{sin^{2}x\cos^{2}x}dx$

$= \int (\dfrac{sin x}{\cos^{2}x}+\dfrac{\cos x}{\sin^{2}x})dx = \int (\sec x\tan x+cosec\,xcotx)dx$

$= sec x-cosec x+c$

$= \dfrac{1}{\cos x}-\dfrac{1}{\sin x}+c$

$= \dfrac{\sin x-\cos x}{\sin x\,\cos x}+c$

Evaluate: $\displaystyle\frac { 1 } { 15 } \int \frac { d t } { 4 + t ^ { 2 } } + \frac { 4 } { 15 } \int \frac { 6 d t } { 1 + 4 t ^ { 2 } }$

$\dfrac{1}{15}\displaystyle\int \dfrac{dt}{4+t^2}+\dfrac{4}{15}\displaystyle\int \dfrac{6dt}{1+4t^2}$

$=\dfrac{1}{15}\cdot \dfrac{1}{2}\tan^{-1}\left(\dfrac{t}{2}\right)+\dfrac{24}{15\times 4}\displaystyle\int \dfrac{dt}{\left(\dfrac{1}{2}\right)^2+t^2}$

$=\dfrac{1}{30}\tan^{-1}\left(\dfrac{t}{2}\right)+\dfrac{6}{15}\cdot\dfrac{1}{1/2}\tan^{-1}\left(\dfrac{t}{\dfrac{1}{2}}\right)+c$

$=\dfrac{1}{30}\tan^{-1}\left(\dfrac{t}{2}\right)+\dfrac{4}{5}\tan^{-1}(2t)+c$ .

1183002_1246156_ans_c59baba0189249bbb5baf5f52b8b457e.jpg

Prove that
$\displaystyle \int _ { 0 } ^ { 1 } \cfrac { 2 n + 3 } { 5 n ^ { 2 } + 1 } d x$

$\int_{0}^{1} \frac{2n+3}{5n^{2}+1} dn$

$= \frac{1}{5} \int_{0}^{1} \frac{10n}{5n^{2}+1} dn+\frac{1}{5}\int \frac{3}{n^{2}+\frac{1}{5}}dn$

$5n^{2}+1=t$

10n dn = dt

$= \frac{1}{5} \int_{0}^{1}\frac{1}{t} dt + \frac{3}{5} \int \frac{1}{n^{2} + \left ( \frac{1}{\sqrt{5}} \right )^{2}}dn$

$= \frac{1}{5}\left [ log(5n^{2} + 1) \right ]_{0}^{1} + \frac{3}{5} (\sqrt{5})\left [ tan^{-1}n\sqrt{5} \right ]_{0}^{1}$

$= \frac{log6}{5} + \frac{3}{\sqrt{5}} tan^{-1}\sqrt{5}$

1141750_1246934_ans_214ace6fe7fe48768dbf5402745b2290.jpg

Evaluate : $\int \frac{dx}{a+be^{cx}}$

$I=\displaystyle\int \dfrac{dx}{a+be^{cx}}$

$I=\displaystyle\int\dfrac{dx}{e^{cx}[ae^{-cx}+b]}$

$I=\displaystyle\int \dfrac{e^{-cx}dx}{ae^{-cx}+b}$

$[$ Let

$ae^{-cx}+b=u$

$a(-c)e^{-cx}dx=du, e^{-cx}dx=\dfrac{du}{-ac}]$

$=\displaystyle\int \dfrac{du/-ac}{u}$

$I=\dfrac{-1}{ac}\displaystyle\int \dfrac{du}{u}$

$I=\dfrac{-1}{ac}(ln u+\alpha)$

$I=\dfrac{-1}{ac}(ln (ae^{-cx}+b)+\alpha)$

Constant of integration.

1179247_1244894_ans_5eb955f86a774acbb345acb7cc151d21.jpg

Evaluate :
$\int \dfrac{(x^{4}-x)^{1/4}}{x^{5}}dx$

$\int \dfrac{(x^{4}-x)^{1/4}}{x^{5}}dx$

$= \int \frac{(x^{4})^{1/4}(1-\frac{x}{x^{4}})^{1/4}}{x^{5}}.dx$

$= \int \frac{x(1-\frac{1}{x^{3}})^{1/4}}{x^{5}}.dx$

$= \int \frac{(1-\frac{1}{x^{3}})^{1/4}}{x^{4}}. dx$

Let

$1-\frac{1}{x^{3}} = t$

$-(\frac{-3}{x^{4}}) = \frac{dt}{dx}$

$\frac{3}{x^{4}} = \frac{dt}{dx}$

$\frac{dt}{3}= \frac{dx}{x^{4}}$

$\rightarrow = \int t^{2/4} (\frac{dt}{3})$

$= \frac{1}{3}\int t^{\frac{1}{4}}.dt$

$= \frac{1}{3}(\frac{t^{5/4}}{5/4})+C$

$= \frac{1\times 4}{3\times 5}t^{5/4}+C$

$= \frac{4}{15}t^{5/4}+C$

$= \frac{4}{15}\times (1-\frac{1}{x^{3}})^{5/4}+C$

1167729_1248163_ans_669a64dd50324f2ea75b05587fb33023.jpg

Evaluate: $\displaystyle\int_{0}^{\frac{\pi}{4}}{\ln{\left(1+\tan{x}\right)}dx}$

1353890_1245863_ans_d78a55f2200343678ce98fc48cf4f128.PNG

Evaluate: $\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{x\,dx}{1+\sin{x}}}$

1353892_1245979_ans_1d08f294213247a1aa32b7277f5e8e21.PNG

Evaluate: $\int _ { 0 } ^ { \pi } \dfrac { x \sin x } { 1 + \cos ^ { 2 } x } d x$

Let

$I=\displaystyle\int^{\pi}_0\dfrac{x\sin x}{1+\cos^2x}dx$ ………

$(1)$

$I=\displaystyle\int^{\pi}_0\dfrac{(\pi -x)\sin(\pi-x)}{1+\cos^2(\pi-x)}dx$

$=\displaystyle\int^{\pi}_0\dfrac{(\pi -x)\sin x}{1+\cos^2x}dx$ ……..

$(2)$

Adding both

$(1)$ and

$(2)$ , we get

$2I=\displaystyle\int^{\pi}_0\dfrac{\pi\sin x}{1+\cos^2x}dx$

Let

$t=\cos x$

$\cos 0=1$

$dt=-\sin xdx$

$\cos\pi =-1$

So,

$2I=\displaystyle\int^{-1}_1\dfrac{-\pi dt}{1+t^2}$

$2I=\pi\displaystyle\int^1_{-1}\dfrac{dt}{1+t^2}=\pi\left[\tan^{-1}t\right]^1_{-1}$

$2I=\pi\left[\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4}\right)\right]$

$2I=\dfrac{\pi^2}{2}$

$I=\dfrac{\pi^2}{4}$

$I=\left(\dfrac{\pi}{2}\right)^2$ .

1203823_1246362_ans_0406e36d7fec4fb99b436f48eb90249a.jpg

Evaluate : $\displaystyle \int \dfrac{x}{x-\sqrt{x^2-1}}dx$

1344724_1254663_ans_72379671e09f4ff6ba3496a50531eaf3.jpg

Evaluate: $\displaystyle \int e^{2x} \sin 3x \ dx$

1347405_1258480_ans_466cdfcf1dac42c693d2d43f496fc9c1.jpg

Evaluate:
$\displaystyle \int_{0}^{\dfrac {\pi}{4}}\dfrac {\sin x+\cos x}{9+16\sin 2x}dx$

1170006_1248471_ans_d2acf109a8a945ffaef8d50e24d52891.jpg

Evaluate: $\displaystyle\int \dfrac { 1 } { 7 + 5 \cos x } d x =$

$I=\displaystyle \int \dfrac{1}{7+5\cos x}dx$

$\cos x=\dfrac{1-\tan^2\left(\dfrac{x}{2}\right)}{1+\tan^2\left(\dfrac{x}{2}\right)}$

Let

$t=\tan \left(\dfrac{x}{2}\right)$

$\therefore \cos x=\dfrac{1-t^2}{1+t^2}$

$\dfrac{dt}{dx}=\dfrac{1}{2}\sec^2\left(\dfrac{x}{2}\right)$

$\left[\because \sec^2\dfrac{x}{2}-\tan ^2\dfrac{x}{2}=1\right]$

$\dfrac{dt}{dx}=\dfrac{1}{2}[1+t^2]$

$I=\displaystyle \int\dfrac{1}{7+5\left(\dfrac{1-t^2}{1+t^2}\right)}\times \dfrac{dt}{\dfrac{1}{2}[1+t^2]}$

$I=2\displaystyle \int \dfrac{(1+t^2)dt}{[7+7t^2+5-5t^2][(1+t^2)]}$

$I=\displaystyle \dfrac{2dt}{2t^2+12}=\int \dfrac{dt}{t^2+6}$

$\left[\because\displaystyle \int \dfrac{dx}{a^2+x^2}=\dfrac{1}{a}tan^{-1}\left(\dfrac{x}{a}\right)+c\right]$

$I=\displaystyle \dfrac{1}{(\sqrt 6)^2+(t)^2}=\dfrac{1}{\sqrt 6}tan^{-1}\left(\dfrac{t}{\sqrt 6}\right)+c$

$I=\dfrac{1}{\sqrt 6}tan^{-1}\left(\dfrac{tan \left(\dfrac{x}{2}\right)}{\sqrt 6}\right)+c$

Evaluate:
$\displaystyle\int_{-\pi}^{\pi}{\dfrac{{\cos}^{2}{x}}{1+{a}^{x}}dx}$ where $a>0$

1353895_1248638_ans_37d6b98273c94dba9377b540eef859a7.PNG

$\int \sqrt { \sec x - 1 } d x$ is equal to

$I=\displaystyle \int (\sqrt{\sec x - 1}) dx$

$I=\displaystyle \int \sqrt{\dfrac{1}{\cos x} -1} = \int \sqrt{\dfrac{1 -\cos x}{\cos x}} dx$

$1 - cos 2x = 2 \sin^2 (x)$

$\cos 2x = 2 \cos^2 x - 1$

$I=\displaystyle \int \sqrt{\dfrac{2 \sin^2 (x/2)}{2 \cos^2 (x/2) - 1}} dx$

$u = \cos x/2 \,\,\, du = -\sin (x/2) \times \dfrac{1}{2} dx$

$I=\displaystyle \int \sqrt{\dfrac{2 \times (\sin^2 (x /2))}{2u^2 - 1}} \times \dfrac{du}{\dfrac{-1}{2} \times \sin (x/2)}$

$I=-2\sqrt{2} \displaystyle \int \dfrac{1}{\sqrt{2u^2 - 1}} du$

$I=\dfrac{-2 \sqrt{2}}{\sqrt{2}} \displaystyle \int \dfrac{1}{\sqrt{u^2 - \dfrac{1}{2}}} du$

$I=\displaystyle \int \dfrac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + c$

$I=-2 \log |u + \sqrt{u^2 - 1/2} | + c$

$I=-2 \log |\cos x/2 + \sqrt{\cos^2 \dfrac{x}{2} - \dfrac{1}{2}} | + c$ .

Integrate:
$\frac { 1 } { \sqrt { 16 - 9 x ^ { 2 } } }$

$\\I=\int (\frac{1}{\sqrt{16-9x^2}})dx\\=(\frac{1}{3})\int (\frac{1}{\sqrt{(\frac{16}{9})-x^2}})dx\\=(\frac{1}{3})\int(\frac{1}{\sqrt{(\frac{4}{3})^2-x^2}})dx\\=(\frac{1}{3})sin^{-1}\left((\frac{x}{(\frac{4}{3})})\right)+C\\=(\frac{1}{3})sin^{-1}(\frac{3x}{4})+C$

Solve:
$\displaystyle \int_{0}^{T/2}{I_{0}\sin(wt)dt}$

$\int_{0}^{T/2}I_{0} sin(wt).dt$

$=I_{0}\int_{0}^{T/2} sin (wt).dt$

$=I_{0}[\frac{-cos\, wt}{w}]_{0}^{T/2}$

$= \frac{I_{0}}{w}[- cos\, w.\frac{T}{2}-(-cos 0)]$

$= \frac{I_{0}}{w}[-cos(\frac{2\pi }{T}.\frac{T}{2})+cos 0]$

$= \frac{I_{0}}{w}[- cos\pi + cos0]$

$= \frac{I_{0}}{w}[I+1]$

$= \frac{2I_{0}}{w}$

$= \frac{2I_{0}}{(\frac{2\pi }{T})}$

$= \frac{T.I_{0}}{\pi }$

1174750_1249584_ans_a18ab767f06b44c2a48d0db6a0d3c670.jpg

Evaluate: $\displaystyle\int \dfrac { \cos ^ { 3 } x } { \sin ^ { 2 } x + \sin x }$

$I=\displaystyle\int\dfrac{\cos^3x}{\sin^2x+\sin x}dx$

$I=\displaystyle\int\dfrac{(\cos^2x).\cos x}{\sin^2x+\sin x}dx$

$I=\displaystyle\int\dfrac{(1-\sin^2x).\cos x}{\sin^2x+\sin x}dx$

Let

$u=\sin x$

$du=\cos xdx$

$\displaystyle=\int \dfrac{(1-u^2)\cos x}{u^2+u}.\dfrac{du}{\cos x}$

$\displaystyle=\int \dfrac{(1+u)(1-u)}{u(u+1)}du$

$\displaystyle=\int \dfrac{1}{u}du-\int \dfrac{u}{u}du$

$=ln|u|-u+c$

$=ln|\sin x|-\sin x+c$ .

Evaluate: $\displaystyle\int{\dfrac{\left(x+1\right)dx}{x{\left(1+x{e}^{x}\right)}^{2}}}$

1354282_1263302_ans_bdcd41b0ed874f2e829005519824d274.PNG

Evaluate:
$\displaystyle\int { x\log { \left( x+1 \right) } dx }$

Let

$I=\int x \log \left( 1+x\right) dx$

$\Rightarrow I=\log \left( 1+x\right) \int x dx -\int \dfrac{1}{x+1}\dfrac{x^2}{2} dx$

$=\dfrac{x^2}{2}\log \left( x+1\right) -\dfrac{1}{2} \int\dfrac {x^2}{x+1} dx$

$=\dfrac{x^2}{2}\log \left( x+1\right) -\dfrac{1}{2}\int \dfrac{x^2}{x+1}dx$

$=\dfrac{x^2}{2}\log \left( x+1\right) -\dfrac{1}{2}\int \dfrac{x^2-1+1}{x+1}dx$

$=\dfrac{x^2}{2}\log \left( x+1\right) -\dfrac{1}{2}\int \left( \dfrac{x^2-1}{x+1}+\dfrac{1}{x+1}\right)dx$

$=\dfrac{x^2\log \left( x+1\right)}{2}-\dfrac{1}{2}\int \left( \dfrac{ \left( x+1\right) \left( x-1\right) }{x+1}+\dfrac{1}{x+1}\right) dx$

$=\dfrac{x^2\log \left( x+1\right)}{2}-\dfrac{1}{2}\int \left( x-1+\dfrac{1}{x+1}\right) dx$

$=\dfrac{x^2\log \left( x+1\right)}{2}-\dfrac{1}{2}\int x dx -\int 1 dx+\int \dfrac{1}{x+1} dx$

$=\dfrac{x^2\log \left( x+1\right) }{2}-\dfrac{1}{2}\left[ \dfrac{x^2}{2}-x+\log \left( x+1\right) \right] +c$

$=\dfrac{x^2\log \left( x+1\right)}{2}-\dfrac{x^2}{4}+\dfrac{x}{2}-\dfrac{-\log \left( x+1\right) }{2}+c$

Hence, the answer is

$\dfrac{x^2\log \left( x+1\right)}{2}-\dfrac{x^2}{4}+\dfrac{x}{2}-\dfrac{-\log \left( x+1\right) }{2}+c.$

Evaluate: $\displaystyle \int {\dfrac {dx } { \sqrt { 1 - {e} ^ { 2 x } } } }$

$\displaystyle\int \dfrac{dx}{\sqrt{1-e^{2x}}}$

$=\displaystyle\int \dfrac{t\cdot dt}{t(t^2-1)}$

$=\displaystyle\int \dfrac{dt}{t^2-1}$

$=\dfrac{1}{2}log\left|\dfrac{t-1}{t-1}\right|+c$

$=\dfrac{1}{2}log\left|\dfrac{\sqrt{1-e^{2x}}-1}{\sqrt{1-e^{2x}}+1}\right|+c$

Let

$t=\sqrt{1-e^{2x}}$

$t^2=1-e^{2x}$

$1-t^2=e^{2x}$

$-2tdt=2e^{2x}dx$

$\dfrac{-t}{e^{2x}}=dx$

$\dfrac{t}{t^2-1}dt=dx$ .

1213990_1259095_ans_3938eb198b13458c9136cef1b3935caf.jpg

Integrate: $\sin ^ { 3 } x \cos ^ { 3 } x$

$I=\int \sin^{3}x \cos^{3}x dx$

$I=\int \sin^{3}x \cdot \cos^{2}x \cdot \cos x\ dx$

$I=\int \sin^{3}x (1 -\sin^{2}x) \cdot \cos x\ dx$

Let

$u = \sin x$

$du = \cos x\ dx$

$I=\int u^{3} (1 - u^{2})du$

$I=\int (u^{3} - u^{5})dx$

$I=\dfrac {u^{4}}{4} - \dfrac {u^{6}}{6} + c$

$I=\dfrac {1}{4} \sin^{4}x - \dfrac {1}{6} \sin^{6}x + C$ .

Evaluate: $\displaystyle\int{\left({e}^{\ln{x}}+\sin{x}\right)\cos{x}dx}$

1354268_1263048_ans_36c5d2fe7b3f40a4a0389e4e38b0c81f.PNG

Evaluate the following:
$\displaystyle\int \dfrac { 2 x + 7 } { ( x - 4 ) ^ { 2 } } d x$

$I=\displaystyle\int\dfrac{2x+7}{(x-4)^2}dx$

$I=\displaystyle\int\dfrac{2x-8+15}{(x-4)^2}dx=\displaystyle\int\dfrac{2x-8}{(x-4)^2}dx+\displaystyle\int\dfrac{15}{(x-4)^2}dx$

$I=\displaystyle\int\dfrac{2(x-4)}{(x-4)^2}dx+15\displaystyle\int\dfrac{1}{(x-4)^2}dx$

Let

$x-4=t\Rightarrow dx=dt$

$I=\displaystyle2\int \dfrac{dx}{t}+15\int \dfrac{1}{t^2}dt$

$I=2\log(t)+15\dfrac{t^{-1}}{-1}$

$I=2log(t)-\dfrac{15}{t}$

$I=2log (x-4)-\dfrac{15}{(x-4)}+c$ .

Evaluate: $\displaystyle\int{\sqrt{1+\csc{x}}dx}$

1354264_1259025_ans_a14c65aec28047bbafaebc6ad2c0e49e.PNG

Evaluate :
$\displaystyle \int \dfrac{x^2}{\sqrt{1-x^2}}dx$ .

1345923_1261276_ans_45f97fb6e79847f1b5f51ffec5a77e2f.jpg

Evaluate the following definite integral:
$\displaystyle \int _{\pi /6}^{\pi /4} \csc x \ dx$

$I=\displaystyle \int _{\pi /6}^{\pi /4} \csc dx=\left [\log (\csc x-\cot x)]_{\pi /6}^{\pi /4} =\log \left (\csc \dfrac {\pi}{4} -\cot \dfrac {\pi}{4}\right) -\log (\csc \dfrac {\pi}{6}-\cot \dfrac {\pi}{6} \right)$

Integrate $\dfrac{x^{2}+5x+5}{x^{2}+3x+2}$ with respect to $x$ .

$I=\int \frac{x^{2}+3x+2}{x^{2}+3x+2}+\frac{2x+3}{x^{2}+3x+2}dx$

$= \int 1+\frac{f'(x)}{f(x)}dx(f(x)=x^{2}+3x+2)$

$= x+log (x^{2}+3x+2)+c$

1188860_1279499_ans_7a53f73374fe4ade9e97a524fb86b4bf.jpg

solve the following.
$\int\limits_0^1 {\frac{{\log (1 + t)}}{{(1 + {t})^2}}} dt$

1212175_1277494_ans_907c31647bb64d09b6a866ec81ecfebe.JPG

Evaluate $\int {\dfrac{{{{\cos }^2}x}}{{{{\sin }^3}x{{\left( {{{\sin }^5}x + {{\cos }^5}x} \right)}^{\frac{3}{5}}}}}dx}$

$I=\int { \dfrac { { { { \cos }^{ 4 } }x } }{ { { { \sin }^{ 3 } }x{ { \left( { { { \sin }^{ 5 } }x+{ { \cos }^{ 5 } }x } \right) }^{ \frac { 3 }{ 5 } } } } } dx }$

$I =\int { \dfrac { { { { \sec }^{ 2 } }xdx } }{ { { { \tan }^{ 6 } }x{ { \left( { 1+\frac { 1 }{ { { { \tan }^{ 5 } }x } } } \right) }^{ \frac { 3 }{ 5 } } } } } }$

Let

$\tan x=p\, \, \, \, { \sec ^{ 2 } }xdx=dp$

Therefore,

$\int { \dfrac { { dp } }{ { { p^{ 6 } }{ { \left( { 1+\frac { 1 }{ { { p^{ 5 } } } } } \right) }^{ \frac { 3 }{ 5 } } } } } } \\ 1+\dfrac { 1 }{ { { p^{ 5 } } } } =k \\ \dfrac { { dk } }{ { dp } } =\dfrac { { -5 } }{ { { p^{ 6 } } } } \\$

Therefore,

$I=-\dfrac { { 1 } }{ 5 } \int { { { \left( k \right) }^{ \frac { { -3 } }{ 5 } } }dk } \\ =\dfrac { { -1 } }{ 5 } \left( { \dfrac { 5 }{ 2 } } \right) { k^{ \frac { 2 }{ 5 } } }+C \\ =\dfrac { { -1 } }{ 2 } { \left[ { \dfrac { { { p^{ 5 } }+1 } }{ { { p^{ 5 } } } } } \right] ^{ \frac { 2 }{ 5 } } } \\ =-\dfrac { 1 }{ 2 } { \left[ { \dfrac { { { { \tan }^{ 5 } }x+1 } }{ { { { \tan }^{ 5 } }x } } } \right] ^{ \frac { 2 }{ 5 } } }+C \\ =-\dfrac { 1 }{ 2 } { \left( { 1+{ { \cot }^{ 5 } }x } \right) ^{ \frac { 2 }{ 5 } } }+C$

Hence, this is the answer.

$\int_0^{\pi /2} {\dfrac{1}{{1 + {{\tan }^3}x}}} dx$

We have,

$I=\int _{ 0 }^{ \pi /2 }{ \dfrac { 1 }{ { 1+{ { \tan }^{ 3 } }x } } } dx........\left( 1 \right) \\$

Apply property, we get

$I=\int _{ 0 }^{ \pi /2 }{ \dfrac { 1 }{ { 1+{ { \cot }^{ 3 } }x } } } dx \\ I=\int _{ 0 }^{ \pi /2 }{ \dfrac { { { { \tan }^{ 3 } }x } }{ { { { \tan }^{ 3 } }x+1 } } } dx.........\left( 2 \right)$

On adding equation (1) and (2), we get

$\\ 2I=\int _{ 0 }^{ \pi /2 }{ dx } \\ I=\dfrac { \pi }{ 2 } \\ I=\dfrac { \pi }{ 4 }$

Hence, this is the answer.

Evaluate :
$\displaystyle\int { \dfrac { \sin { \left( x-a \right) } }{ \sin { \left( x+a \right) } } dx }$

$\int \frac{sin (x-a)}{sin (x+a)}dx$

Let

$x+a = \theta$

$dx = d\theta$

$x-a = \theta -2a$

$\int \frac{sin (\theta -2a)}{sin \theta }d\theta$

$\int \frac{sin \theta cos2a-sin2a cos\theta }{sin \theta }d\theta$

$\int (cos2a-sin2a . tan\theta )d\theta$

$= \int cos2a d\theta -\int sin2a . tan\theta .d\theta$

$= \theta cos2a-sin2a\int tan\theta\, d\theta$

$= \theta cos2a-sin2a (-log \,cos\theta )+c$

$= \theta cos2a+sin2a . log\, cos\theta$

$= (x+a)cos2a+sin2a\, log\, cos(x+a)$

1203878_1282996_ans_fe55ce0f7b8b476d8d59152200087216.jpg

Evaluate: $\displaystyle \int{\dfrac{dx}{(2x-7)\sqrt{(x-3)(x-4)}}}$ .

$\displaystyle\int \dfrac{dx}{(2x-7)\sqrt{(x-3)(x-4)}}$

$\displaystyle\int \dfrac{dx}{(2x-7)\sqrt{x^2-7x+12}}$

$\displaystyle\int \dfrac{dx}{2(2x-7)\sqrt{x^2-7x+12}}$

$\displaystyle\int \dfrac{2dx}{(2x-7)\sqrt{4x^2-28x+48}}$

$\displaystyle\int \dfrac{2dx}{(2x-7)\sqrt{4x^2-28x+49-1}}$

$\displaystyle\int \dfrac{2dx}{(2x-7)\sqrt{(2x-7)^2-1}}$

Let

$2x-7=t$

$2dx=dt$

$\displaystyle\int \dfrac{dt}{t\sqrt{t^2-1}}$

$=\sec^{-1}(t)+c$

$=\sec^{-1}(2x-7)+c$ .

1201572_1281214_ans_5fbab3db20df45af9b2c8a271cc1dad1.jpg

Prove that
$\displaystyle \int \dfrac {x^{2}}{x^{6}+1}dx$

We have,

$\begin{array}{l} \int { \dfrac { { { x^{ 2 } } } }{ { { x^{ 6 } }+1 } } } dx \\ put\, \, \, { x^{ 3 } }=t\, \, \, 3{ x^{ 2 } }dx=dt \\ =\int { \dfrac { { \dfrac { { dt } }{ 3 } } }{ { { t^{ 2 } }+1 } } } \\ =\dfrac { 1 }{ 3 } { \tan ^{ -1 } }t+c \\ =\dfrac { 1 }{ 3 } { \tan ^{ -1 } }{ x^{ 3 } }+c \end{array}$

Hence, this is the answer.

Solve:
$\displaystyle \int{\dfrac{dx}{\cos(x-a)\cos (x-b)}}$

$\int \frac{dx}{cos (x-a)cos (x-b)}$

$\int \frac{sin (a-b)}{sin (a-b)}\times \frac{1}{cos (x-a)cos (x-b)}dx$

$=\frac{1}{sin (a-b)}\int \frac{sin (a-b)}{cos (x-a)cos (x-b)}dx$

$=\frac{1}{sin (a-b)}\int \frac{sin a-b+x-x}{cos (x-a)cos (x-b)}dx$

$=\frac{1}{sin (a-b)}\int \frac{sin (x-b)+(a-x)}{cos (x-a)cos (x-b)}dx$

$=\frac{1}{sin (a-b)}\int \frac{sin (x-b)-(x-a)}{cos (x-a)cos (x-b)}dx$

$=\frac{1}{sin (a-b)}\int \frac{sin (x-b)cos (x-a)-cos (x-b)sin(x-a)}{cos (x-a)cos (x-b)}dx$

$=\frac{1}{sin (a-b)}\int \frac{sin(x-b)cos (x-a)}{cos (x-a)cos (x-b)}-\frac{cos (x-b)sin (x-a)}{cos (x-a) cos (x-b)}dx$

$=\frac{1}{sin (a-b)}[\int tan (x-b).dx-\int tan (x-a).dx]$

$=\frac{1}{sin (a-b)}[-log \left | cos (x-b) \right |+log\left | cos (x-a) \right |]+c$

$=\frac{1}{sin (a-b)}[log\left | cos (x-a) \right |-log\left | cos (x-b) \right |]+c$

$=\frac{1}{sin (a-b)}log[\frac{cos (x-a)}{cos (x-b)}]+c$

1205582_1283406_ans_44db2d55bb1b442d8cb3f00049dd6627.jpg

Solve:
$\displaystyle\int { \sqrt { ax+b } } dx$

$\int \sqrt{ax+b}.dx$

Let

$ax+b = t$

diff w.r.t 'x'

$a(1)+0=\frac{dt}{dx}$

$dx=\frac{dt}{a}$

now

$\int \sqrt{t}.\frac{dt}{a}$

$\frac{1}{a}\int \sqrt{t}.dt$

$\frac{1}{a}\int t^{1/2}.dt$

$\frac{1}{a}[\frac{t^{3/2}}{3/2}+C]$

$\frac{1}{a}.[\frac{2}{3}t^{3/2}+C]$

$\frac{2}{3a}t^{3/2}+C$

$\frac{2}{3a}(ax+b)^{3/2}+C$

1211719_1285291_ans_134d1f1e34e3413ab38fe63f1fee31e1.jpg

Evaluate the integral $\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$ .

$I=\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$ . Let

$x+2=t^{2}$ . Then ,

$dx=2t dt$

Also

$x=0\Rightarrow t^{2}=2\Rightarrow t=\sqrt{2}$ and,

$x=2\Rightarrow t^{2}=4\Rightarrow t=2$

$\therefore I=\displaystyle\int_{\sqrt{2}}^{2}(t^{2}-2)\sqrt{t^{2}}2t dt=2\displaystyle\int_{\sqrt{2}}^{2}(t^{4}-2t^{2})dt=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$

$\Rightarrow I=2\left[\left(\dfrac{32}{5}-\dfrac{16}{3}\right)-\left(\dfrac{4\sqrt{2}}{5}-\dfrac{4\sqrt{2}}{3}\right)\right]=2\left(\dfrac{16}{15}+\dfrac{8\sqrt{2}}{15}\right)=\dfrac{32+16\sqrt{2}}{15}$

$\int _{ 0 }^{{ \pi /2 } }{ { x }^{ 2 } } { cos }^{ 2 }$ x dx

We have,

$\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}{{\cos }^{2}}}xdx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \cos 2\theta =2{{\cos }^{2}}\theta -1$

$\Rightarrow \int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}}\dfrac{\left( 1+\cos 2x \right)}{2}dx$

$\Rightarrow \int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{x}^{2}}+{{x}^{2}}\cos 2x}{2}dx}$

$\Rightarrow \dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}+\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{{{x}^{2}}\cos 2x}dx}$

On integration and we get,

$\dfrac{1}{2}{{\left[ \dfrac{{{x}^{3}}}{3} \right]}_{0}}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ {{x}^{2}}\int_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}-\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{d}{dx}{{x}^{2}}\int_{0}^{\dfrac{\pi }{2}}{\cos 2x}dx \right)dx} \right]$

$\Rightarrow \dfrac{1}{2}\left[ \dfrac{\dfrac{{{\pi }^{3}}}{8}-{{0}^{3}}}{3} \right]+\dfrac{1}{2}\left[ {{\left\{ {{x}^{2}}\left( \dfrac{\sin 2x}{2} \right) \right\}}_{0}}^{\dfrac{\pi }{2}}-\int_{0}^{\dfrac{\pi }{2}}{\left( 2x\left( \dfrac{\sin 2x}{2} \right) \right)dx} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}+\dfrac{1}{2}\left[ \left( \dfrac{\pi }{2}-0 \right)\dfrac{\left( \sin 2\times \dfrac{\pi }{2}-\sin 0 \right)}{2} \right]-\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{\left( x\sin 2x \right)dx}$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}+\dfrac{1}{2}\left[ \left( \dfrac{\pi }{2} \right)\dfrac{\left( 0-0 \right)}{2} \right]-\dfrac{1}{2}\left[ x\int_{0}^{\dfrac{\pi }{2}}{\sin 2xdx}-\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{d}{dx}x\int_{0}^{\dfrac{\pi }{2}}{\sin 2xdx} \right)dx} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}+0-\dfrac{1}{2}{{\left[ x\left( \dfrac{-\cos 2x}{2} \right) \right]}_{0}}^{\dfrac{\pi }{2}}+\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( -\cos 2x \right)}{2}dx}$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{1}{2}\left[ \left\{ \left( \dfrac{\pi }{2}-0 \right)\left( \dfrac{-\cos 2\times \dfrac{\pi }{2}+\cos 0}{2} \right) \right\} \right]-\dfrac{1}{2}{{\left[ \dfrac{\sin 2x}{2} \right]}_{0}}^{\dfrac{\pi }{2}}$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{1}{2}\left[ \left\{ \left( \dfrac{\pi }{2} \right)\left( \dfrac{1+1}{2} \right) \right\} \right]-\dfrac{1}{2}\left[ \dfrac{\sin 2\times \dfrac{\pi }{2}-\sin 0}{2} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{\pi }{4}-\dfrac{1}{2}\left[ \dfrac{0-0}{2} \right]$

$\Rightarrow \dfrac{{{\pi }^{3}}}{48}-\dfrac{\pi }{4}$

$\Rightarrow \dfrac{\pi }{4}\left( \dfrac{{{\pi }^{2}}}{12}-1 \right)$

Hence, this is the answer.

$\displaystyle \int \dfrac{\tan^{ - 1}x}{1+x^2}\ dx$

We have,

$\displaystyle I=\int \dfrac{\tan^{ - 1}x}{1+x^2}\ dx$

Let

$t=\tan^{-1}x$

$dt=\dfrac{1}{1+x^2}dx$

Therefore,

$\displaystyle I=\int t\ dt$

$I=\dfrac{t^2}{2}+C$

$I=\dfrac{(\tan^{-1}x)^2}{2}+C$

Hence, this is the answer.

Integrate $\int {({{\sin }^{ - 1}}} x{)^2}dx$

$\begin{array}{l} I=\int { { { \left( { { { \sin }^{ -1 } }x } \right) }^{ 2 } }dx } & \\ I=\int { \left( { { { \sin }^{ -1 } }x } \right) .\left( { { { \sin }^{ -1 } }x } \right) dx } & \left[ \begin{array}{l} Put\, \, { \sin ^{ -1 } }=t \\ x=\sin t \\ dx=\cos tdt \end{array} \right] \\ =\int { { t^{ 2 } }.\cos tdt } & \end{array}$

Using

$ILATE$ formulae

$\begin{array}{l} ={ t^{ 2 } }\sin t-\int { 2t\sin tdt } \\ ={ t^{ 2 } }\sin t-2\int { t\sin tdt } \to Apply\, \, ILATE\, \, formulae \\ ={ t^{ 2 } }\sin t-2\left[ { -t\cos t+\int { \cos tdt } } \right] \\ ={ t^{ 2 } }\sin t+2t\cos t-2\sin t+c \\ ={ \left( { { { \sin }^{ -1 } }x } \right) ^{ 2 } }\sin .{ \sin ^{ -1 } }x+2{ \sin ^{ -1 } }x\sqrt { 1-{ x^{ 2 } } } -2\sin .{ \sin ^{ -1 } }x+c \\ ={ \left( { { { \sin }^{ -1 } }x } \right) ^{ 2 } }.x+2{ \sin ^{ -1 } }x\sqrt { 1-{ x^{ 2 } } } -2x+c \end{array}$

$\displaystyle\int ^{\pi/2}_{\pi/4} \sqrt{1-\sin 2 x} d x=?$

We have,

$\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\sqrt{1-\sin 2x}}dx$

$\Rightarrow \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x-\sin 2x}}dx$

$\Rightarrow \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\sqrt{{{\left( \sin x-\cos x \right)}^{2}}}}dx$

$\Rightarrow \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\left( \sin x-\cos x \right)}dx$

$\Rightarrow \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\sin xdx-\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cos xdx}}$

On integrating and we get,

$\Rightarrow {{\left[ -\cos x \right]}_{\dfrac{\pi }{4}}}^{\dfrac{\pi }{2}}-{{\left[ \sin x \right]}_{\dfrac{\pi }{4}}}^{\dfrac{\pi }{2}}$

$\Rightarrow \left[ -\cos \dfrac{\pi }{2}+\cos \dfrac{\pi }{4} \right]-\left[ \sin \dfrac{\pi }{2}-\sin \dfrac{\pi }{4} \right]$

$\Rightarrow \left[ -0+\dfrac{1}{\sqrt{2}} \right]-\left[ 1-\dfrac{1}{\sqrt{2}} \right]$

$\Rightarrow \dfrac{2}{\sqrt{2}}-1$

$\Rightarrow \sqrt{2}-1$

Hence, this is the answer.

$\displaystyle\int\dfrac{1}{3\sin x-4\sin^3x}dx$ .

We have,

$I=\int \dfrac{1}{3\sin x -4\sin ^3x}dx$

$I=\int \dfrac{1}{\sin 3x}dx$

$I=\int \csc 3x\;dx$

$I=-\dfrac{\log \left| \csc 3x +\cot 3x \right|}{3}$

Hence, the answer is

$-\dfrac{\log \left| \csc 3x +\cot 3x \right|}{3}.$

Solve $\int \dfrac{cos^{9}x}{sin x} dx$

$\int { \cfrac { \cos ^{ 9 }{ x } }{ \sin { x } } dx } =\int { \cfrac { \cos ^{ 8 }{ x } \cos { x } }{ \sin { x } } dx } \\ =\int { \cfrac { { (1-\sin ^{ 2 }{ x } ) }^{ 4 }\cos { x } }{ \sin { x } } dx } \quad \quad (\because \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } =1)$

Let

$t=\sin { x } \quad \quad dt=\cos { x } dx$

$=\int { \cfrac { { (1-{ t }^{ 2 }) }^{ 4 } }{ t } dt } \\ =\int { \cfrac { { (1+{ t }^{ 4 }-2{ t }^{ 2 }) }^{ 2 } }{ t } dt } \\ =\int { \cfrac { 1+{ t }^{ 8 }+4{ t }^{ 4 }+2{ t }^{ 4 }-4{ t }^{ 2 }-4{ t }^{ 6 } }{ t } dt } \\ =\int { \cfrac { 1 }{ t } dt } +\int { { t }^{ 7 }dt } +4\int { { t }^{ 3 }dt } +2\int { { t }^{ 3 }dt } -4\int { tdt } -4\int { { t }^{ 5 }dt } \\ =\int { \cfrac { 1 }{ t } dt } +\int { { t }^{ 7 }dt } +6\int { { t }^{ 3 }dt } -4\int { tdt } -4\int { { t }^{ 5 }dt } \\ =\ln { \left| t \right| } +\cfrac { { t }^{ 8 } }{ 8 } +\cfrac { 6{ t }^{ 4 } }{ 4 } -\cfrac { 4{ t }^{ 2 } }{ 2 } -\cfrac { 4{ t }^{ 6 } }{ 6 } +C\\ =\ln { \left| t \right| } +\cfrac { { t }^{ 8 } }{ 8 } +\cfrac { 3{ t }^{ 4 } }{ 2 } -2{ t }^{ 2 }-\cfrac { 2 }{ 3 } { t }^{ 6 }+C\\ =\ln { \left| \sin { x } \right| } +\cfrac { \sin ^{ 8 }{ x } }{ 8 } +\cfrac { 3\sin ^{ 4 }{ x } }{ 2 } -2\sin ^{ 2 }{ x } -\cfrac { 2 }{ 3 } \sin ^{ 6 }{ x } +C$

Evaluate $\displaystyle\int \dfrac {x^3+4x^2+9x}{x^2+4x+9} dx$

$\displaystyle\int \dfrac {x^3+4x^2+9x}{x^2+4x+9} dx\\\displaystyle \int \dfrac{x(x^2+4x+9)}{x^2+4x+9}dx\\\displaystyle \int x dx\\\dfrac{x^2}{2}+c$

Using integration, find the area of the triangle formed by negative $x - axis$ and tangent and normal to the circle ${x^2} + {y^2} = 9$ at $( - 1,2\sqrt 2 )$ .

Therefore, Area of triangle = 9√5/4 units^2.

1325607_1350442_ans_2edde1f0dd3d40c6873cae6065108de3.jpg

Evaluate: $\displaystyle \int{\dfrac{(2x+3)dx}{x^2+3x+6}}$

$\displaystyle \int{\dfrac{(2x+3)dx}{x^2+3x+6}}\\t=x^2+3x+6\implies dt=2x+3dx\\\displaystyle \int \dfrac{dt}{t}\\\log t+c\\\log(x^2+3x+6)+c$

$\displaystyle \int{ \sin }^{ 4 }x \cos x dx$

$\displaystyle \int \sin^4x\cos xdx$

Let

$\sin x=t\implies \cos xdx=dt$

$\displaystyle\int t^4dt\\\dfrac{t^5}{5}+c\\\dfrac{\sin^5x}{5}+c$

$\int _{ 0 }^{ 1 }{ \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } } dx$

$\displaystyle\int_{0}^{1}{\dfrac{xdx}{\sqrt{1+{x}^{2}}}}$

$=\dfrac{1}{2}\displaystyle\int_{0}^{1}{\dfrac{2xdx}{\sqrt{1+{x}^{2}}}}$

$=\dfrac{1}{2}\displaystyle\int_{0}^{1}{{\left(1+{x}^{2}\right)}^{\frac{-1}{2}}}$

$=\dfrac{1}{2}\left[\dfrac{{\left(1+{x}^{2}\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{1}$

$=\dfrac{1}{2}\left[\dfrac{{\left(1+{x}^{2}\right)}^{\frac{3}{2}}}{\dfrac{3}{2}}\right]_{0}^{1}$

$=\dfrac{1}{2}\left[\dfrac{2{\left(1+{x}^{2}\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$

$=\dfrac{1}{2}\left[\dfrac{2{\left(1+{1}^{2}\right)}^{\frac{3}{2}}}{3}-\dfrac{2{\left(1+0\right)}^{\frac{3}{2}}}{3}\right]$

$=\dfrac{1}{2}\left[\dfrac{{2}^{\frac{5}{2}}}{3}-\dfrac{2}{3}\right]$

$=\dfrac{1}{6}\left(2\left({2}^{\frac{3}{2}}-1\right)\right)$

$=\dfrac{{2}^{\frac{3}{2}}-1}{3}$

$=\dfrac{2\sqrt{2}-1}{3}$

Evaluate :
$\int x^2 (1-\dfrac{1}{x^2})dx$

$I=\int x^2(1-\dfrac{1}{x^2})dx$

$I=\int (x^2-1)dx$

$I=\dfrac{x^3}{3}-x+C$

Evaluate $\displaystyle\int^\pi_0 \dfrac{x}{a^2cos^2x+ b^2sin^2x} dx$

$I=\int^\pi_0\cfrac{x}{a^2\cos^2 x+b^2\sin^2 x}dx$

$\implies$

$I=\int^\pi_0\cfrac{\pi-x}{a^2\cos^2 x+b^2\sin^2 x}dx$

Adding above two equations

$I+I=\int^\pi_0\cfrac{x}{a^2\cos^2 x+b^2\sin^2 x}dx$

$+\int^\pi_0\cfrac{\pi-x}{a^2\cos^2 x+b^2\sin^2 x}dx$

$2I=\int^\pi_0\cfrac{\pi}{a^2\cos^2 x+b^2\sin^2 x}dx$

$I=\cfrac{\pi}{2}\int^\pi_0\cfrac{sec^{2} x}{a^2+b^2\tan^2x}$

Now, let

$tan$

$x$

$=$

$t$

Therefore,

$sec^{2} x dx = dt$

$I=\dfrac{2\pi}{2}\int^\infty_{0}\dfrac{dt}{a^{2} + b^{2}t^{2}}$

$I=\dfrac{\pi}{b^{2}}\int^\infty_{0}\dfrac{dt}{(\dfrac{a}{b})^{2} + t^{2}}$

$I=\dfrac{\pi}{b^{2}}\times \dfrac{b}{a} [tan^{-1}\dfrac{at}{b}]^\infty_{0}$

$I=\dfrac{\pi}{ab}\times \dfrac{\pi}{2}$

$I=\dfrac{\pi^{2}}{2ab}$

Evaluate: $\displaystyle \int (x^3+5x^2+6)(3x^2+5x)d x$

$\displaystyle \int (x^3+5x^2+6)(3x^2+5x)d x \\t=x^3+5x^2+6\implies 3x^2+5x dx =dt \\\displaystyle\int tdt\\\dfrac{t^2}{2}+c\\\dfrac{x^3+5x^2+6}{2}+c$

Evaluate: $\displaystyle \int (x^2+5x+6)(2x+5)d x$

$I=\displaystyle \int (x^2+5x+6)(2x+5)d x$

$t=x^2+5x+6\implies 2x+5 dx =dt \displaystyle\int tdt$

$=\dfrac{t^2}{2}+c$

$=\dfrac{(x^2+5x+6)^{2}}{2}+c$

Find:
$\int _{ 1 }^{ 2 }{ \dfrac { \ell nx }{ { x }^{ 2 } } } dx$

$u=\ln{x}\Rightarrow du=\dfrac{1}{x}dx$

$dv=\dfrac{1}{{x}^{2}}\Rightarrow v=\dfrac{-1}{x}$

$\displaystyle\int_{1}^{2}{\dfrac{\ln{x}}{{x}^{2}}dx}=\left[\ln{x}\times \dfrac{-1}{x}\right]_{1}^{2}-\displaystyle\int_{1}^{2}{\dfrac{-1}{x}\times \dfrac{1}{x}dx}$

$=\left[\dfrac{-\ln{x}}{x}\right]_{1}^{2}-\displaystyle\int_{1}^{2}{\dfrac{-1}{{x}^{2}}dx}$

$=\left[\dfrac{-\ln{2}}{2}-\dfrac{-\ln{1}}{1}\right]+\displaystyle\int_{1}^{2}{\dfrac{1}{{x}^{2}}dx}$

$=\dfrac{-\ln{2}}{2}+0+\left[\dfrac{{x}^{-2+1}}{-2+1}\right]_{1}^{2}$

$=\dfrac{-\ln{2}}{2}+\left[\dfrac{-1}{x}\right]_{1}^{2}$

$=\dfrac{-\ln{2}}{2}-\left[\dfrac{1}{2}-\dfrac{1}{1}\right]$

$=\dfrac{-\ln{2}}{2}-\dfrac{1-2}{2}$

$=\dfrac{-\ln{2}}{2}+\dfrac{1}{2}$

$=\dfrac{1-\ln{2}}{2}$

Integrate: $\displaystyle\int{{\tan}^{3}{x}dx}$

$\displaystyle\int{{\tan}^{3}{x}dx}$

$=\displaystyle\int{\tan{x}{\tan}^{2}{x}dx}$

$=\displaystyle\int{\tan{x}\left({\sec}^{2}{x}-1\right)dx}$

$=\displaystyle\int{\tan{x}{\sec}^{2}{x}dx}+\displaystyle\int{\tan{x}dx}$

Let

${I}_{1}=\displaystyle\int{\tan{x}{\sec}^{2}{x}dx}$ and

${I}_{2}=\displaystyle\int{\tan{x}dx}$

Consider

${I}_{1}=\displaystyle\int{\tan{x}{\sec}^{2}{x}dx}$

Let

$t=\tan{x}\Rightarrow dt={\sec}^{2}{x}dx$

$\Rightarrow \displaystyle\int{tdt}=\dfrac{{t}^{2}}{2}+c=\dfrac{{\tan}^{2}{x}}{2}+c$

$\therefore {I}_{1} =\dfrac{{\tan}^{2}{x}}{2}+c$

Consider

${I}_{2}=\displaystyle\int{\tan{x}dx}$

$=\displaystyle\int{\dfrac{\sin{x}}{\cos{x}}dx}$

Let

$t=\cos{x}\Rightarrow dt=-\sin{x}dx$

$\Rightarrow {I}_{2}=-\displaystyle\int{\dfrac{dt}{t}}=-\log{t}=\log{\dfrac{1}{t}}=\log{\dfrac{1}{\cos{x}}}=\log{\sec{x}}+c$

$\therefore \displaystyle\int{{\tan}^{3}{x}dx}={I}_{1}+{I}_{2}$

$=\dfrac{{\tan}^{2}{x}}{2}+\log{\sec{x}}+c$

Find :
$\int { xsinxdx }$ .

Let

$u=x\Rightarrow du=dx$ and

$dv=\sin{x}dx\Rightarrow v=-\cos{x}$

Integrating by parts, we get

$\displaystyle\int{u.dv}=uv-\displaystyle\int{v.du}$

$=-x\cos{x}-\displaystyle\int{-\cos{x}dx}$

$=-x\cos{x}+\displaystyle\int{\cos{x}dx}$

$=-x\cos{x}+\sin{x}+c$ where

$c$ is the integration constant.

Evaluate: $\displaystyle \int \dfrac{2 x}{\left(x^{2}+4\right)} d x$

$I=\displaystyle \int \dfrac {2x}{x^2+4} dx$

$t=x^2+4\implies dt=2xdx$

$=\displaystyle \int \dfrac 1tdt$

$=\displaystyle \log t$

$=\log (x^2+4)+c$

Simplify:
$\int { \dfrac { secx }{ secx+tanx } }$

$\displaystyle\int{\dfrac{\sec{x}dx}{\sec{x}+\tan{x}}}$

$=\displaystyle\int{\dfrac{\sec{x}dx}{\sec{x}+\tan{x}}\times\dfrac{\sec{x}-\tan{x}}{\sec{x}-\tan{x}}}$

$=\displaystyle\int{\dfrac{\sec{x}\left(\sec{x}-\tan{x}\right)dx}{{\sec}^{2}{x}-{\tan}^{2}{x}}}$

$=\displaystyle\int{\sec{x}\left(\sec{x}-\tan{x}\right)dx}$ since

${\sec}^{2}{x}-{\tan}^{2}{x}=1$

$=\displaystyle\int{{\sec}^{2}{x}dx}-\displaystyle\int{\sec{x}\tan{x}dx}$

$=\tan{x}-\sec{x}+c$ since

$\displaystyle\int{{\sec}^{2}{x}dx}=\tan{x}+c$ and

$\displaystyle\int{\sec{x}\tan{x}dx}=\sec{x}+c$

$\int _{ -5 }^{ 5 }{ \left(x+2\right)} dx$

$\displaystyle\int_{5}^{-5}{\left(x+2\right)dx}$

$=\left[\dfrac{{x}^{2}}{2}+2x\right]_{-5}^{5}$

$=\dfrac{1}{2}\left[{5}^{2}-{\left(-5\right)}^{2}\right]+2\left(5-\left(-5\right)\right)$

$=\dfrac{1}{2}\left[25-25\right]+2\left(5+5\right)$

$=\dfrac{1}{2}\left(0\right)+2\times 10=20$

$\int _{ 1 }^{ 3 }{ a^{ x } } dx$

$\displaystyle\int_{1}^{3}{{a}^{x}dx}$

$=\left[\dfrac{{a}^{x}}{\ln{a}}\right]_{1}^{3}$

$=\dfrac{{a}^{3}}{\ln{a}}-\dfrac{{a}^{1}}{\ln{a}}+c$

$=\dfrac{{a}^{3}}{\ln{a}}-\dfrac{a}{\ln{a}}+c$

Evaluate: $I =\displaystyle \int \log \sin x dx$

$=log\sin x\cdot \int dx - \int(\int dx)\cdot \dfrac{d}{dx}(log\sin x) dx + c$

$= log\sin x\cdot x - \int x \cdot \dfrac{1}{\sin x} \cdot \cos x dx + c$

$=xlog\sin x - \int x \cdot \dfrac{\cos x}{\sin x}dx + c$

$=xlog\sin x - \int x \cot x dx + c$

$=xlog\sin x - x \cdot \int \cot x dx - \int (\int \cot x dx) \dfrac{d}{dx}(x) dx +c +$

$=xlog \sin x - x \cdot log |\sin x| - \int log|\sin x|dx + c'$

$\because I = \int log |\sin x|$

$I = -I + c'$

$2I = c'$

$I = c''$

Evaluate :
$\int { \log x } dx$

$\displaystyle\int{\log{x}dx}$

let

$u=\log{x}$

$v=1$

Integrating by parts,

$=x\log{x}-\displaystyle\int{x\times\dfrac{1}{x}dx}$

$=x\log{x}-\displaystyle\int 1{dx}$

$=x\log{x}-x=x\left(\log{x}-1\right)+c$

Find :
$\int { \dfrac { { sec }^{ 2 }x }{ { tan }^{ 2 }x+4 } dx }$

$\displaystyle\int{\dfrac{{\sec}^{2}{x}dx}{{\tan}^{2}{x}+4}}$

Let

$t=\tan{x}\Rightarrow dt={\sec}^{2}{x}dx$

$=\displaystyle\int{\dfrac{dt}{{t}^{2}+4}}$

$=\dfrac{1}{2}{\tan}^{-1}{\dfrac{x}{2}}+c$ using

$\displaystyle\int{\dfrac{dx}{{x}^{2}+{a}^{2}}}=\dfrac{1}{a}{\tan}^{-1}{\dfrac{x}{a}}+c$

$\therefore \displaystyle\int{\dfrac{{\sec}^{2}{x}dx}{{\tan}^{2}{x}+4}}=\dfrac{1}{2}{\tan}^{-1}{\dfrac{x}{2}}+c$ where

$c$ is the constant of integration.

Find :
$\int { xcosxdx }$ .

Let

$u=x\Rightarrow du=dx$ and

$dv=\cos{x}dx\Rightarrow v=\sin{x}$

Integrating by parts, we get

$\displaystyle\int{u.dv}=uv-\displaystyle\int{v.du}$

$=x\sin{x}-\displaystyle\int{\sin{x}dx}$

$=x\sin{x}-\left(-\cos{x}\right)+c$

$=x\sin{x}+\cos{x}+c$ where

$c$ is the integration constant.

Evaluate : $\int sin4x . sin8x dx$ .

$\displaystyle\int \sin 4 x \sin 8 x d x=\dfrac{1}{2}\int (2\sin 4 x\sin 8 x )d x=\dfrac{1}{2}\int (\cos 4 x-\cos 12 x)d x=\dfrac{1}{2}\int \cos 4 x d x-\dfrac{1}{2}\int \cos 12 x d x$

$=\dfrac{\sin 4 x}{8}-\dfrac{\sin 12 x}{24}+C$

Evaluate:
$\int { { x }^{ 2 }.{ e }^{ -2 } } dx$

$\displaystyle\int{{x}^{2}{e}^{-2}dx}$

$={e}^{-2}\displaystyle\int{{x}^{2}dx}$

$={e}^{-2}\left(\dfrac{{x}^{3}}{3}\right)+c$

$=\dfrac{{x}^{3}}{3{e}^{2}}+c$

Evaluate: $\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{dx}{5+4 cos x}$

$\displaystyle\int^{\pi/4}_0\dfrac{dx}{5+4\cos x}=\displaystyle\int^{\pi/4}_0\dfrac{dx}{5+4\left(\dfrac{1-\tan^2x/2}{1+\tan^2x/2}\right)}$

$=\displaystyle\int^{\pi/4}_0\dfrac{dx(1+\tan^2x/2)}{5+5\tan^2x/2+4-4\tan^2x/2}$

$=\displaystyle\int^{\pi/4}_0\dfrac{\sec^2x/2dx}{9+\tan^2x/2}$ Let

$t=\tan x/2$

$dt=\dfrac{1}{2}\sec^2x/2dx$

$=\displaystyle\int^{\pi/4}_0\dfrac{2dt}{9+t^2}$

When

$x/2=0, t=0$ ; when

$x/2=\pi/4$

$\Rightarrow t=1$

$=\displaystyle\int^1_0\dfrac{2dt}{9+t^2}$

$=\left.2\times \dfrac{1}{3}\tan^{-1}\left(\dfrac{t}{3}\right)\right]^1_0$

$=\dfrac{2}{3}\left[\tan^{-1}\left(1/3\right)-\tan^{-1}(0)\right]$

$=\dfrac{2}{3}\left[\tan^{-1}\left(1/3\right)-0\right]$

$=\dfrac{2}{3}\tan^{-1}\left(1/3\right)$ .

1235278_1502002_ans_df5a6a2837f444ac8e8964cae3ab95bc.PNG

Evaluate : $\displaystyle \int { \log } x{dx}$

${\textbf{Step -1: First solve for}}$

${\textbf{log x,}}$

${\textbf{ and then solve for}}$

${\mathbf{1}}$

$.$

$\int {\log xdx}$

$= \int {\left( {\log x} \right).1dx}$

${\text{Using by parts,}}$

${1^{st}}$

${\text{function}}$

$= f\left( x \right) = \log x$

${\text{and}}$

${2^{nd}}$

${\text{function}}$

$= g\left( x \right) = 1$

${\textbf{Step -2: Use formula}}$

${\mathbf{\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left( {f'\left( x \right)\int {g\left( x \right)dx} } \right)dx} } }}}$

${\text{Placing value of}}$

$f\left( x \right)$

${\text{and}}$

$g\left( x \right)$

$.$

$\int {\left( {\log x} \right).1dx} = \log x\int {1.dx - \int {\left( {\dfrac{{d\left( {\log x} \right)}}{{dx}}\int {1.dx} } \right)dx} }$

$= \left( {\log x} \right)x - \int {\dfrac{1}{x}} .x.dx$

$= x\log x - \int {1.dx}$

$= x\log x - x + C$

${\textbf{ Hence, the value of}}$

${\mathbf{\int {log xdx}}}$

${\textbf{is}}$

${\mathbf{x{\textbf{log}} x - x + C}}$

$.$

Evaluate the following definite integrals:

$\displaystyle \int _{0}^{1/3} \dfrac {1}{\sqrt {1-x^2}}dx$

$\displaystyle\int_{0}^{\frac{1}{3}}{\dfrac{dx}{\sqrt{1-{x}^{2}}}}$

Let

$x=\sin{\theta}\Rightarrow\,dx=\cos{\theta}d\theta$

$=\displaystyle\int{\dfrac{\cos{\theta}d\theta}{\sqrt{1-{\sin}^{2}{\theta}}}}$

$=\displaystyle\int{\dfrac{\cos{\theta}d\theta}{\sqrt{{\cos}^{2}{\theta}}}}$

$=\displaystyle\int{\dfrac{\cos{\theta}d\theta}{\cos{\theta}}}$

$=\displaystyle\int{d\theta}$

$=\left[\theta\right]$

$=\left[{\sin}^{-1}{x}\right]_{0}^{\frac{1}{3}}$ where

$\theta=\sin^{-1}{x}$

$={\sin}^{-1}{\dfrac{1}{3}}-{\sin}^{-1}{0}$

$={\sin}^{-1}{\dfrac{1}{3}}-n\pi$

Evaluate $\displaystyle\int_{1}^{2}\dfrac{3x}{9x^{2}-1}dx$

$\displaystyle\int_{1}^{2}{\dfrac{3x}{9{x}^{2}-1}dx}$

$=\dfrac{1}{6}\displaystyle\int_{1}^{2}{\dfrac{18x}{9{x}^{2}-1}dx}$

Let

$t=9{x}^{2}-1\Rightarrow\,dt=18x\,dx$

$=\dfrac{1}{6}\displaystyle\int_{1}^{2}{\dfrac{dt}{t}}$

$=\dfrac{1}{6}\left[\log{t}\right]_{1}^{2}$

$=\dfrac{1}{6}\left[\log{\left(9{x}^{2}-1\right)}\right]_{1}^{2}$

$=\dfrac{1}{6}\left[\log{35}-\log{8}\right]$

$=\dfrac{1}{6}\log{\left(\dfrac{35}{8}\right)}$

Solve : $\displaystyle \int _{ 0 }^{ \pi }{ \dfrac { x\tan{ x } }{ \sec { x } +\tan { x } } } dx$

$I$ =

$\displaystyle \int _{ 0 }^{ \pi }{ \dfrac { x\tan{ x } }{ \sec { x } +\tan { x } } } =\displaystyle \int_0^{\pi}{\dfrac{( \pi -x) \tan ( \pi -x)}{\sec ( \pi -x) +\tan ( \pi -x)}}dx$

$\displaystyle \int _{ 0 }^{ \pi }{ \dfrac { \pi\tan{ (\pi - x) } }{ -(\sec { x } +\tan { x }) } } -\dfrac{x\tan ( \pi -x)}{-(\sec x+\tan x)}$

$I=\displaystyle \int_0^{\pi}{\dfrac{\pi \tan x}{\sec x+\tan x}dx -\dfrac{x\tan x}{\sec x\tan x}dx}$

$2I =\displaystyle \int_0^{\pi}{\dfrac{\pi \tan x}{\sec x +\tan x}dx}=\displaystyle \int_0^{\pi}{\pi \times \left( \dfrac{\sin x}{1+\sin x}\right)}dx$

$=\displaystyle \int_0^{\pi}{\pi }({\dfrac{\sin x (1-\sin x)}{1-\sin^2 x}})=\displaystyle \int_0^{\pi}{\pi}\dfrac{\sin x (1-\sin x)}{\cos^2 x}$

$=\pi \displaystyle \int_0^{\pi}{( \sec x \tan x-\tan ^2x)}dx=\displaystyle \int_0^{\pi}{\pi ( \sec x \tan x-\sec^2 x+1)}dx$

$=(\left.\pi \sec x \right]_0^{\pi}-\pi \tan x ]_0^{\pi} +x]_0^{\pi})=(-1-1)-(0-0)+\pi )\pi$

$=(\pi -2)\pi$

$I=\dfrac{\pi}{2}( \pi -2)$

Evaluate the following definite integrals :

$\displaystyle \int _{0}^1 \dfrac {1}{1+x^2}dx$

Let

$x=\tan{\theta}\Rightarrow\,dx={\sec}^{2}{\theta}d\theta$

When

$x=0\Rightarrow\,\theta=0$

When

$x=1\Rightarrow\,\theta=\dfrac{\pi}{4}$

$\displaystyle\int_{0}^{1}{\dfrac{dx}{1+{x}^{2}}}$

$=\displaystyle\int_{0}^{\frac{\pi}{4}}{\dfrac{{\sec}^{2}{\theta}d\theta}{1+{\tan}^{2}{\theta}}}$

$=\displaystyle\int_{0}^{\frac{\pi}{4}}{\dfrac{{\sec}^{2}{\theta}d\theta}{{\sec}^{2}{\theta}}}$

$=\displaystyle\int_{0}^{\frac{\pi}{4}}{d\theta}$

$=\left[\theta\right]_{0}^{\frac{\pi}{4}}$

$=\dfrac{\pi}{4}-0=\dfrac{\pi}{4}$

$\displaystyle\int_{0}^{1}\dfrac{e^{x}}{1+e^{2x}}dx$

Let

${e}^{x}=\tan{\theta}\Rightarrow\,{e}^{x}dx={\sec}^{2}{\theta}d\theta$

When

$x=0\Rightarrow\,1=\tan{\theta}\therefore\,\theta=\dfrac{\pi}{4}$

When

$x=1\Rightarrow\,e=\tan{\theta}\therefore\,\theta={\tan}^{-1}{e}$

$\displaystyle\int_{0}^{1}{\dfrac{{e}^{x}}{1+{e}^{2x}}dx}$

$=\displaystyle\int_{\frac{\pi}{4}}^{{\tan}^{-1}{e}}{\dfrac{{\sec}^{2}{\theta}d\theta}{1+{\tan}^{2}{\theta}}}$

$=\displaystyle\int_{\frac{\pi}{4}}^{{\tan}^{-1}{e}}{\dfrac{{\sec}^{2}{\theta}d\theta}{{\sec}^{2}{\theta}}}$

$=\displaystyle\int_{\frac{\pi}{4}}^{{\tan}^{-1}{e}}{d\theta}$

$=\left[\theta\right]_{\frac{\pi}{4}}^{{\tan}^{-1}{e}}$

$={\tan}^{-1}{e}-\dfrac{\pi}{4}$

Find the value of $\int x \tan^2x \, dx$ .

$\int{x \; \tan^{2}{x} \; dx}$

$= \int{x \left( \sec^{2}{x} - 1 \right) dx} \quad \left( \because 1 + \tan^{2}{x} = \sec^{2}{x} \right)$

$= \int{x \; \sec^{2}{x} \; dx} - \int{x \; dx}$

$= \left( x \left( \tan{x} \right) - \int{\tan{x} \; dx} \right) - \cfrac{{x}^{2}}{2} + C$

$= x \tan{x} - \ln{\left| \sec{x} \right|} - \cfrac{{x}^{2}}{2} + C$

Hence

$\int{x \; \tan^{2}{x} \; dx} = x \tan{x} - \ln{\left| \sec{x} \right|} - \cfrac{{x}^{2}}{2} + C$

Evaluate $\displaystyle\int_{2}^{4}\dfrac{x}{x^{2}+1}dx$

$\displaystyle\int_{4}^{2}{\dfrac{x}{{x}^{2}+1}dx}$

$=\dfrac{1}{2} \displaystyle\int_{4}^{2}{\dfrac{2x}{{x}^{2}+1}dx}$

Let

$t={x}^{2}+1\Rightarrow\,dt=2x\,dx$

$=\dfrac{1}{2} \displaystyle\int{\dfrac{dt}{t}}$

$=\dfrac{1}{2}\left[\log{\left|t\right|}\right]$

$=\dfrac{1}{2}\left[\log{\left|{x}^{2}+1\right|}\right]_{4}^{2}$ where

$t={x}^{2}+1$

$=\dfrac{1}{2}\left[\log{\left|{2}^{2}+1\right|}-\log{\left|{4}^{2}+1\right|}\right]$

$=\dfrac{1}{2}\left[\log{\left|4+1\right|}-\log{\left|16+1\right|}\right]$

$=\dfrac{1}{2}\left[\log{\left|5\right|}-\log{\left|17\right|}\right]$

$=\dfrac{1}{2}\log{\left|\dfrac{5}{17}\right|}$

Evaluate $\displaystyle\int_{0}^{\pi/2}\dfrac{\cos x}{1+\sin^{2}x}dx$

$\displaystyle\int_{0}^{\frac{\pi}{2}}{\dfrac{\cos{x}}{1+{\sin}^{2}{x}}dx}$

Let

$t=\sin{x}\Rightarrow\,dt=\cos{x}dx$

$x\rightarrow\,0\Rightarrow\,t\rightarrow\,0$

$x\rightarrow\,\dfrac{\pi}{2}\Rightarrow\,t\rightarrow\,1$

$\displaystyle\int_{0}^{\frac{\pi}{2}}{\dfrac{\cos{x}}{1+{\sin}^{2}{x}}dx}$

$=\displaystyle\int_{0}^{1}{\dfrac{dt}{1+{t}^{2}}}$

$=\left[{\tan}^{-1}{t}\right]_{0}^{1}$

$={\tan}^{-1}{1}-{\tan}^{-1}{0}$

$=\dfrac{\pi}{4}-0$

$=\dfrac{\pi}{4}$

Evaluate:

$\displaystyle\int_{0}^{\pi/2}\dfrac{\sin\theta}{\sqrt{1+\cos\theta}}d\theta$

$\displaystyle\int_{0}^{\frac{\pi}{2}}{\dfrac{\sin{\theta}}{\sqrt{1+\cos{\theta}}}d\theta}$

Let

$t=\cos{\theta}\Rightarrow\,dt=-\sin{\theta}d\theta$

$\theta\rightarrow\,0\Rightarrow\,t\rightarrow\,1$

$\theta\rightarrow\,\dfrac{\pi}{2}\Rightarrow\,t\rightarrow\,0$

Now,

$\displaystyle\int_{0}^{\frac{\pi}{2}}{\dfrac{\sin{\theta}}{\sqrt{1+\cos{\theta}}}d\theta}$

$=-\displaystyle\int_{1}^{0}{\dfrac{dt}{\sqrt{1+t}}}$

$=\displaystyle\int_{0}^{1}{{\left(t+1\right)}^{-\frac{1}{2}}dt}$

$=\left[\dfrac{{\left(t+1\right)}^{-\frac{1}{2}+1}}{-\dfrac{1}{2}+1}\right]_{0}^{1}$

$=\left[\dfrac{{\left(t+1\right)}^{\frac{1}{2}}}{\dfrac{1}{2}}\right]_{0}^{1}$

$=\left[2\sqrt{t+1}\right]_{0}^{1}$

$=2\left(\sqrt{2}-\sqrt{1}\right)=2\left(\sqrt{2}-1\right)$

Evaluate the following definite integrals:
$\displaystyle \int _{\pi /6}^{\pi /4} cosec \ x \ dx$

$I=\displaystyle \int _{\pi /6}^{\pi /4} cosec \ x \ dx$

$=[\log (cosec \ x-\cot x)]_{\pi /6}^{\pi /4}$

$=\log \left (cosec \dfrac {\pi}{4} -\cot \dfrac {\pi}{4}\right) -\log \left (cosec \dfrac {\pi}{6}-\cot \dfrac {\pi}{6} \right)$

$=\log (\sqrt 2-1)-\log (2-\sqrt 3)$

$=\log \left (\dfrac {\sqrt 2-1}{2-\sqrt 3}\right )$

Evaluate $\displaystyle\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$

$I=\displaystyle\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$

Let

$x=a\sin t$

Then,

$dx=a\cos t dt$ [ Differentiating both sides ]

When

$x=0,$

$0=a\sin t$

$t=\sin^{-1}(0)$

$\therefore$

$t=0$

When,

$x=a,$

$a=a\sin t$

$1=\sin t$

$t=\sin^{-1}(1)$

$\therefore$

$t=\dfrac{\pi}{2}$

$I=\displaystyle\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$

$\Rightarrow$

$I=\displaystyle\int_0^{\dfrac{\pi}{2}}\sqrt{(a^2-a^2\sin^2t)}a\cos t dt$

$=\displaystyle\int_0^{\dfrac{\pi}{2}}\sqrt{a^2(1-\sin^2t)}a\cos t dt$

$=\displaystyle\int_0^{\dfrac{\pi}{2}}\sqrt{a^2\cos^2 t}a\cos t dt$

$=\displaystyle\int_0^{\dfrac{\pi}{2}}a\cos t.a\cos t dt$

$=\displaystyle\int_0^{\dfrac{\pi}{2}}a^2\cos^2 t dt$

$=a^2\displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{1+\cos 2t}{2}dt$

$=\dfrac{a^2}{2}\left[t+\dfrac{\sin 2t}{2}\right]^{\dfrac{\pi}{2}}_0$

$=\dfrac{a^2}{2}\left(\dfrac{\pi}{2}-0\right)$

$=\dfrac{a^2\pi}{4}$

Evaluate the following definite integrals :
$\displaystyle \int _{0}^{\pi} \dfrac {1}{1+\sin x}dx$

$I=\displaystyle \int _{0}^{\pi} \dfrac {1}{1+\sin x}dx$

$=\displaystyle \int _{0}^{\pi}\dfrac {1-\sin x}{(1+\sin x)(1-\sin x)}dx$

$=\displaystyle \int _{0}^{\pi} \dfrac {1-\sin x}{\cos^2 x}dx$

$\Rightarrow \ I=\displaystyle \int _{0}^{\pi} (\sec^2 x-\sec x\tan x)dx$

$= [\tan x-\sec x]_{0}^{\pi}$

$\Rightarrow \ I=(\tan \pi -\sec \pi )-(\tan 0-\sec 0)=0+1+1=2$

Evaluate the integral $\displaystyle\int_{0}^{1/2}\dfrac{x\sin^{-1}x}{\sqrt{1-x^{2}}}dx$ .

$\int ^{\dfrac{1}{2}}_{0}\dfrac{xsin^{-1}x}{\sqrt{1-x^2}}dx$

Using byparts

$=\dfrac{x(sin^{-1}x)^2}{2}|^{\dfrac{1}{2}}_{0}-\int^{\dfrac{1}{2}}_{0}\dfrac{(sin^{-1}x)^2}{2}dx$

Substituting x=sinu

$=\dfrac{(\pi)^2}{144}-\int^{\dfrac{\pi}{6}}_{0}\dfrac{u^2}{2}du$

$=\dfrac{(\pi)^2}{144}-\dfrac{(\pi)^3}{648}$

Evaluate $\displaystyle\int_{0}^{\pi/4}(\sqrt{\tan}x+\sqrt{\cot}x)dx$

Let

$I=\displaystyle\int_{0}^{\pi/4}(\sqrt{\tan}x+\sqrt{\cot}x)dx$

$I=\displaystyle\int_{0}^{\pi/4}(\sqrt{\dfrac{\sin x}{\cos x}}+\sqrt{\dfrac{\cos x}{\sin x}})dx$

$=\displaystyle\int_0^{\pi/4}\dfrac{\sin x+\cos x}{\sqrt{\sin x \cos x}}dx$

$=\sqrt{2}\displaystyle\int_0^{\pi/4}\dfrac{\sin x+\cos x}{\sqrt{2\sin x \cos x}}dx$

$=\sqrt{2}\displaystyle\int_0^{\pi/4}\dfrac{\sin x+\cos x}{\sqrt{1-(\sin x -\cos x)^2}}dx$

Let

$\sin x-\cos x=t.$ Then

$\Rightarrow$

$\cos x+\sin x dx =dt$ [ Differentiating both sides ]

When

$x=0,$

$\sin 0-\cos 0=t\Rightarrow t=1$

When

$x=\dfrac{\pi}{4},$

$\sin\dfrac{\pi}{4}-\cos\dfrac{\pi}{4}=t\Rightarrow$ t=0$$

$\therefore$

$I=\sqrt{2}\displaystyle\int_{-1}^0\dfrac{dt}{\sqrt{1-t^2}}$

$=\sqrt{2}\left[\sin^{-1}t\right]_{-1}^0$

$=\sqrt{2}\left[\sin^{-1}(0)-\sin^{-1}(-1)\right]$

$=\dfrac{\pi}{\sqrt{2}}$

Evaluate $\displaystyle\int_{0}^{2}x\sqrt{x+2} \ dx$

$I=\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$ .

Let

$x+2=t^{2}$ .

Differentiating with respect to

$x$ , we get

$dx=2t dt$

Also

$x=0\Rightarrow t^{2}=2\Rightarrow t=\sqrt{2}$ and,

$x=2\Rightarrow t^{2}=4\Rightarrow t=2$

So, lower limit is

$\sqrt 2$ and upper limit is

$2$

$\therefore I=\displaystyle\int_{\sqrt{2}}^{2}(t^{2}-2)\sqrt{t^{2}} \ \cdot 2t dt=2\displaystyle\int_{\sqrt{2}}^{2}(t^{4}-2t^{2})dt=2\left[\dfrac{t^{5}}{5}-\dfrac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}$

Evaluate the following definite integral.

$\displaystyle \int _{2}^3 \dfrac {x}{x^2+1}dx$

$\displaystyle\int_{2}^{3}{\dfrac{x}{{x}^{2}+1}dx}$

$=\dfrac{1}{2}\displaystyle\int_{2}^{3}{\dfrac{2x}{{x}^{2}+1}dx}$

Let

$t={x}^{2}+1\Rightarrow\,dt=2x\,dx$

When

$x=2\Rightarrow,t=4+1=5$

When

$x=3\Rightarrow,t=9+1=10$

$=\dfrac{1}{2}\displaystyle\int_{5}^{10}{\dfrac{dt}{t}}$

$=\dfrac{1}{2}\left[\log{\left|t\right|}\right]_{5}^{10}$

$=\dfrac{1}{2}\left[\log{10}-\log{5}\right]$

$=\dfrac{1}{2}\log{\dfrac{10}{5}}$

$=\dfrac{1}{2}\log{2}$

Evaluate $\displaystyle\int_{0}^{1}\dfrac{2x}{1+x^{4}}dx$

$\displaystyle\int_{0}^{1}{\dfrac{2x}{1+{x}^{4}}dx}$

$=\displaystyle\int_{0}^{1}{\dfrac{2x}{1+{\left({x}^{2}\right)}^{2}}dx}$

Let

$t={x}^{2}\Rightarrow\,dt=2x\,dx$

$=\displaystyle\int_{0}^{1}{\dfrac{dt}{1+{t}^{2}}}$

$=\left[{\tan}^{-1}{t}\right]_{0}^{1}$

$={\tan}^{-1}{1}-{\tan}^{-1}{0}$

$=\dfrac{\pi}{4}-0$

$=\dfrac{\pi}{4}$

Evaluate the definite integral :
$\displaystyle \int_{0}^{1} \dfrac {2x+3}{5x^2 +1}dx$

$I=\displaystyle \int_0^1 \dfrac {2x+3}{5x^2 +1}dx =\displaystyle \int_0^1 \dfrac {2x}{5x^2 +1}dx +\displaystyle \int_0^1 \dfrac {3}{5x^2 +1}dx =\dfrac {1}{5} \displaystyle \int_0^1 \dfrac {10x}{5x^2 +1}dx +3\displaystyle \int_0^1 \dfrac {1}{(\sqrt {5} x)^2 +1} dx$

$\Rightarrow \ I=\dfrac {1}{5}[\log (5x^2 +1)]_0^1 +\dfrac {3}{\sqrt 5}\left [\tan^{-1} \dfrac {\sqrt 5 x}{1}\right]_0^1$

$\Rightarrow \ I=\dfrac {1}{5}(\log 6-\log 1)+\dfrac {3}{\sqrt 5}\tan^{-1}\sqrt 5 =\dfrac {1}{5}\log 6+\dfrac {3}{\sqrt 5}\tan^{-1}\sqrt 5$

Evaluate the definite integral :
$\displaystyle \int_{0}^{1} (1+x)^5 dx$

$\displaystyle \int_{0}^{1} (1+x)^5 dx$

Let

$t=1+x \implies dx=dt$

$\displaystyle \int _0^1 t^5 dt$

$=\dfrac {t^6}6$

$=\dfrac {(1+x)^6}{6}$

Evaluate the definite integral:
$\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

Let

$I=\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

and

Let

$\sin 2t=u$ .

Then,

$d(\sin 2t)=du\Rightarrow 2\cos 2t\ dt=du\Rightarrow \cos 2t dt=\dfrac{1}{2}du$

Also,

$t=0\Rightarrow u=\sin 0=0$

$t=\dfrac{\pi}{4}\Rightarrow u=\sin\dfrac{\pi}{2}=1$

Replacing we get,

$\therefore I=\dfrac{1}{2}\displaystyle\int_{0}^{1}u^{3}du=\dfrac{1}{8}\left[u^{4}\right]_{0}^{1}=\dfrac{1}{8}$

Evaluate: $\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$ .

Let

$I=\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

and let

$\sin 2t=u$ .

Then,

$du=d(\sin 2t)= 2\cos 2t\ dt$

$\Rightarrow du= \cos 2t \ dt$

$=\dfrac{1}{2}du$

Also,

$t=0\Rightarrow u=\sin 0=0$

$t=\dfrac{\pi}{4}\Rightarrow u=\sin \left (2\times \dfrac{\pi}{4}\right )=\sin\dfrac{\pi}{2}=1$

$\therefore I=\dfrac{1}{2}\displaystyle\int_{0}^{1}u^{3}du=\dfrac{1}{8}\left[u^{4}\right]_{0}^{1}=\dfrac{1}{8}$

Evaluate:
$\displaystyle\int \sec x\tan x dx$

$\displaystyle\int \sec x\tan x dx$

$\displaystyle \int \dfrac{\sin x}{\cos ^2 x} dx$

$t=\cos x \implies -\sin x dx =dt$

Replacing we get,

$\displaystyle \int \dfrac {-1}{t^2} dt$

$=\dfrac 1t$

$=\dfrac 1{\cos x}=\sec x$

Evaluate the following definite integral:
$\displaystyle \int_{1}^{2} e^{2x} \left (\dfrac {1}{x}-\dfrac {1}{2x^2}\right)dx$

1616369_1661513_ans_b62a9da0a1d049d8b34cab9b002b7ee5.jpeg

Evaluate:
$\displaystyle\int_{0}^{\pi}\log (1-\cos x)\ dx$

$I = \int_0^{\pi} \log (1-\cos x) dx$ ...(i)

$I = \int_0^{\pi} \log (1-\cos (\pi -x)) dx$

$I = \int_0^{\pi} \log (1 + \cos x) dx$ ...(ii)

Add (i) & (ii)

$2I = \int_0^{\pi} [\log (1 - \cos x) + \log (1 + \cos x)]dx$

$2I = \int_0^{\pi} \log (1-\cos ^2x)dx$

$2I = \int_0^{\pi} \log \sin^2 x\, dx$

$\Rightarrow 2I = 2\int_0^{\pi} \log \sin x$

$I = \int _0^{\pi} \log \sin x \, dx$

$I = -\pi \log 2$ ............................(as

$\int _0^{\pi} \log \sin x \, dx= -\pi \log 2$ )

$\displaystyle \int \dfrac 1{\sqrt {3-x^2}}dx$

$\because \displaystyle \int \dfrac 1{\sqrt {a^2-x^2}}dx=\sin ^{-1} \left(\dfrac xa\right)$

$\therefore \displaystyle \int \dfrac 1{\sqrt {\left({\sqrt 3}\right)^2-x^2}}=\sin ^{-1}\dfrac x{\sqrt 3}$

Evaluate $\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x+\cos x}{3+\sin 2x}dx$

Let

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{3+\sin 2x}\ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{3+1-1+\sin 2x}\ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{4-(1-\sin 2x)} \ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{4-(\sin ^2x+\cos ^2 x-\sin 2x)} \ dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{4-(\sin x-\cos x)^{2}}\ dx$

Let

$(\sin x-\cos x)=t \Rightarrow (\cos x+\sin x)=\dfrac {dt}{dx}$

$\therefore I_t=\displaystyle\int\dfrac{1}{4-(t)^{2}}\ dt$

$\therefore I=\dfrac{1}{4}\left[\log\left|\dfrac{2+\sin x-\cos x}{2-\sin x+\cos x}\right|\right]_{0}^{\pi/4}=-\dfrac{1}{4}\log\dfrac{1}{3}=\dfrac{1}{4}\log_{e}3$

Evaluate:
$\displaystyle\int_{0}^{\pi/4}\dfrac{\tan^{3}x}{1+\cos 2x}dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\tan^{3}x}{2\cos^{2}x}dx$

$=\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\tan^{3}x\sec^{2}x\ dx$

$t=\tan x$

differentiating with respect to

$x$ , we get

$\dfrac {dt}{dx}=\sec^2{x}\Rightarrow dt=\sec^2{x}\cdot dx$

now,

$\tan \dfrac{\pi}{4}=1$ and

$\tan 0=0$

Replacing in the equation, we get

$\dfrac{1}{2}\displaystyle\int_{0}^{1}t^{3}dt$ , where

$t=\tan x$

$\Rightarrow I=\dfrac{1}{2}\left[\dfrac{t^{4}}{4}\right]_{0}^{1}=\dfrac{1}{2}\left(\dfrac{1}{4}-0\right)=\dfrac{1}{8}$

Evaluate the following integral :
$\displaystyle\int_{0}^{\pi/2}\dfrac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\ dx$

We have,

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\ dx$ ....(i)

As we know

$\displaystyle \int_{a}^b f(x) \ dx= \int_{a}^b f(a+b-x) \ dx$

$\Rightarrow I=\displaystyle\int_{0}^{\pi/2}\dfrac{\left(\dfrac{\pi}{2}-x\right)\cos x\sin x}{\cos^{4}x+\sin^{4}x}dx$ .....(ii)

Adding

$(i)$ and

$(ii)$ , we get

$2I=\dfrac{\pi}{2}\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x\cos x}{\cos^{4}x+\sin^{4}x}dx$

$\Rightarrow 2I=\dfrac{\pi}{4}\displaystyle\int_{0}^{1}\dfrac{1}{(1-t^{2})+t^{2}}dt$ , where

$t=\sin^{2}x\Rightarrow \dfrac{dt}{2}=sinx\,cosx\,dx$

$2I=\dfrac{\pi}{8}\dfrac{\pi}{8}\displaystyle\int_{0}^{1}\dfrac{1}{\left(t-\dfrac{1}{2}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}}dt=\dfrac{\pi}{8}\times 2\left[\tan^{-1} (2t-1)\right]_{0}^{1}$

$\Rightarrow I=\dfrac{\pi}{8}\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)=\dfrac{\pi^{2}}{16}$

Evaluate the definite integral :
$\displaystyle \int_{1}^{2}\dfrac {1}{\sqrt {(x-1) (2-x)}}dx$

Using,

$\displaystyle \int \dfrac 1{\sqrt {a^2-x^2}}=\sin ^{-1}\dfrac xa$

Replacing, we get

$\displaystyle \int _1^2\dfrac 1{\sqrt {\left(\dfrac 12\right)^2-\left(x-\dfrac 32\right)^2}}$

$=\left| \sin ^{-1}\left(\dfrac {x-\dfrac {3}{2}}{\dfrac 12}\right)\right|_1^2$

$\left| \sin^{-1} (2x-3)\right|_1^2=\sin ^{-1}(1)-\sin ^{-1}(-1)$

$=\dfrac {\pi}2+\dfrac {\pi}2$

$=\pi$

Evaluate the integral $\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$ .

Let

$I=\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

and let

$\sin 2t=u$ . Then,

$d(\sin 2t)=du\Rightarrow 2\cos 2t\ dt=du\Rightarrow \cos 2t dt=\dfrac{1}{2}du$

Also,

$t=0\Rightarrow u=\sin 0=0$

and

$t=\dfrac{\pi}{4}\Rightarrow u=\sin\dfrac{\pi}{2}=1$

$\therefore I=\dfrac{1}{2}\displaystyle\int_{0}^{1}u^{3}du$

$=\dfrac{1}{8}\left[u^{4}\right]_{0}^{1}$

$=\dfrac{1}{8}$

$\displaystyle\int_{0}^{\pi/4}\sec ^2 x \ dx$

$=\left[\tan x\right]_0^{\pi/4}$

$=1-0$

$=1$

Evaluate $\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$

Consider,

$I=\displaystyle\int_{0}^{1}\dfrac{\tan^{-1}x}{1+x^{2}}dx$

Let,

$\tan^{-1}x=t$ .

Then,

$d(\tan^{-1}x)=dt\Rightarrow dx=(1+x^{2})dt$

Also,

$x=0\Rightarrow t=\tan^{-1}0=0$ and

$x=1\Rightarrow t=\tan^{-1}1=\dfrac{\pi}{4}$

$\therefore I=\displaystyle\int_{0}^{\pi/4}t\ dt$

$I=\left[\dfrac{t^{2}}{2}\right]_{0}^{\pi/4}$

$I=\dfrac{\pi^{2}}{32}$

Evaluate the following definite integral:
$\displaystyle \int_{-1}^{1}\dfrac {1}{x^2 +2x+5}dx$

Consider,

$I=\displaystyle \int_{-1}^1\dfrac {1}{x^2+2x+5}dx$

$I =\displaystyle \int_{-1}^1\dfrac {1}{(x+1)^2 +2^2}dx$

$I=\dfrac {1}{2}\left [\tan^{-1} \dfrac {x+1}{2}\right]_{1-}^1$

$I=\dfrac {1}{2}(\tan^{-1} 1-\tan^{-1}0)$

$I=\dfrac {\pi}{8}$

$\displaystyle\int_{0}^{1}\dfrac{1-x^{2}}{(1+x^{2})^{2}}dx$

Consider,

$I=\displaystyle\int_{0}^{1}\dfrac{{\dfrac{1}{x^2}-1}}{\left(x+\dfrac{1}{x}\right)^{2}}\ dx$

Let

$x+\dfrac{1}{x}=t$ . Then,

$d\left(x+\dfrac{1}{x}\right)=dt\Rightarrow \left(1-\dfrac{1}{x^{2}}\right)dx=dt$

Clearly,

$x=0\Rightarrow t=\infty$ and

$x=1\Rightarrow t=2$

$\therefore I =\displaystyle\int_{\infty}^{2}-\dfrac{1}{t^{2}}dt=\left[\dfrac{1}{t}\right]_{\infty}^{2}=\dfrac{1}{2}$

Evaluate the following definite integrals :
$\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$I=\displaystyle \int _{0}^{\pi /2} \cos^2 x\ dx$

$=\displaystyle \int _{0}^{\pi /2} \dfrac {1+\cos 2x}{2} dx$ ................

$\because \left (\cos 2x=2 \cos^2 x-1\right )$

$=\dfrac {1}{2} \left [x +\dfrac {\sin 2x}{2} \right]_0^{\pi /2}$

$=\dfrac {1}{2}\left [\left (\dfrac {\pi}{2}+\dfrac {\sin \pi}{2} \right) -\left (0+\dfrac {\sin 0}{2}\right) \right]$

$=\dfrac {\pi}{4}$

Evaluate the integral $\displaystyle\int_{-1}^{1}5x^{4}\sqrt{x^{5}+1}\ dx$ .

Consider,

$I=\displaystyle\int_{-1}^{1}5x^{4}\sqrt{x^{5}+1}\ dx$

Let

$x^{5}+1=t$ .

Then,

$d(x^{5}+1)=dt\Rightarrow 5x^{4}dx=dt$

Also,

$x=-1\Rightarrow t=0$ and

$x=1\Rightarrow t=2$

$\therefore I =\displaystyle\int_{0}^{2}\sqrt{t}dt$

$I=\dfrac{2}{3}\left[t^{3/2}\right]_{0}^{2}$

$I=\dfrac{2}{3}\times 2^{3/2}$

$I=\dfrac{4\sqrt{2}}{3}$

Evaluate the following definite integral:
$\displaystyle \int_{e}^{e^2} \left\{\dfrac {1}{\log x} -\dfrac {1}{(\log x)^2}\right\} dx$

$I=\displaystyle \int_{e}^{e^2} \left\{\dfrac {1}{\log x} -\dfrac {1}{(\log x)^2}\right\} dx$

$I=\displaystyle\int_{e}^{e^2}1\dfrac{1}{\log x}dx-\displaystyle\int_{e}^{e^2}\dfrac{1}{(\log x)^2}dx$

Integrating by parts

$\Rightarrow$

$I=\left\{\left[\dfrac{x}{\log x}\right]_e^{e^2}-\displaystyle\int_e^{e^2}\dfrac{-1}{x(\log x)^2}x dx\right\}-\displaystyle\int_e^{e^2}\dfrac{1}{(\log x)^2}dx$

$\Rightarrow$

$I=\left[\dfrac{x}{\log x}\right]_e^{e^2}+\displaystyle\int_e^{e^2}\dfrac{1}{(\log x)^2} dx-\displaystyle\int_e^{e^2}\dfrac{1}{(\log x)^2}dx$

$\Rightarrow$

$I=\left[\dfrac{x}{\log x}\right]_e^{e^2}+0$

$\Rightarrow$

$I=\dfrac{e^2}{\log e^2}-\dfrac{e}{\log e}$

$\Rightarrow$

$I=\dfrac{e^2}{2\log e}-\dfrac{e}{\log e}$

Since,

$\log e=1$

$\Rightarrow$

$I=\dfrac{e^2}{2}-e$

Evaluate the following integral:
$\displaystyle\int { \cfrac { \cos { x } }{ \cos { 3x } } } dx$

1617665_1664163_ans_c5cfb1ee26114049add48c675b54b440.jpeg

Evaluate the following definite integral:

$\displaystyle\int_{4}^{9}\dfrac{\sqrt{x}}{(30-x^{3/2})^{2}}\ dx$

$I=\displaystyle\int_{4}^{9}\dfrac{\sqrt{x}}{(30-x^{3/2})^{2}}dx$ .

Let

$30-x^{3/2}=t$ .

Then,

$-\dfrac{3}{2}\sqrt{x}dx=dt \Rightarrow \sqrt{x}dx=-\dfrac{2}{3}dt$ .

Now

$x=4\Rightarrow t=22$ and

$x=9\Rightarrow t=3$

$\therefore I=\dfrac{-2}{3}\displaystyle\int_{22}^{3} \dfrac{1}{t}dt=\dfrac{2}{3}\left[\dfrac{1}{t}\right]_{22}^{3}$

$=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{22}\right)=\dfrac{19}{99}$

Evaluate the following integral:
$\displaystyle \int { \cfrac { 1 }{ 2{ x }^{ 2 }-x-1 } } dx\quad$

$\displaystyle\int{\dfrac{dx}{2{x}^{2}-x-1}}$

$=\displaystyle\int{\dfrac{dx}{2\left({x}^{2}-\dfrac{1}{2}x-\dfrac{1}{2}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({x}^{2}-\dfrac{1}{2}x-\dfrac{1}{2}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({x}^{2}-2\times\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{2}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({\left(x-\dfrac{1}{4}\right)}^{2}-\dfrac{1}{16}-\dfrac{8}{16}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({\left(x-\dfrac{1}{4}\right)}^{2}-\dfrac{9}{16}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({\left(x-\dfrac{1}{4}\right)}^{2}-{\left(\dfrac{3}{4}\right)}^{2}\right)}}$

We know that

$\displaystyle\int{\dfrac{dx}{{x}^{2}-{a}^{2}}}=\dfrac{1}{2a}\log{\left|\dfrac{x-a}{x+a}\right|}+c$

Replace

$x\rightarrow\,x-\dfrac{1}{4}$ and

$a\rightarrow\,\dfrac{3}{4}$ we get

$=\dfrac{1}{2}\times\dfrac{1}{2\times\dfrac{3}{4}}\log{\left|\dfrac{x-\dfrac{1}{4}-\dfrac{3}{4}}{x-\dfrac{1}{4}+\dfrac{3}{4}}\right|}+c$

$=\dfrac{1}{2}\times\dfrac{2}{3}\log{\left|\dfrac{x-\dfrac{4}{4}}{x+\dfrac{2}{4}}\right|}+c$

$=\dfrac{1}{2}\times\dfrac{2}{3}\log{\left|\dfrac{x-1}{x+\dfrac{1}{2}}\right|}+c$

$=\dfrac{1}{3}\log{\left|\dfrac{2x-2}{2x+1}\right|}+c$

Evaluate $\displaystyle \int_{0}^{3}(2x^2+3x +5)dx$

$\displaystyle \int_{0}^{3}(2x^2+3x +5)dx$

$\displaystyle \int_0^3 2x^2 +\int_0^3 3x dx+\int_0^3 5 dx$

$=\dfrac {2x^3}3\bigg|_0^3 +\dfrac{3x^2}2\bigg|_0^3 +5x\bigg|_0^3$

$=2\dfrac{3^3}{3}-0+3\dfrac{3^2}{2}-0+5(3)-0$

$=18+\dfrac{27}2+15 =\dfrac{93}2$

Solve :
$\displaystyle \int { \left( x+2 \right) \sqrt { 3x+5 } } dx$

Let

$I=\displaystyle\int{\left(x+2\right)\sqrt{3x+5}\,dx}$

Let

$x+2=\lambda\left(3x+5\right)+\mu$

On equating the coefficients of like powers of

$x$ on both sides, we get

$\Rightarrow\,3\lambda=1\Rightarrow\,\lambda=\dfrac{1}{3}$

$\Rightarrow\,5\lambda+\mu=2\Rightarrow\,5\times\dfrac{1}{3}+\mu=2$

$\Rightarrow\,\mu=2-\dfrac{5}{3}=\dfrac{6-5}{3}=\dfrac{1}{3}$

$\therefore\,I=\displaystyle\int{\left(x+2\right)\sqrt{3x+5}\,dx}$

$=\displaystyle\int{\left\{\lambda\left(3x+5\right)+\mu\right\}\sqrt{3x+5}\,dx}$

$=\displaystyle\int{\left\{\dfrac{1}{3}{\left(3x+5\right)}^{1+\frac{1}{2}}+\dfrac{1}{3}\sqrt{3x+5}\right\}\,dx}$

$=\displaystyle\int{\left\{\dfrac{1}{3}{\left(3x+5\right)}^{\frac{3}{2}}+\dfrac{1}{3}{\left(3x+5\right)}^{\frac{1}{2}}\right\}\,dx}$

We know that

$\displaystyle\int{{\left(ax+b\right)}^{n}}=\dfrac{1}{a\left(n+1\right)}{\left(ax+b\right)}^{n+1}$

$=\dfrac{1}{3}\times\dfrac{1}{3}\dfrac{{\left(3x+5\right)}^{\frac{3}{2}+1}}{\dfrac{3}{2}+1}+\dfrac{1}{3}\times\dfrac{1}{3}\dfrac{{\left(3x+5\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+c$

$=\dfrac{1}{9}\dfrac{{\left(3x+5\right)}^{\frac{5}{2}}}{\dfrac{5}{2}}+\dfrac{1}{9}\dfrac{{\left(3x+5\right)}^{\frac{3}{2}}}{\dfrac{3}{2}}+c$

$=\dfrac{2}{45}{\left(3x+5\right)}^{\frac{5}{2}}+\dfrac{2}{27}{\left(3x+5\right)}^{\frac{3}{2}}+c$

Evaluate the following integral:
$\displaystyle \int { \cfrac { 1 }{ \left( \sin { x } -2\cos { x } \right) \left( 2\sin { x } +\cos { x } \right) } } dx$

1617768_1664171_ans_fc238d5b297f4563be12b83a3fd8b36e.jpeg

Evaluate the following integrals:
$\displaystyle\int { \cfrac { 1 }{ 4\sin ^{ 2 }{ x } +5\cos ^{ 2 }{ x } } } dx$

1617560_1664159_ans_bbea829cb4e342cb9895284fdc600443.jpeg

Evaluate the following integral:
$\displaystyle\int { \cfrac { 1 }{ 3+2\cos ^{ 2 }{ x } } } dx\quad$

1617708_1664169_ans_bd11870e73d84f2aa811ac73a2395997.jpeg

Evaluate the following integrals:
$\displaystyle \int { \cfrac { 2 }{ 2+\sin { 2x } } } dx\quad$

1617624_1664162_ans_e08f42580e064f0a93a6ece3cfe09bf6.jpeg

Evaluate the following definite integral:

$\displaystyle \int_{e}^{e^2} \left\{\dfrac {1}{\log x} -\dfrac {1}{(\log x)^2}\right\} dx$

$\displaystyle I = \int_{e}^{{e}^{2}}{\left( \cfrac{1}{\log{x}} - \cfrac{1}{{\left( \log{x} \right)}^{2}} \right) dx}$

Let

$\log{x} = t$

$\Rightarrow x = {e}^{t}$

$\Rightarrow dx = {e}^{t} dt$

Therefore,

$\displaystyle I = \int_{1}^{2}{\left( \cfrac{1}{t} - \cfrac{1}{{t}^{2}} \right) {e}^{t} dt}$

As we know that,

$\displaystyle \int{{e}^{x} \left( f{\left( x \right)} - f^{\prime}{\left( x \right)} \right) dx} = {e}^{x} f{\left( x \right)}$

Therefore,

$\displaystyle I = \int_{1}^{2}{\left( \cfrac{1}{t} - \cfrac{1}{{t}^{2}} \right) {e}^{t} dt} = \left[ \cfrac{{e}^{t}}{t} \right]_{1}^{2} = \left( \cfrac{{e}^{2}}{2} - e \right)$

Evaluate the following integrals:
$\int { \cfrac { 1 }{ 5-4\sin { x } } } dx\quad$

Let

$\text{I}=\displaystyle\int \dfrac{1}{5-4\sin x} d x$

$=\displaystyle\int \dfrac{1}{5-4\bigg(\dfrac{2\tan (x/2)}{1+\tan^2 (x/2)}\bigg)}d x$

$\bigg(\because \sin 2 A=\dfrac{2\tan A}{1+\tan^2 A}\bigg)$

$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{5\bigg(1+\tan^2 \dfrac{x}{2}\bigg)-8\tan \dfrac{x}{2}}d x$

$=\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{5-8\tan \dfrac{x}{2}+5\tan^2 \dfrac{x}{2}}d x$

$(\because \text{sec}^2 A-\tan^2 A=1)$

Let

$t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$

$I=\displaystyle\int \dfrac{2 d t}{5 t^2-8 t+5}$

$=\dfrac{2}{5}\displaystyle\int \dfrac{d t}{t^2-2(t)\bigg(\dfrac{4}{5}\bigg)+1}$

$=\dfrac{2}{5}\displaystyle\int \dfrac{d t}{t^2-2(t)\bigg(\dfrac{4}{5}\bigg)+\bigg(\dfrac{4}{5}\bigg)^2+1-\bigg(\dfrac{4}{5}\bigg)^2}$

$=\dfrac{2}{5}\displaystyle\int \dfrac{d t}{\bigg(t-\dfrac{4}{5}\bigg)^2+\dfrac{9}{25}}$

$(\because (a-b)^2=a^2-2 a b+b^2\ )$

$=\dfrac{2}{5}\displaystyle\int \dfrac{d t}{\bigg(t-\dfrac{4}{5}\bigg)^2+\bigg(\dfrac{3}{5}\bigg)^2}$

$=\dfrac{2}{5}\times \dfrac{1}{\bigg(\dfrac{3}{5}\bigg)}\text{tan}^{-1} \dfrac{t-\dfrac{4}{5}}{\dfrac{3}{5}}+c$

$\bigg(\because \displaystyle\int \dfrac{d x}{x^2+a^2}=\dfrac{1}{a}\text{tan}^{-1} \dfrac{x}{a}+c\ \bigg)$

$=\dfrac{2}{3}\text{tan}^{-1} \bigg(\dfrac{1}{3}\bigg(5\tan \dfrac{x}{2}-4\bigg)\ \bigg)+c$

$\bigg(\because t=\tan \dfrac{x}{2}\bigg)$

$=\dfrac{2}{3}\text{tan}^{-1} \bigg(\dfrac{5}{3}\tan \dfrac{x}{2}-\dfrac{4}{3}\bigg)+c$

Evaluate the following integral:
$\displaystyle \int { \cfrac { 1 }{ \sin ^{ 2 }{ x } +\sin { 2x } } } dx\quad$

1618630_1664180_ans_0a5cbb6709504e5ab062bc68f9c9bcbb.jpeg

Evaluate the following integrals:
$\int { \cfrac { 1 }{ 1-2\sin { x } } } dx\quad$

Let

$\text{I}=\displaystyle\int \dfrac{1}{1-2\sin x} d x$

$=\displaystyle\int \dfrac{1}{1-2\bigg(\dfrac{2\tan (x/2)}{1+\tan^2 (x/2)}\bigg)}d x$

$\bigg(\because \sin 2 A=\dfrac{2\tan A}{1+\tan^2 A}\bigg)$

$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{\bigg(1+\tan^2 \dfrac{x}{2}\bigg)-4\tan \dfrac{x}{2}}d x$

$=\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{1-4\tan \dfrac{x}{2}+\tan^2 \dfrac{x}{2}}d x$

$(\because \text{sec}^2 A-\tan^2 A=1)$

Let

$t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$

$I=\displaystyle\int \dfrac{2 d t}{ t^2-4 t+1}$

$=2\displaystyle\int \dfrac{d t}{t^2-2(t)(2)+1}$

$=2\displaystyle\int \dfrac{d t}{t^2-2(t)(2)+(2)^2+1-(2)^2}$

$=2\displaystyle\int \dfrac{d t}{(t-2)^2-3}$

$(\because (a-b)^2=a^2-2 a b+b^2\ )$

$=2\displaystyle\int \dfrac{d t}{(t-2)^2-(\sqrt{3})^2}$

$=2\times \dfrac{1}{2\sqrt{3}}\log \bigg| \dfrac{(t-2)-\sqrt{3}}{(t-2)+(\sqrt{3})}\bigg|+c$

$\bigg(\because \displaystyle\int \dfrac{d x}{x^2-a^2}=\dfrac{1}{2 a}\log \bigg|\dfrac{x-a}{x+a}\bigg|+c\ \bigg)$

$=\dfrac{1}{\sqrt{3}}\log \bigg|\dfrac{\tan \dfrac{x}{2}-(2+\sqrt{3})}{\tan \dfrac{x}{2}-2+\sqrt{3}}\bigg|+c$

$\bigg(\because t=\tan \dfrac{x}{2}\bigg)$

Evaluate the following integral:
$\displaystyle\int { \cfrac { 1 }{ \cos { 2x } +3\sin ^{ 2 }{ x } } } dx$

1618637_1664182_ans_18c0bc545a724ab2b3d4e9ba899101a1.jpeg

Evaluate $\displaystyle \int_{1}^{3}(2x^2+5x)dx$

$\displaystyle \int_{1}^{3}(2x^2+5x)dx$

$=\dfrac{2x^3}3+\dfrac{5x^2}2\bigg|_1^3$

$=\left( \dfrac{2\ . 3^3}3+\dfrac{5\ . 3^2}2\right )-\left( \dfrac{2\ . 1^3}3+\dfrac{5\ . 1^2}2\right )$

$=18+\dfrac{45}2-\dfrac 23-\dfrac 52=38-\dfrac23=37\dfrac 13$

Evaluate the following integrals
$\int { \cfrac { 1 }{ \sin { x } -\sqrt { 3 } \cos { x } } } dx\quad$

1657150_1664625_ans_7e55bbfacce04892835dea94e7c4c79d.jpg

Evaluate the following integrals
$\int { \cfrac { 1 }{ 5+7\cos { x } +\sin { x } } } dx$

1657175_1664626_ans_80c213ddbdb842458e3f581a606e1a30.png

Evaluate the following integrals
$\int { \cfrac { 1 }{ 1-\cot { x } } } dx\quad$

1657120_1664627_ans_11ad955e9c2e45629cabc27fe116fdc0.jpg

Evaluate the following integrals
$\int { \cfrac { 1 }{ \sin { x } +\cos { x } } } dx$

Let

$\text{I}=\displaystyle\int \dfrac{1}{\sin x+\cos x} d x$

$=\displaystyle\int \dfrac{1}{\bigg(\dfrac{2\tan (x/2)}{1+\tan^2 (x/2)}\bigg)+\bigg(\dfrac{1-\tan^2 x/2}{1+\tan^2 x/2}\bigg)}d x$

$\bigg(\because \sin 2 A=\dfrac{2\tan A}{1+\tan^2 A},\cos 2 A=\dfrac{1-\tan^2 A}{1+\tan^2 A}\bigg)$

$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{\bigg(1-\tan^2 \dfrac{x}{2}\bigg)+2\tan \dfrac{x}{2}}d x$

$=-\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{-1-2\tan \dfrac{x}{2}+\tan^2 \dfrac{x}{2}}d x$

$(\because \text{sec}^2 A-\tan^2 A=1)$

Let

$t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$

$I=-\displaystyle\int \dfrac{2 d t}{ t^2-2 t-1}$

$=-2\displaystyle\int \dfrac{d t}{t^2-2(t)(1)-1}$

$=-2\displaystyle\int \dfrac{d t}{t^2-2(t)(1)+(1)^2-1-(1)^2}$

$=-2\displaystyle\int \dfrac{d t}{(t-1)^2-2}$

$(\because (a-b)^2=a^2-2 a b+b^2\ )$

$=-2\displaystyle\int \dfrac{d t}{(t-1)^2-(\sqrt{2})^2}$

$=-2\times \dfrac{1}{2\sqrt{2}}\log \bigg| \dfrac{(t-1)-\sqrt{2}}{(t-1)+(\sqrt{2})}\bigg|+c$

$\bigg(\because \displaystyle\int \dfrac{d x}{x^2-a^2}=\dfrac{1}{2 a}\log \bigg|\dfrac{x-a}{x+a}\bigg|+c\ \bigg)$

$=-\dfrac{1}{\sqrt{2}}\log \bigg|\dfrac{\tan \dfrac{x}{2}-1-\sqrt{2})}{\tan \dfrac{x}{2}-1+\sqrt{2}}\bigg|+c$

$\bigg(\because t=\tan \dfrac{x}{2}\bigg)$

Evaluate the following integrals:
$\int { \cfrac { 1 }{ 4\cos { x } -1 } } dx\quad$

Let

$\text{I}=\displaystyle\int \dfrac{1}{4\cos x-1} d x$

$=\displaystyle\int \dfrac{1}{4\bigg(\dfrac{1-\tan^2 (x/2)}{1+\tan^2 (x/2)}\bigg)-1}d x$

$\bigg(\because \cos 2 A=\dfrac{1-\tan^2A}{1+\tan^2 A}\bigg)$

$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{-\bigg(1+\tan^2 \dfrac{x}{2}\bigg)+4\bigg(1-\tan^2 \dfrac{x}{2}\bigg)}d x$

$=\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{3-5\tan^2 \dfrac{x}{2}}d x$

$(\because \text{sec}^2 A-\tan^2 A=1)$

Let

$t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$

$I=\displaystyle\int \dfrac{2 d t}{3-5 t^2}$

$=-\dfrac{2}{5}\displaystyle\int \dfrac{d t}{t^2-\bigg(\sqrt{\dfrac{3}{5}}\bigg)^2}$

$=-\dfrac{2}{5}\times \dfrac{1}{2\sqrt{\dfrac{3}{5}}}\log \bigg| \dfrac{t-\sqrt{\dfrac{3}{5}}}{t+\sqrt{\dfrac{3}{5}}}\bigg|+c$

$\bigg(\because \displaystyle\int \dfrac{d x}{x^2-a^2}=\dfrac{1}{2 a}\log \bigg|\dfrac{x-a}{x+a}\bigg|+c\ \bigg)$

$=-\dfrac{\sqrt{5}}{5\sqrt{3}}\log \bigg| \dfrac{\sqrt{5} t-\sqrt{3}}{\sqrt{5} t+\sqrt{3}}\bigg|+c$

$=-\dfrac{1}{\sqrt{15}}\log \bigg|\dfrac{\sqrt{5}\tan \dfrac{x}{2}-\sqrt{3}}{\sqrt{5}\tan \dfrac{x}{2}+\sqrt{3}}\bigg|+c$

$\bigg(\because t=\tan \dfrac{x}{2}\bigg)$

Evaluate the given integral.
$\displaystyle \int { \cfrac { \log { \left( \log { x } \right) } }{ x } } dx$

$I=\displaystyle\int{\dfrac{\log{\left(\log{x}\right)}}{x}dx}$

Let

$t=\log{x}\Rightarrow\,dt=\dfrac{1}{x}dx$

$\Rightarrow\,I=\displaystyle\int{\log{t}dt}$

Integrating by parts, we get

Let

$u=\log{t}\Rightarrow\,du=\dfrac{1}{t}dt$ and

$dv=dt\Rightarrow\,v=t$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=t\log{t}-\displaystyle\int{t\times\dfrac{dt}{t}}$

$\Rightarrow\,I=t\log{t}-\displaystyle\int{dt}$

$\Rightarrow\,I=t\log{t}-t+c$

$\Rightarrow\,I=\log{x}\log{t}-\log{x}+c$ where

$t=\log{x}$

$\Rightarrow\,I=\log{x}\log{\left(\log{x}\right)}-\log{x}+c$ where

$t=\log{x}$

$\therefore\,I=\log{x}\left[\log{\left(\log{x}\right)}-1\right]+c$

Evaluate the following integrals
$\int { \cfrac { 2\sin { x } +3\cos { x } }{ 3\sin { x } +4\cos { x } } } dx\quad$

1687934_1664634_ans_a48a0fc7bec844aab1897be0437ec292.jpg

Evaluate the following integrals
$\int { \cfrac { 1 }{ 3+4\cot { x } } } dx\quad$

1688144_1664635_ans_355f7428b43a4125b4e72e08a2d20099.jpg

Evaluate the following integral:

$\displaystyle \int { \cfrac { { x }^{ 2 } }{ { x }^{ 4 }-{ x }^{ 2 }-12 } } dx$

$\displaystyle\int{\dfrac{{x}^{2}}{{x}^{4}-{x}^{2}-12}dx}$

$=\displaystyle\int{\dfrac{{x}^{2}}{{x}^{4}-4{x}^{2}+3{x}^{2}-12}dx}$

$=\displaystyle\int{\dfrac{{x}^{2}}{{x}^{2}\left({x}^{2}-4\right)+3\left({x}^{2}-4\right)}dx}$

$=\displaystyle\int{\dfrac{{x}^{2}}{\left({x}^{2}-4\right)\left({x}^{2}+3\right)}dx}$

$=12\displaystyle\int{\dfrac{{x}^{2}}{\left(3{x}^{2}-12\right)\left(4{x}^{2}+12\right)}dx}$

$=\dfrac{12}{7}\displaystyle\int{\dfrac{\left[\left(4{x}^{2}+12\right)+\left(3{x}^{2}-12\right)\right]}{\left(3{x}^{2}-12\right)\left(4{x}^{2}+12\right)}dx}$

$=\dfrac{12}{7}\displaystyle\int{\dfrac{\left(4{x}^{2}+12\right)dx}{\left(3{x}^{2}-12\right)\left(4{x}^{2}+12\right)}}+\dfrac{12}{7}\displaystyle\int{\dfrac{\left(3{x}^{2}-12\right)dx}{\left(3{x}^{2}-12\right)\left(4{x}^{2}+12\right)}}$

$=\dfrac{12}{7}\displaystyle\int{\dfrac{dx}{\left(3{x}^{2}-12\right)}}+\dfrac{12}{7}\displaystyle\int{\dfrac{dx}{\left(4{x}^{2}+12\right)}}$

$=\dfrac{12}{7}\times\dfrac{1}{3}\displaystyle\int{\dfrac{dx}{\left({x}^{2}-4\right)}}+\dfrac{12}{7}\times\dfrac{1}{4}\displaystyle\int{\dfrac{dx}{\left({x}^{2}+3\right)}}$

$=\dfrac{4}{7}\displaystyle\int{\dfrac{dx}{\left({x}^{2}-4\right)}}+\dfrac{3}{7}\displaystyle\int{\dfrac{dx}{\left({x}^{2}+3\right)}}$

We know that

$\displaystyle\int{\dfrac{dx}{\left({x}^{2}-{a}^{2}\right)}}=\dfrac{1}{2a}\log{\left|\dfrac{x-a}{x+a}\right|}$ and

$\displaystyle\int{\dfrac{dx}{\left({x}^{2}+{a}^{2}\right)}}=\dfrac{1}{a}{\tan}^{-1}{\dfrac{x}{a}}$

$=\dfrac{4}{7}\times\dfrac{1}{2\times 2}\log{\left|\dfrac{x-2}{x+2}\right|}+\dfrac{3}{7}\times\dfrac{1}{\sqrt{3}}{\tan}^{-1}{\dfrac{x}{\sqrt{3}}}+c$

$=\dfrac{4}{7}\times\dfrac{1}{4}\log{\left|\dfrac{x-2}{x+2}\right|}+\dfrac{3}{7}\times\dfrac{1}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}{\tan}^{-1}{\dfrac{x}{\sqrt{3}}}+c$

$=\dfrac{1}{7}\log{\left|\dfrac{x-2}{x+2}\right|}+\dfrac{3\sqrt{3}}{3\times 7}{\tan}^{-1}{\dfrac{x}{\sqrt{3}}}+c$

$=\dfrac{1}{7}\log{\left|\dfrac{x-2}{x+2}\right|}+\dfrac{\sqrt{3}}{7}{\tan}^{-1}{\dfrac{x}{\sqrt{3}}}+c$

Evaluate the given integral.
$\int { \log { (x+1) } } dx$

$I=\displaystyle\int{\log{\left(x+1\right)}dx}$

Integrating by parts, we get

Let

$u=\log{\left(x+1\right)}\Rightarrow\,du=\dfrac{1}{x+1}dx$

and

$dv=dx\Rightarrow\,v=x$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=x\log{\left(x+1\right)}-\displaystyle\int{\dfrac{x\,dx}{x+1}}$

$\Rightarrow\,I=x\log{\left(x+1\right)}-\displaystyle\int{\dfrac{\left(x+1-1\right)dx}{x+1}}$

$\Rightarrow\,I=x\log{\left(x+1\right)}-\displaystyle\int{dx}+\displaystyle\int{\dfrac{dx}{x+1}}$

$\Rightarrow\,I=x\log{\left(x+1\right)}-x+\log{\left(x+1\right)}+c$

$\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$ .

Let

$I=\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

Let

$\sin 2t=u$ .

Then,

$d(\sin 2t)=du$

$\Rightarrow 2\cos 2t\ dt=du$

$\Rightarrow \cos 2t dt=\dfrac{1}{2}du$

Also,

$t=0\Rightarrow u=\sin 0=0$

$t=\dfrac{\pi}{4}$

$\Rightarrow u=\sin\dfrac{\pi}{2}=1$

$\therefore I=\dfrac{1}{2}\displaystyle\int_{0}^{1}u^{3}du$

$\dfrac{1}{8}\left[u^{4}\right]_{0}^{1}=\dfrac{1}{8}$

Evaluate the following integrals
$\int { \cfrac { 1 }{ p+q\tan { x } } } dx\quad$

1657138_1664631_ans_08119546279a46fcabe4864bdfcbf6ab.jpg

Evaluate the given integral.
$\int { { x }^{ 3 }\log { x } } dx$

$I=\displaystyle\int{{x}^{3}\log{x}dx}$

Integrating by parts, we get

Let

$dv={x}^{3}dx\Rightarrow\,v=\dfrac{{x}^{4}}{4}$

$u=\log{x}\Rightarrow\,du=\dfrac{1}{x}dx$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=\dfrac{{x}^{4}\log{x}}{4}-\displaystyle\int{\dfrac{{x}^{4}}{4}\times\dfrac{1}{x}dx}$

$\Rightarrow\,I=\dfrac{{x}^{4}\log{x}}{4}-\dfrac{1}{4}\displaystyle\int{{x}^{3}dx}$

$\Rightarrow\,I=\dfrac{{x}^{4}\log{x}}{4}-\dfrac{1}{4}\dfrac{{x}^{4}}{4}+c$

$\therefore\,I=\dfrac{{x}^{4}\log{x}}{4}-\dfrac{{x}^{4}}{16}+c$

Evaluate the following integrals
$\int { \cfrac { 1 }{ 1-\tan { x } } } dx\quad$

1657130_1664628_ans_a2305661d9f94dcd93d5e1b5c8bf5b18.jpg

Evaluate: $\displaystyle\int \dfrac{-\sin x }{5+\cos x}dx$

$\Rightarrow\displaystyle\int \dfrac{-\sin x }{5+\cos x}dx$

Let

$t=5+\cos x \implies dt=-\sin x dx\\$

$\Rightarrow \displaystyle \int \dfrac 1t dt$

$\Rightarrow\log t$

$\Rightarrow\log(5+ \cos x)$

Evaluate: $\displaystyle\int \sec x\tan x dx$

$I=\displaystyle\int \sec x\tan x dx$

$I=\displaystyle \int \dfrac{\sin x}{\cos ^2 x} dx$

$t=\cos x \implies -\sin x dx =dt$

$I=\displaystyle \int \dfrac {-1}{t^2} dt$

$I=\dfrac 1t$ .............

$\left [\because \int x^n=\dfrac{x^{n+1}}{n+1}\right ]$

$I=\dfrac 1{\cos x}$

$I=\sec x$

Evaluate: $\displaystyle \int \cot x dx$ .

$I=\displaystyle \int \dfrac {\cos x}{\sin x} dx$

Let

$t=\sin x \implies dt=\cos x dx$

$I=\displaystyle \int \dfrac 1t dt$

$I=\log t$

$I=\log \sin x$

Solve : $\displaystyle\int_{0}^{3}|3x-1|\ dx$

Let

$I=\displaystyle \int_{0}^3 |3x-1|dx$

we know

$|2x-1|=\begin{cases} 1-3x & x<1/3 \\ 3x-1 & x\ge 1/3 \end{cases}$

$I=\displaystyle \int_{0}^{1/3} 1-3x\ dx +\displaystyle \int_{1/3}^{3} 3x-1\ dx$

$=x -\dfrac {3x^2}{2}|_0^{1/3}+\dfrac {3x^2}{2}-x|_{1/3}^3$

$=\dfrac {1}{3}-\dfrac {1}{6}+\left (\dfrac {27}{2}-2\right)-\left (\dfrac {1}{6}-\dfrac {1}{3}\right)$

$=\dfrac {2}{3}-\dfrac {1}{3}+\dfrac {27}{3}-3=\dfrac {1}{3}+\dfrac {21}{2}$

$=\dfrac {65}{6}$

Evaluate the following definite integral:
$\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{3/2} x}{\sin^{3/2}x+\cos^{3/2} x}\ dx$

Let

$I=\displaystyle\int_{\pi/2}^{0}\dfrac{\sin^{3/2}}{\sin^{3/2} x+\cos^{3/2}x}\ dx$

Then,

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{3/2(\pi/2-x)}}{\sin^{3/2}(\pi/2-x)+\cos^{3/2}(\pi/2-x)}\ dx$

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{3/2}}{\cos^{3/2}x+\sin^{3/2}x}\ dx$

Adding

$(1)$ and

$(2)$ , we get

$2I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{3/2}+\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\ dx$

$=\displaystyle\int_{0}^{\pi/2}1.dx$

$=[x]_{0}^{\pi/2}$

$=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

Evaluate the following definite integral:
$\displaystyle\int_{-1}^{1}5x^{4}\sqrt{x^{5}+1}\ dx$

Let

$I=\displaystyle\int_{-1}^{1}5x^{4}\sqrt{x^{5}+1}\ dx$

let

$x^{5}+1=t$ .

Then,

$d(x^{5}+1)=dt$

$\Rightarrow 5x^{4}dx=dt$

Also,

$x=-1\Rightarrow t=0$

$x=1\Rightarrow t=2$

$\therefore I=\displaystyle\int_{0}^{2}\sqrt{t}dt$

$I=\dfrac{2}{3}\left[t^{3/2}\right]_{0}^{2}$

$I=\dfrac{2}{3}\times 2^{3/2}$

$I=\dfrac{4\sqrt{2}}{3}$

Evaluate the following integrals :
$\displaystyle\int_{0}^{\pi} x\ dx$

$\displaystyle\int_{0}^{\pi} x\ dx$

Using

$\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$ , we get

$\Rightarrow \left[\dfrac {x^2}2\right]_0^\pi$

$\Rightarrow \dfrac {\pi^2}2$

Solve : $\displaystyle\int_{-2}^{2}|2x+3|\ dx$

Let

$I=\displaystyle \int_{-2}^{2}|2x+2|dx$

we know

$|2x+3|=\begin{cases} -\left( 3x+3 \right) & x<\dfrac { -3 }{ 2 } \\ 2x+3 & x\ge \dfrac { 3 }{ 2 } \end{cases}$

then

$I=I=\displaystyle \int_{-2}^{-3/2} -2x-3\ dx +I=\displaystyle \int_{-3/2}^{2} 2x+3\ dx$

$=-x^2 -3x|_{-2}^{-3/2} +x^2 +3x|_{-3/2}^2$

$=\left (-\dfrac {9}{4}+\dfrac {9}{2}\right)-(5+6)+(4+6)-\left (\dfrac {9}{4}-\dfrac {9}{2}\right)$

$=\dfrac {9}{2}-2+10+\dfrac {9}{4}=8+\dfrac {9}{2}$

$=\dfrac {25}{2}$

Evaluate $\displaystyle \int _0^2 \dfrac x 3 \, dx$

$\displaystyle \int _0^2 \dfrac x 3 dx$

$= \dfrac 13 \left. \dfrac {x^2}2\right|_0^2$

$=\dfrac 13 \left(\dfrac {2^2}2 -0 \right)=\dfrac 46-0=\dfrac 23$

Evaluate the following integrals
$\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{3/2} x}{\sin^{3/2}x+\cos^{3/2} x}\, dx$

Let

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{3/2}}{\sin^{3/2} x+\cos^{3/2}x}\, dx$

Then,

Formula:

$I=\displaystyle \int_{0}^{a}f(x)=f(a-x)$

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{3/2}{(\pi/2-x)}}{\sin^{3/2}(\pi/2-x)+\cos^{3/2}(\pi/2-x)}\, dx$

$I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{3/2}}{\cos^{3/2}x+\sin^{3/2}x}\, dx$

Adding

$(1)$ and

$(2)$ , we get

$2I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{3/2}+\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\, dx$

$2I=\displaystyle\int_{0}^{\pi/2}1.dx$

$2I=[x]_{0}^{\pi/2}$

$2I=\dfrac{\pi}{2}$

$I=\dfrac{\pi}{4}$

Evaluate $\displaystyle\int^{16}_0x^{3/4}dx$ .

1544031_1706273_ans_6165f485c08a41d6b63e1ca96e2c7068.jpg

Evaluate $\displaystyle\int^4_1\sqrt{x}dx$ .

1544005_1706271_ans_1dae91e4d5ce4e1daa71c4fb2f695295.jpg

Evaluate $\displaystyle\int^3_1x^4dx$ .

1543995_1706270_ans_52e7aeb96034498c8f6556f859d2bf7b.jpg

Evaluate $\displaystyle\int^4_1\dfrac{dx}{\sqrt{x}}$ .

1544573_1706275_ans_8ef09d3175c34a8e919ad885eef85695.png

Evaluate : $\displaystyle \int \dfrac{dx}{x^2+4x+8}$

$\displaystyle I=\int \dfrac{dx}{x^2+4x+8}$

$\displaystyle I=\int \dfrac{dx}{x^2+4x+4+4}$

$\displaystyle I=\int \dfrac{dx}{(x+2)^2+2^2}$

Let

$(x+2)=t$

diff w.to

$x$

$dx=dt$

$\displaystyle I=\int \dfrac{dt}{t^2+2^2}$

$I=\dfrac{1}{2}\tan^{-1}\dfrac{t}{2}+c$

$I=\dfrac{1}{2}\tan^{-1}\left(\dfrac{x+2}{2}\right)+c$

Evaluate $\displaystyle\int^1_0\dfrac{dx}{\sqrt [3]{x}}$ .

1544582_1706276_ans_08b2a03f16984fcf976645b10c747427.png

Evaluate $\displaystyle\int^{-1}_{-4}\dfrac{dx}{x}$ .

1544048_1706274_ans_c6ce75ba46ac43a2b390ffb65d93f292.jpg

Evaluate $\displaystyle\int^2_1x^{-5}dx$ .

1544015_1706272_ans_f2c4092dcd07470c9f8fede74145683c.jpg

Evaluate $\displaystyle\int^{1/2}_{1/4}\dfrac{dx}{\sqrt{x-x^2}}$ .

1546833_1706320_ans_779074f89fb447c0be8772f711f97609.jpg

Evaluate $\displaystyle\int^8_1\dfrac{dx}{x^{2/3}}$ .

1544590_1706277_ans_0a3eb7741bbc4976992fa12ef4e22df9.png

Evaluate $\displaystyle\int^2_1\dfrac{dx}{\sqrt{x^2+4x+3}}$ .

1545765_1706313_ans_5ac321e966754f2a8f1b6afb444d5a7f.png

Evaluate $\displaystyle\int^4_3\dfrac{dx}{(x^2-4)}$ .

1545755_1706311_ans_cb09e72dfa3d4d78a9a9216800f2195f.png

Evaluate $\displaystyle\int^1_0\dfrac{dx}{\sqrt{1-x^2}}$ .

1544604_1706281_ans_873bdaaade96412e96169317b6b34ac6.png

Evaluate $\displaystyle\int^1_0\dfrac{dx}{(1+x^2)}$ .

1544598_1706279_ans_ac2125ed45294585a8fdd3759eb8edb2.png

Evaluate $\displaystyle\int^4_23dx$ .

1544594_1706278_ans_cbab95fe7b0d4a35bcfe4934db2f7484.png

Evaluate $\displaystyle\int^1_0\dfrac{dx}{(1+x+2x^2)}$ .

1545780_1706314_ans_e1114a7d040b4172959615860498fd36.png

Evaluate $\displaystyle\int^4_0\dfrac{dx}{\sqrt{x^2+2x+3}}$ .

1545763_1706312_ans_799a35ee31444677a3377cf9d094c193.png

Evaluate $\displaystyle\int^{\infty}_0\dfrac{dx}{(1+x^2)}$ .

1544602_1706280_ans_5bebdf27431041ee9f899b25a1b82858.png

Evaluate the following integral:
$\displaystyle\int^1_0\dfrac{e^x}{(1+e^{2x})}dx$ .

1545824_1706393_ans_4c0abd2ba9054710a421801328bcfc31.png

Evaluate $\displaystyle\int^2_1\dfrac{5x^2}{(x^2+4x+3)}dx$ .

1548005_1706377_ans_39916ade41a747f58a7c9949a0c42685.jpg

Evaluate $\displaystyle\int^3_1\dfrac{dx}{x^2(x+1)}$ .

1547869_1706323_ans_bc53d8ee5c4b4f5ea35f2379d83ebd06.jpg

Evaluate the following integral:
$\displaystyle\int^1_0\dfrac{dx}{(2x-3)}$ .

1545792_1706383_ans_4375863414a847f8bf24f90e3c8bfdbb.png

Evaluate the following integral:
$\displaystyle\int^1_0\dfrac{2x}{(1+x^2)}dx$ .

1545800_1706387_ans_4cf53635ecfc4641bbb3e58844d68d66.png

Evaluate $\displaystyle\int^1_0\sqrt{x(1-x)}dx$ .

1546837_1706321_ans_d4ab8022029448308edfb30fe17ab55d.jpg

Evaluate the following integral:
$\displaystyle\int^2_1\dfrac{3x}{(9x^2-1)}dx$ .

1545805_1706391_ans_96d742c53d3949bc97d28d3696593182.png

Evaluate $\displaystyle\int^2_1\dfrac{dx}{x(1+2x)^2}$ .

1547870_1706325_ans_4f1813b61d8745629268d8029ad801ac.jpg

Evaluate the following integral:
$\displaystyle\int^1_0\dfrac{2x}{(1+x^4)}dx$ .

1545829_1706394_ans_7b1bfe96abc14ecfb5cdaadc4ba4e8b4.png

Evaluate $\displaystyle \int_{a}^{b} \sqrt{\dfrac{x-a}{b-x}}dx$

1644426_1775662_ans_c9c03f794c4a44248c15c989fa82206f.jpg

$\text { Evaluate: } \displaystyle \int_{0}^{1} \dfrac{d x}{1+x^{2}}$

$\displaystyle \int_{0}^{1} \dfrac{\mathrm{d} x}{1+x^{2}}=\left[\tan ^{-1} x\right]_{0}^{1} \\\quad=\tan ^{-1} 1-\tan ^{-1} 0=\dfrac{\pi}{4}-0=\dfrac{\pi}{4}$

$\text { Evaluate: } \displaystyle \int_{0}^{1 / \sqrt{2}} \dfrac{\mathrm{d} \mathrm{x}}{\sqrt{1-x^{2}}}$

Let

$I=\displaystyle \int_{0}^{1 / \sqrt{2}} \dfrac{\mathrm{d} \mathrm{x}}{\sqrt{1-x^{2}}}$

$=\left[\sin ^{-1} x\right]_{0}^{1 / \sqrt{2}}$ ........ Using standard formula

$\displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}dx =\sin^{-1}\left(\dfrac {x}{a}\right)+c$

$=\sin ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)-\sin ^{-1}(0)\\$

$=\dfrac{\pi}{4}-0=\dfrac{\pi}{4}$

Evaluate $\displaystyle \int_{a}^{b} \dfrac{1}{\sqrt{(x-a)(b-x)}}dx,b>a$

1644448_1775658_ans_a7d6d301a3064823800b23068bce8f9a.jpg

$\displaystyle \int_{1}^{2}\dfrac{dx}{\sqrt{x}-\sqrt{x-1}}$

1715072_1775651_ans_7870c13f3381463988db2ac0e3341bdc.jpg

$\text { Evaluate: } \displaystyle \int \dfrac{d x}{x+x \log x}$

Let

$I=\int \dfrac{d x}{x(1+\log x)}$

Put

$1+\log {x}=z,$ we get

$\begin{array}{l}\dfrac{1}{x} d x=d z \\\therefore \quad I=\displaystyle \int \dfrac{\mathrm{d} z}{z}=\log z+C=\log (1+\log x)+C\end{array}\\$

If $\int_{0}^{1}\left(3 x^{2}+2 x+k\right) \mathrm{dx}=0,$ then find the value of $k$ .

$\text { Given, } \int_{0}^{1}\left(3 x^{2}+2 x+k\right) \mathrm{d} \mathrm{x}=0$

$\Rightarrow\left[\dfrac{3 x^{3}}{3}+\dfrac{2 x^{2}}{2}+{kx}\right]_{0}^{1}=0$

$\begin{array}{l}\Rightarrow\left[x^{3}+x^{2}+{kx}\right]_{0}^{1} \Rightarrow(1+1+k)-(0)=0 \\\begin{array}{l}k=-2\end{array}\end{array}$

Find the value of the following integrals:
$\displaystyle \int^{1}_{0} \frac{x^2}{1 + x^2} dx$

$\displaystyle \int^{1}_{0} \frac{x^2}{1 + x^2} dx = \int^{1}_{0} \frac{(1 + x^2) - 1}{(1 + x^2)} dx$

$\displaystyle \int^{1}_{0} dx - \int^{1}_{0} \frac{1}{1 + x^2} dx$

$= [x]^{1}_{0} - [\tan^{-1} x]^{1}_{0}$

$= (1 - 0) - (\tan^{-1} 1 - \tan^{-1} 0)$

$= 1 - \left ( \frac{\pi}{4} - 0 \right )$

$= 1 - \frac{\pi}{4}$

$\text { Evaluate: } \displaystyle \int_{0}^{1} \dfrac{d x}{\sqrt{2 x+3}} d x$

$\text { Let } I=\displaystyle \int_{0}^{1} \dfrac{\mathrm{d} x}{\sqrt{2 x+3}}=\int_{0}^{1}(2 x+3)^{1 / 2} \mathrm{dx} \\$

$=\left[\dfrac{(2 x+3)^{1 / 2+1}}{\left(-\dfrac{1}{2}+1\right) \times 2}\right]_{0}^{1}=\left[\dfrac{(2 x+3)^{1 / 2}}{\dfrac{1}{2} \times 2}\right]_{0}^{1}$

$=5^{1 / 2}-3^{1 / 2}=\sqrt{5}-\sqrt{3}$

$\displaystyle \int_{0}^{\pi} log \;(1 - cos \;x) dx$

Let

$\displaystyle I = \int_{0}^{\pi} log \;(1 - cos \;x) dx$ ....(i)

$\displaystyle \int_{0}^{\pi} log \;[1 + cos (\pi - x) ]dx$

$\displaystyle I = \int_{0}^{\pi} log \;(1 + cos \;x) dx$ .....(ii)

Adding equation (i) and (ii),

$\displaystyle 2I = \int_{0}^{\pi} log \;(1 - cos \;x) dx$

$\displaystyle + \int_{0}^{\pi} log \;(1 + cos \;x) dx$

$\displaystyle = \int_{0}^{\pi} log \;(1 - cos \;x)(1 + cos \;x) dx$

$\displaystyle = \int_{0}^{\pi} log \;(1 - cos^{2} \;x) dx$

$\displaystyle = \int_{0}^{\pi} log \;sin^{2} x \;dx$

$\displaystyle = 2 \int_{0}^{\pi} log \;sin x \;dx$

$\displaystyle I = \int_{0}^{\pi} log \;sin \;x \;dx$

$\displaystyle = 2\int_{0}^{\pi/2} log \;sin \;x \;dx = 2I_{1}$

$\displaystyle \left [ \because \int_{0}{2a} f(x) \;dx \right ]$

When

$f(2a - x ) = f(x)$

Now

$\displaystyle I_{1} = \int_{0}^{\pi/2} log \;sin \;x \;dx$ ....................(iv)

$\displaystyle I_{1} = \int_{0}^{\pi/2} log \;sin \left ( \frac{\pi}{2} - x \right )dx$

$\displaystyle I_{1} = \int_{0}^{\pi/2} log \;cos \;x \;dx$ ....................(v)

$\displaystyle \left [ \because \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a - x)dx \right ]$

Adding equation (iv) and (v),

$\displaystyle 2I_{1} = \int_{0}^{\pi/2} log \;sin \;x \;dx + \int_{0}^{\pi/2} log \;cos \;x \;dx$

$\displaystyle = \int_{0}^{\pi/2} [ log \;sin \;x \;dx + log \;cos \;x ] \;dx$

$\displaystyle = \int_{0}^{\pi/2} log (sin \;x \;cos \;x ] \;dx$

$\displaystyle = \int_{0}^{\pi/2} log \left ( \frac{2sin \;x \;cos \;x}{2} \right ) \;dx$

$\displaystyle = \int_{0}^{\pi/2} [ log(2sin \;x \;cos \;x) - log2] \;dx$

$\displaystyle 2 I_{1} = \int_{0}^{\pi/2} log \;sin \;2x \;dx - log 2 \int_{0}^{\pi/2} dx$

$\displaystyle 2 I_{1} = \int_{0}^{\pi/2} log \;sin \;2x - log 2 [x]_{0}^{\pi/2}$

$\displaystyle = \int_{0}^{\pi/2} log \;sin \;2x \;dx - log 2 \left [ \frac{\pi}{2} - 0 \right ]$

$\displaystyle 2 I_{1} = \int_{0}^{\pi/2} log \;sin \;2x - \frac{\pi}{2} log 2$ .............(vi)

$\displaystyle 2I_{1} = I_{2} - \frac{\pi}{2} log 2$

Now

$\displaystyle I_{2} = \int_{0}^{\pi/2} log \;sin \;2x \;dx$

Putting

$2x = 1$

$2dx = dt$ or

$dx = \frac{1}{2} dt$

When

$x = 0$ then

$t = 0,$ when

$x= \frac{\pi}{2}$ then

$t = \pi$

$\displaystyle \therefore I_{2} = \int_{0}^{\pi} log \;sin \;t \frac{dt}{2}$

$\displaystyle = \frac{1}{2} \int_{0}^{\pi} log \;sin \;t \;dt$

$\displaystyle = \frac{1}{2} \times 2 \int_{0}^{\pi/2} log (sin \;t) dt$

$\displaystyle = \int_{0}^{pi/2} log\;sin \;x \;dx = I_{1}$

$I_{2} = I_{1}$

From equation (vi)

$\displaystyle 2I_{1} = I_{1} - \frac{\pi}{2} log 2$

$(\because I_{2} = I_{1}$

$\displaystyle \therefore I_{1} = - \frac{\pi}{2} log 2$ Putting value of

$I_{1}$ in equation (iii)

$\displaystyle I = 2I_{1} = 2 \left ( - \frac{\pi}{2} log 2 \right ) = -\pi log 2$

$\displaystyle I = - \pi \;log 2$

$\displaystyle \Rightarrow I = - \pi \;log (2-1)$

$\displaystyle \Rightarrow I = - \pi \;log \frac{1}{2}$

$\displaystyle \int_{0}{\pi} log (1 - cos x) dx = \pi \;log \frac{1}{2}$

Find the value of the following integrals:
$\displaystyle \int^{3}_{0} \sqrt{\frac{x}{3 - x}} dx$

$\displaystyle \int^{3}_{0} \sqrt{\frac{x}{3 - x}} dx$
Let

$x = 3 \sin^2 \theta$

$dx = 3 \sin 2 \theta d \theta$
When

$x = 3$ , then

$\theta = \frac{\pi}{2}$
When

$x = 0$ , then

$\theta = 0$

$\therefore \displaystyle \int^{3}_{0} \sqrt{\frac{x}{3 - x}} dx = \displaystyle \int^{\pi/2}_{0} \sqrt{\frac{3 \sin^2 \theta}{3 (1 - \sin^2 \theta)}} \cdot 3 \sin 2 \theta d \theta$

$= 3 \displaystyle \int^{\pi/2}_{0} 2 \sin^2 \theta d \theta$

$= 3 \int^{\pi/2}_{0} (1 - \cos 2 \theta) d \theta$

$= 3 \left ( \theta - \frac{\sin 2 \theta}{2} \right )^{\pi/2}_{0}$

$= 3 \left ( \frac{\pi}{2} - \frac{1}{2} \sin \pi \right ) - (0 - 0)$

$= 3 \left ( \frac{\pi}{2} - 0 \right ) = \frac{3 \pi}{2}$

If $\displaystyle \int _a^x t \ y (t)dt = x^2 + y(x)$ , then find $y(x)$ .

$\int_a^x t \ y(t) dt = x^2 + y(x)$
Differentiating both sides w.r.t.

$x$ , we get

$x y (x) = 2x + y' (X)$
hence

$\dfrac{dy}{dx} - xy = -2x$ (linerar)

$I.F. = e^{\int -x \ dx} = e^{-x^2/2}$

$\Rightarrow$ solution is

$ye^{-x^2/2}= \int -2xe^{-x^2/2} dx$

$\Rightarrow y = 2 + ce^{-x^2/2}$
if

$x = a \Rightarrow a^2 + y = 0$

$\Rightarrow y = -a^2$
hence

$-a^2 = 2 + ce^{-a^2/2}$

$\Rightarrow ce^{-a^2/2} = -(2+a^2)$

$\Rightarrow c = - (2 + a^2)e^{a^2/2}$

$\Rightarrow y = 2 - (2+a^2)e^{(a^2 - x^2)/2}$

$\text { Evaluate: } \int x \sin ^{-1} x d x$

Let

$I=\int x \sin ^{-1} x d x$

$=\sin ^{-1} x \cdot \dfrac{x^{2}}{2}-\displaystyle \int \dfrac{x^{2}}{2 \sqrt{1-x^{2}}} \mathrm{d} \mathrm{x} \quad[\text { By using integration by part }]$

$\begin{array}{l}=\dfrac{x^{2}}{2} \sin ^{-1} x+\dfrac{1}{2} \displaystyle \int \dfrac{1-x^{2}-1}{\sqrt{1-x^{2}}} \mathrm{d} \mathrm{x}=\dfrac{x^{2}}{2} \sin ^{-1} x+\dfrac{1}{2} \int \sqrt{1-x^{2}} \mathrm{dx}-\dfrac{1}{2} \int \dfrac{\mathrm{d} \mathrm{x}}{\sqrt{1-x^{2}}} \\=\dfrac{x^{2}}{2} \sin ^{-1} x-\dfrac{1}{2} \sin ^{-1} x+\dfrac{1}{2}\left[\dfrac{x}{2} \sqrt{1-x^{2}}+\dfrac{1}{2} \sin ^{-1} x\right]+C \\=\dfrac{x^{2}}{2} \sin ^{-1} x-\dfrac{1}{2} \sin ^{-1} x+\dfrac{x}{4} \sqrt{1-x^{2}}+\dfrac{1}{4} \sin ^{-1} x+C \\=\dfrac{x^{2}}{2} \sin ^{-1} x-\dfrac{1}{4} \sin ^{-1} x+\dfrac{x}{4} \sqrt{1-x^{2}}+C\end{array}\\$

$\text{Note:} \displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}dx =\sin^{-1}\left(\dfrac {x}{a}\right)+c$

$\left [ \int \sqrt{x^2+a^2} dx = \dfrac {x}{2} \sqrt{x^2+a^2} +\dfrac {a^2}{2} sin^{-1} \left ( \dfrac {x}{a} \right )+C\right ]$

$\text { Evaluate: } \displaystyle \int_{2}^{3} \dfrac{1}{x} \mathrm{d} \mathrm{x}$

$\displaystyle \int_{2}^{3} \dfrac{1}{x} \mathrm{d} \mathrm{x}$

$=[\log x]_{2}^{3}$

$=\log 3-\log 2$

$=\log \dfrac{3}{2}$

Find the value of the following integrals:
$\displaystyle \int^{2}_{1} \frac{dx}{(x + 1)(x + 2)}$

$\displaystyle \int^{2}_{1} \frac{dx}{(x + 1)(x + 2)} = \int^{2}_{1} \frac{(x + 2 - x - 1)}{(x + 1)(x + 2)} dx$

$= \displaystyle \int^{2}_{1} \frac{(x + 2)}{(x + 1)(x + 2)} dx - \int^{2}_{1} \frac{(x + 1)}{(x + 2)} dx$

$= \displaystyle \int^{2}_{1} \frac{1}{x + 1} dx - \int^{2}_{1} \frac{1}{x + 2} dx$

$[\log (x + 1)]^2_1 - [\log(x+2)]^2_1$

$= [log (2 + 1) - \log (1 + 1)]$

$= [\log (2 + 2) \cdot (1 + 2)]$

$= \log 3 - \log 2 - \log 4 + \log 3$

$= 2 \log 3 - (\log 2 + \log 4)$

$= 2 \log 3 - \log 8$

$= \log^{ \frac{3^2}{8}} = \log^{\frac{9}{8}}$

If $f(x)$ is a function satisfying, $\displaystyle f\left(\frac{1}{x}\right)+x^2f(x)=0$ for all non-zero $x$ ,
then evaluate $\displaystyle\int_{\displaystyle\sin{\theta}}^{\displaystyle\csc{\theta}}{f(x)dx}$ .

We have,

$\displaystyle f\left(\dfrac{1}{x}\right)+x^2f(x)=0$

$\displaystyle\Rightarrow f(x)=-\dfrac{1}{x^2}f\left(\dfrac{1}{x}\right)$

Let

$\displaystyle I=\int_{\displaystyle\sin{\theta}}^{\displaystyle\csc{\theta}}{f(x)dx}$

$=\int_{\displaystyle\sin{\theta}}^{\displaystyle\csc{\theta}}{-\dfrac{1}{x^2}f\left(\displaystyle \frac{1}{x}\right)dx}$

Put

$\displaystyle\dfrac{1}{x}=t$

$\Rightarrow -\displaystyle \frac{1}{x^2}dx=dt$

$\displaystyle=\int_{\displaystyle\csc{\theta}}^{\displaystyle\sin{\theta}}{f(t)dt}$

$\displaystyle=-\int_{\displaystyle\sin{\theta}}^{\displaystyle\csc{\theta}}{f(t)dt}=-I$ (Integration is independent of change of variable)

$\therefore 2I=0\implies I=0$

If $\displaystyle\int_{0}^{\pi /2}\frac{dx}{3+4\sin x}dx=\frac{1}{\sqrt{\left ( k \right )}}\log \left ( \frac{4+\sqrt{7}}{3} \right )$ then find the value of $k$ .

Let

$\displaystyle I=\int _{ 0 }^{ \pi /2 } \frac { dx }{ 3+4\sin x } =\int _{ 0 }^{ \pi /2 } \frac { dx }{ 3+8\sin \left( x/2 \right) \cos \left( x/2 \right) }$

$\displaystyle =\int _{ 0 }^{ \pi /2 } \frac { \sec ^{ 2 } \left( x/2 \right) dx }{ 3\left( 1+\tan ^{ 2 } \left( x/2 \right) \right) +8\tan \left( x/2 \right) } =\int _{ 0 }^{ \pi /2 } \frac { \sec ^{ 2 } \left( x/2 \right) dx }{ 3\tan ^{ 2 } \left( x/2 \right) +8\tan \left( x/2 \right) +3 }$

Substitute

$\displaystyle \tan \left( x/2 \right) =t\Rightarrow \left( 1/2 \right) \sec ^{ 2 } \left( x/2 \right) dx=dt$

Therefore

$\displaystyle I=\int _{ 0 }^{ 1 } \dfrac { 2dt }{ 3t^{ 2 }+8t+3 } =\dfrac { 2 }{ 3 } \int _{ 0 }^{ 1 } \dfrac { dt }{ t^{ 2 }+\dfrac { 8 }{ 3 } t+1 } =\dfrac { 2 }{ 3 } \int _{ 0 }^{ 1 } \dfrac { dt }{ \left( t+\dfrac { 4 }{ 3 } \right) ^{ 2 }-\left[ \sqrt { \left( 7 \right) /3 } \right] ^{ 2 } }$

$\displaystyle =\dfrac { 2 }{ 3 } .\dfrac { 1 }{ 2.\left[ \sqrt { \left( 7 \right) /3 } \right] } \left[ \log \dfrac { t+\dfrac { 4 }{ 3-\left( \sqrt { 7/3 } \right) } }{ t+\dfrac { 4 }{ 3 } +\left[ \sqrt { \left( 7 \right) /3 } \right] } \right]$

$\displaystyle =\dfrac { 1 }{ \sqrt { \left( 7 \right) } } \left[ \log \dfrac { 1+\dfrac { 4 }{ 3 } -\left( \sqrt { 7/3 } \right) }{ 1+\dfrac { 4 }{ 3 } +\left[ \sqrt { \left( 7 \right) /3 } \right] } -\log \dfrac { \dfrac { 4 }{ 3 } -\left( \sqrt { 7/3 } \right) }{ \dfrac { 4 }{ 3 } +\left[ \sqrt { \left( 7 \right) /3 } \right] } \right]$

$\displaystyle =\dfrac { 1 }{ \sqrt { \left( 7 \right) } } \log \dfrac { 7-\sqrt { 7 } }{ 7+\sqrt { \left( 7 \right) } } -\log \dfrac { -4\sqrt { 7 } }{ 4+\sqrt { \left( 7 \right) } } =\dfrac { 1 }{ \sqrt { \left( 7 \right) } } \log \left( \dfrac { \sqrt { 7 } -1 }{ \sqrt { \left( 7 \right) } +1 } \right) .\left( \dfrac { 4+\sqrt { 7 } }{ 4-\sqrt { \left( 7 \right) } } \right)$

$\displaystyle =\dfrac { 1 }{ \sqrt { \left( 7 \right) } } \log \left[ \dfrac { \left( \sqrt { 7 } -1 \right) ^{ 2 } }{ 7-1 } \times \dfrac { \left( 4+\sqrt { 7 } \right) ^{ 2 } }{ 16-7 } \right] =\dfrac { 1 }{ \sqrt { \left( 7 \right) } } \log \left[ \dfrac { 4-\sqrt { 7 } }{ 3 } \times \dfrac { 23+8\sqrt { 7 } }{ 9 } \right]$

$\displaystyle =\dfrac { 1 }{ \sqrt { \left( 7 \right) } } \log \left( \dfrac { 36+9\sqrt { 7 } }{ 27 } \right) =\dfrac { 1 }{ \sqrt { \left( 7 \right) } } \log \left( \dfrac { 4+\sqrt { 7 } }{ 3 } \right)$

For $x > 0$ , let $f\left ( x \right )= \displaystyle \int_{1}^{x}\displaystyle \frac{\log t}{t+1}dt$ . The value of $98\left ( f\left ( e \right )+f\left ( 1/e \right ) \right )$ is

$f\left( x \right) =\int _{ 1 }^{ x } \dfrac { \log t }{ t+1 } dt\\ \Rightarrow f\left( \cfrac { 1 }{ x } \right) =\int _{ 1 }^{ x } \dfrac { \log t }{ t\left( t+1 \right) } dt\\ \therefore f\left( x \right) +f\left( \cfrac { 1 }{ x } \right) =\int _{ 1 }^{ x } \left( \dfrac { \log t }{ t+1 } +\dfrac { \log t }{ t\left( t+1 \right) } \right) dt=\cfrac { { \left( \log { x } \right) }^{ 2 } }{ 2 } \\ \Rightarrow 98\left( f\left( e \right) +f\left( \cfrac { 1 }{ e } \right) \right) =\cfrac { 98 }{ 2 } =49$

Let $\displaystyle I= \int_{3n\pi }^{\left ( n+\frac{1}{n} \right )3\pi }\frac{4x dx}{\left [ \left ( a^{2}+b^{2} \right )+\left ( a^{2}-b^{2} \right )\cos \frac{2nx}{3} \right ]^{2}}$ (where a,b>0)
prove that $\displaystyle I=\frac{9\left ( 2n^{2}+1 \right )\pi }{n^{2}}\frac{a^{2}-b^{2}}{a^{3}b^{3}}$

Substitue

$\displaystyle \frac { nx }{ 3 } =t\Rightarrow \frac { n }{ 3 } dx=dt$

$\displaystyle \therefore I=\int _{ n^{ 2 }\pi }^{ \left( n^{ 2 }+1 \right) \pi } \frac { \left( 4.\frac { 3t }{ n } \right) \left( \frac { 3 }{ n } dt \right) }{ \left[ a^{ 2 }\left( 1+\cos 2t \right) +b^{ 2 }\left( 1-\cos 2t \right) \right] ^{ 2 } } =\frac { 9 }{ n^{ 2 } } \int _{ n^{ 2 }\pi }^{ \left( n^{ 2 }+1 \right) \pi } \frac { 4tdt }{ 4\left[ a^{ 2 }\cos ^{ 2 } t+b^{ 2 }\sin ^{ 2 } t \right] ^{ 2 } }$

$\displaystyle =\frac { 9 }{ n^{ 2 } } \int _{ n^{ 2 }\pi }^{ \left( n^{ 2 }+1 \right) n } \frac { tdt }{ \left[ a^{ 2 }\cos ^{ 2 } t+b^{ 2 }\sin ^{ 2 } t \right] ^{ 2 } }$ ...(1)
Using

$\int _{ a }^{ b } f\left( x \right) dx=\int _{ a }^{ b } f\left( a+b-x \right) dx$

$\displaystyle \therefore I=\frac { 9 }{ n^{ 2 } } \int _{ n^{ 2 }\pi }^{ \left( n^{ 2 }+1 \right) \pi } \frac { \left( 2n^{ 2 }+1 \right) \pi -t }{ \left[ a^{ 2 }\cos ^{ 2 } t+b^{ 2 }\sin ^{ 2 } t \right] ^{ 2 } }$ ...(2)
Adding (1) and (2)

$\displaystyle \therefore 2I=\frac { 9 }{ n^{ 2 } } \left( 2n^{ 2 }+1 \right) \pi .\int _{ n^{ 2 }\pi }^{ \left( n^{ 2 }+1 \right) \pi } \frac { dt }{ \left[ a^{ 2 }\cos ^{ 2 } t+b^{ 2 }\sin ^{ 2 } t \right] ^{ 2 } }$

$\displaystyle \therefore I=\frac { 9\left( 2n^{ 2 }+1 \right) \pi }{ n^{ 2 } } \frac { a^{ 2 }+b^{ 2 } }{ a^{ 3 }b^{ 3 } }$

If $\displaystyle \int_{0}^{1}\frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}}=\frac{\pi A}{\sqrt{6}(\sqrt{3}+1) (157)}$ , then $A$ is equal to

Since

$x(1-x)=\dfrac14 -\left(x-\dfrac12\right)^{2}$ ,

so put

$x=\dfrac12+\dfrac12 \sin \theta$

Therefore,

$\displaystyle \int_{0}^{1}\frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}}$

$\displaystyle \int_{-\frac{\pi} 2}^{\frac{\pi}2}\frac{\dfrac12\cos \theta}{\left(\dfrac{1}{4}(1+\sin \theta)^{2}+\dfrac{3}{2}(1+\sin \theta )+2\right) \dfrac{1}{2}\cos \theta}d\theta$

$\displaystyle =4 \int_{0}^{\pi}\frac{d \phi}{\cos^{2}\phi-8\cos \phi +15}$ , where

$\phi=\theta +\dfrac{\pi}2$

$\displaystyle =\frac{4}{2} \left[ \int_{0}^{\pi}\frac{d \phi}{\cos \phi-(4+1)}\int_{0}^{\pi} \frac{d\phi}{\cos \phi -(4-1)}\right]$

Let,

$\displaystyle I_{1}= \int_{0}^{\pi}\frac{d\phi}{\cos \phi -a}$ , where

$a>1$

Put

$\tan\dfrac{\phi}2= t\Rightarrow d\phi =2\dfrac{dt}{(1+t^{2})}$

$\displaystyle\therefore I_{1} =\int_{0}^{\infty}\frac{\dfrac2{(1+t^{2})}}{\dfrac{(1-t^{2})}{(1+t^{2})}-a}dt=\frac{-2}{1+a}\int_{0}^{\infty} \frac{dt}{t^{2}+\dfrac{(a-1)}{(a+1)}}$

$\displaystyle =\frac{-2}{1+a}\sqrt{\frac{a+1}{a-1}}\left(\tan^{-1} \left. \frac{t\sqrt{a+1}}{\sqrt{a-1}}\right) \right]_{0}^{\infty}$

$\displaystyle = \frac{-2}{\sqrt{a^{2}-1}}\cdot \frac{\pi}{2}=\frac{-\pi}{\sqrt{a^{2}-1}}$

Thus,

$\displaystyle I=2 \left[ -\frac{\pi}{\sqrt{24}}+\frac{\pi}{\sqrt{8}}\right]$

$\displaystyle =\pi \left( \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}\right) =\pi \frac{(\sqrt{3}-1)}{\sqrt{6}}$

Hence

$A=314$

Evaluate: $\displaystyle\int_{0}^{\dfrac{\pi}{2}}\sqrt{sin\,\phi} \, cos^5\,\phi\,d\,\phi$

1442644_694114_ans_f65131209d6a4e15bf511a1438884d4d.jpeg

Evaluate $\int_{0}^{\pi}e^{|\cos x|}\left (2\sin \left (\dfrac {1}{2}\cos x\right ) + 3\cos \left (\dfrac {1}{2}\cos x\right )\right )\sin x dx$ .

Given :

$\int _{ 0 }^{ \pi }{ { e }^{ \left| \cos { x } \right| } } \left[ 2\sin { \left( \cfrac { 1 }{ 2 } \cos { x } \right) +3\cos { \left( \cfrac { 1 }{ 2 } \cos { x } \right) } } \right] \sin { x } dx$

$\int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x)dx } \\ I=\int _{ 0 }^{ \pi }{ { e }^{ \left| \cos { x } \right| } } \left[ 2\sin { \left( \cfrac { 1 }{ 2 } \cos { x } \right) +3\cos { \left( \cfrac { 1 }{ 2 } \cos { x } \right) } } \right] \sin { x } dx.......(1)\\ I=\int _{ 0 }^{ \pi }{ { e }^{ \left| \cos { (\pi -x) } \right| } } \left[ 2\sin { \left( \cfrac { 1 }{ 2 } \cos { (\pi -x) } \right) +3\cos { \left( \cfrac { 1 }{ 2 } \cos { (\pi -x) } \right) } } \right] \sin { (\pi -x) } dx\\ I=\int _{ 0 }^{ \pi }{ { e }^{ \left| \cos { x } \right| } } \left[ -2\sin { \left( \cfrac { 1 }{ 2 } \cos { x } \right) +3\cos { \left( \cfrac { 1 }{ 2 } \cos { x } \right) } } \right] \sin { x } dx..........(2)$

Adding

$(1) and (2)$

$2I=\int _{ 0 }^{ \pi }{ { e }^{ \left| \cos { x } \right| } } 6\cos { \left( \cfrac { 1 }{ 2 } \cos { x } \right) \sin { x } dx } \\ I=3\int _{ 0 }^{ \pi }{ { e }^{ \left| \cos { x } \right| } } \cos { \left( \cfrac { 1 }{ 2 } \cos { x } \right) \sin { x } dx }$

Put

$t=\cos { x } \quad x\rightarrow 0\quad t\rightarrow 1$

$dt=-\sin { x } dx\quad x\rightarrow \pi \quad t\rightarrow -1\\ I=-3\int _{ 1 }^{ -1 }{ e^{ \left| t \right| }\cos { \left( \cfrac { t }{ 2 } \right) .dt } } \\ I=3\left[ \int _{ -1 }^{ 0 }{ { e }^{ -t }\cos { \left( \cfrac { t }{ 2 } \right) .dt } } +\int _{ 0 }^{ 1 }{ { e }^{ +1 }\cos { \left( \cfrac { t }{ 2 } \right) .dt } } \right] \\ { I }_{ 1 }=\int _{ -1 }^{ 0 }{ { e }^{ -t }\cos { \left( \cfrac { t }{ 2 } \right) .dt } } =-\cos { \left( \cfrac { t }{ 2 } \right) { e }^{ -t } } -\int { \sin { \cfrac { t }{ 2 } } \cfrac { 1 }{ 2 } { e }^{ -t } } dt\\ { I }_{ 1 }=-\cos { \left( \cfrac { t }{ 2 } \right) { e }^{ -t } } -\cfrac { 1 }{ 2 } \left[ -\left( \sin { \cfrac { t }{ 2 } } \right) { e }^{ -t }+\int { \cos { \cfrac { t }{ 2 } \cfrac { 1 }{ 2 } { e }^{ -t } } } \right] \\ \cfrac { 5 }{ 4 } { I }_{ 1 }=\left[ -\cot { \cfrac { t }{ 2 } { e }^{ -t } } +\cfrac { 1 }{ 2 } \sin { \left( \cfrac { t }{ 2 } \right) } { e }^{ -t } \right] _{ -1 }^{ 0 }\\ \cfrac { 5 }{ 4 } { I }_{ 1 }=-1+0+\cos { \cfrac { 1 }{ 2 } e } +\cfrac { 1 }{ 2 } \sin { \cfrac { 1 }{ 2 } } e\\ \cfrac { 5 }{ 4 } { I }_{ 1 }=e\left[ \cos { \cfrac { 1 }{ 2 } } +\cfrac { 1 }{ 2 } \sin { \cfrac { 1 }{ 2 } } \right] -1\\ { I }_{ 1 }=\cfrac { 4 }{ 5 } e\left[ \cos { \cfrac { 1 }{ 2 } } +\cfrac { 1 }{ 2 } \sin { \cfrac { 1 }{ 2 } } \right] -\cfrac { 4 }{ 5 }$

${ I }_{ 2 }=\int _{ 0 }^{ 1 }{ { e }^{ t } } \cos { \left( \cfrac { t }{ 2 } \right) dt } =\cos { \left( \cfrac { t }{ 2 } \right) } { e }^{ t }-\int { -\sin { \cfrac { t }{ 2 } } \cfrac { 1 }{ 2 } { e }^{ t } } \\ { I }_{ 2 }=\cos { \left( \cfrac { t }{ 2 } \right) } { e }^{ t }+\cfrac { 1 }{ 2 } \left[ \sin { \cfrac { t }{ 2 } { e }^{ t } } -\cfrac { 1 }{ 2 } \int { \cos { \left( \cfrac { t }{ 2 } \right) } { e }^{ t } } \right] \\ \cfrac { 5 }{ 4 } { I }_{ 2 }=\left[ \cos { \left( \cfrac { t }{ 2 } \right) } { e }^{ t }+\cfrac { 1 }{ 2 } \sin { \cfrac { t }{ 2 } { e }^{ t } } \right] _{ 0 }^{ 1 }\\ { I }_{ 2 }=\cfrac { 4 }{ 5 } \left[ e\cos { \cfrac { 1 }{ 2 } } +\cfrac { 1 }{ 2 } \sin { \cfrac { 1 }{ 2 } } e-1 \right] \\ I=3\left( { I }_{ 1 }+{ I }_{ 2 } \right) \\ =3\times 2\times \cfrac { 4 }{ 5 } \left[ e\cos { \cfrac { 1 }{ 2 } } +\cfrac { 1 }{ 2 } \sin { \cfrac { 1 }{ 2 } } e \right] \\ =\cfrac { 24e }{ 5 } \left[ \cos { \cfrac { 1 }{ 2 } } +\cfrac { 1 }{ 2 } \sin { \cfrac { 1 }{ 2 } } \right]$

Hence the correct answer is

$\cfrac { 24e }{ 5 } \left[ \cos { \cfrac { 1 }{ 2 } } +\cfrac { 1 }{ 2 } \sin { \cfrac { 1 }{ 2 } } \right]$

Evaluate: $\displaystyle\int \dfrac{tan\,x}{(a^2+b^2tan^2x)}dx$

1442648_694116_ans_e0da5361ed4247198d0209efec21191d.jpeg

The value of $\displaystyle \int_{1/3}^{3}\dfrac {\tan \left (x^{2} + \dfrac {1}{x^{2}}\right )}{\sin \left (x + \dfrac {1}{x}\right )} \dfrac {dx}{x}$ is

Let

$I=\displaystyle \int_{1/3}^{3}\dfrac {\tan \left (x^{2} + \dfrac {1}{x^{2}}\right )}{\sin \left (x + \dfrac {1}{x}\right )} \dfrac {dx}{x}$

Let

$x+\dfrac{1}{x}=t$ ............. I

When

$x=\dfrac{1}{3}, t=\dfrac{10}{3}$

When

$x=3, t=\dfrac{10}{3}$

Squaring both side, we get

$\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2$ .............. II

Differentiating both sides of equation I, we get

$\Rightarrow \left ( 1-\dfrac{1}{x^2} \right )dx=dt$

$\Rightarrow \left ( x-\dfrac{1}{x} \right)\dfrac{dx}{x}=dt$ ................. III

Now,

$\left(x-\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}-2$

$=t^2-2-2$

$=t^2-4$

$\Rightarrow\left(x-\dfrac{1}{x}\right)=\sqrt{t^2-4}$ ................. IV

From equation III and IV, we get

$\Rightarrow\dfrac{dx}{x}=\dfrac{dt}{\sqrt{t^2-4}}$ .............. V

Substituting equations I, II and V in the equation given in question and changing the limits accordingly, we get

$I=\displaystyle \int_{10/3}^{10/3}\dfrac{\tan(t^2-2)dt}{\sqrt{t^2-4}\sin{t}}$

$=0$ [

$\because$ upper and lower limits of integration become equal]

Hence, the answer is

$0$

Evaluate $\displaystyle \int { \dfrac { \left( \cos { 5x+\cos { 4x } } \right) }{ 1-2\cos { 3x } } } dx$

Given:

$\int { \cfrac { (\cos { 5x } +\cos { 4x } ) }{ 1-2\cos { 3x } } dx } \\ solution;\\ \cos C+\cos D=2\cos \left( \cfrac { C+D }{ 2 } \right) \cos \left( \cfrac { C-D }{ 2 } \right) \\ \therefore \cos 5x+\cos 4x=2cos\left( \cfrac { 9x }{ 2 } \right) \cos \left( \cfrac { x }{ 2 } \right) \\ 1-2\cos 3x=1-2\left( 2\cos ^{ 2 }{ \cfrac { 3x }{ 2 } -1 } \right) \\ =3-4\cos ^{ 2 }{ \cfrac { 3x }{ 2 } } =\cfrac { 3\cos \cfrac { 3x }{ 2 } -4\cos ^{ 3 }{ \cfrac { 3x }{ 2 } } }{ \cos \cfrac { 3x }{ 2 } } \\ =\cfrac { -\left( 4\cos ^{ 3 }{ \cfrac { 3x }{ 2 } } -3\cos \cfrac { 3x }{ 2 } \right) }{ \cos \cfrac { 3x }{ 2 } } =\cfrac { -\cos \left( 3\times \cfrac { 3x }{ 2 } \right) }{ \cos \cfrac { 3x }{ 2 } } \\ \therefore \cfrac { \cos 5x+\cos 4x }{ 1-2\cos 3x } =\cfrac { 2\cos \cfrac { 9x }{ 2 } \cos \cfrac { x }{ 2 } }{ -\cos \left( \cfrac { 9x }{ 2 } \right) /\cos \left( \cfrac { 3x }{ 2 } \right) } \\ =-2\cos \cfrac { x }{ 2 } \cos \cfrac { 3x }{ 2 } =-\left( \cos \left( \cfrac { 3x }{ 2 } -\cfrac { x }{ 2 } \right) +\cos \left( \cfrac { 3x }{ 2 } +\cfrac { x }{ 2 } \right) \right) \\ =-(\cos x +\cos 2x)\\ \therefore \int { -2 } (\cos +\cos 2x)dx\\ =-(2\sin x+\sin 2x)+C$

Hence the correct answer is

$= -(2\sin x+\sin 2x)+C$

Solve $\displaystyle\int \dfrac {dx}{\sin x+\sec x}$

Consider the following integral.

$I=\int{\dfrac{1}{\sin x+\sec x}dx}$

$=\int{\dfrac{2\cos x}{2+2\sin x\cos x}dx}$

$=\int{\dfrac{\cos x-\sin x+\sin x+\cos x}{1+{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}dx}$

$=\int{\dfrac{\sin x+\cos x}{1+{{\left( \sin x+\cos \right)}^{2}}}}dx+\int{\dfrac{\cos x-\sin x}{1+{{\left( \sin x+\cos \right)}^{2}}}}dx$

$I={{I}_{1}}+{{I}_{2}}$

Integrate $I_1$ and $I_2$ we get,

${{I}_{1}}=\int{\dfrac{\sin x+\cos x}{1+{{\left( \sin x+\cos \right)}^{2}}}}dx$ $$

Let, $t=\sin x+\cos x$

And differentiate both side w.r.t x

$dt=(\cos x-\sin x)dx$

$=\int{\dfrac{\cos x-\sin x}{1+{{\left( t \right)}^{2}}}}*\dfrac{1}{\left( \cos x-\sin x \right)}dt$

$=\int{\dfrac{1}{1+{{\left( t \right)}^{2}}}}dt$

$={{\tan }^{-1}}t+C$

$+{{\tan }^{-1}}\left( \sin x+\cos x \right)+C$

${{I}_{2}}=\int{\dfrac{\sin x+\cos x}{2+\sin 2x}}dx$

$=\int{\dfrac{\sin x+\cos x}{3-(1-\sin 2x)}}dx$

$=\int{\dfrac{\sin x+\cos x}{3-({{\sin }^{2}}x+{{\cos }^{2}}-2\sin x\cos x)}}dx$

$=\int{\dfrac{\sin x+\cos x}{3-(\sin x-\cos x)}}dx$

$u=\sin x-\cos x$

$dx=\dfrac{du}{\sin x+\cos x}$

${{I}_{2}}=\int{\dfrac{\sin x+\cos x}{\sqrt{3}-{{u}^{2}}}}\dfrac{du}{(\sin x+\cos x)}$

$=\int{\dfrac{1}{{{\left( \sqrt{3} \right)}^{2}}-{{u}^{2}}}}du$

$=\dfrac{1}{\sqrt{3}}\ln \left( \dfrac{\sqrt{3}+u}{\sqrt{3}-u} \right)$

$=\dfrac{1}{\sqrt{3}}\ln \left( \dfrac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\left( \sin x-\cos x \right)} \right)$

$I=\int{\dfrac{1}{\sin x+\sec x}dx}=\dfrac{1}{\sqrt{3}}\ln \left( \dfrac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\left( \sin x-\cos x \right)} \right)+{{\tan }^{-1}}\left( \sin x+\cos x \right)+C$

Hence, this is the correct answer.

Evaluate $\int\limits_0^{\frac{\pi }{4}} {{{\cos }^{\frac{3}{2}}}\left( {2x} \right)cos\left( x \right)dx}$

Consider the given integral.

$I=\int_{0}^{\frac{\pi }{4}}{{{\cos }^{3/2}}2x\cos x}dx$

$I=\int_{0}^{\frac{\pi }{4}}{{{\left( \cos 2x \right)}^{3/2}}\cos x}dx$

$I=\int_{0}^{\frac{\pi }{4}}{{{\left( 1-2{{\sin }^{2}}x \right)}^{3/2}}\cos x}dx$

Let $t=\sin x$

$dt=\cos xdx$

Therefore,

$I=\int_{0}^{\frac{1}{\sqrt{2}}}{{{\left( 1-2{{t}^{2}} \right)}^{3/2}}dt}$

$I=\int_{0}^{\frac{1}{\sqrt{2}}}{{{\left( 1-{{\left( \sqrt{2}t \right)}^{2}} \right)}^{3/2}}dt}$

Let $\sqrt{2}t=\sin u$

$\sqrt{2}\dfrac{dt}{du}=\cos u$

$dt=\dfrac{\cos udu}{\sqrt{2}}$

Therefore,

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( 1-{{\sin }^{2}}u \right)}^{3/2}}\cos udu}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( {{\cos }^{2}}u \right)}^{3/2}}\cos udu}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\cos }^{4}}udu}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( {{\cos }^{2}}u \right)}^{2}}du}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( \dfrac{1+\cos 2u}{2} \right)}^{2}}du}$

$I=\dfrac{1}{4\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{\left( 1+{{\cos }^{2}}2u+2\cos 2u \right)du}$

$I=\dfrac{1}{4\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{\left( 1+\left( \dfrac{1+\cos 4u}{2} \right)+2\cos 2u \right)du}$

$I=\dfrac{1}{4\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{\left( 1+\dfrac{1}{2}+\dfrac{\cos 4u}{2}+2\cos 2u \right)du}$

$I=\dfrac{1}{4\sqrt{2}}\left[ u+\dfrac{u}{2}+\dfrac{1}{2}\left( \dfrac{\sin 4u}{4} \right)+2\dfrac{\sin 2u}{2} \right]_{0}^{\frac{\pi }{2}}$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3u}{2}+\dfrac{\sin 4u}{8}+\sin 2u \right]_{0}^{\frac{\pi }{2}}$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3\pi }{4}+\dfrac{\sin 4\left( \dfrac{\pi }{2} \right)}{8}+\sin 2\left( \dfrac{\pi }{2} \right)-\left( 0 \right) \right]$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3\pi }{4}+\dfrac{\sin \left( 2\pi \right)}{8}+\sin \left( \pi \right) \right]$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3\pi }{4}+0+0 \right]$

$I=\dfrac{3\pi }{16\sqrt{2}}$

Hence, this is the answer.

Solve : $\displaystyle\int sec^{-1} \sqrt{x} \, dx$

$\displaystyle\int \sec^{-1}\sqrt{x}dx$

Let

$x=\cos^2t$ here

$t=\cos^{-1}\sqrt{x}$

$dx=-2\cos t\sin tdt$

Put there value

$-\displaystyle\int 2\sec^{-1}\sqrt{\cos^2x}\cos t\sin t dt$

$=-2\displaystyle\int \sec^{-1}\cos x\cos t\sin t dt$

$=-2\displaystyle\int x\cos t\sin tdt$

$=-2\displaystyle\int \cos^2t\cos t\sin tdt$

$=-2\displaystyle\int \cos^3t\sin tdt$

Let

$\cos t=m$

$-\sin tdt=dm$

$=-2\displaystyle\int m^3dm$

$=-2\left[\dfrac{m^4}{4}\right]+c$

$=-\dfrac{\cos^4t}{2}+c=-\dfrac{\cos^4(\cos^{-1}\sqrt{x})}{2}+c$ .

Solve $\int_0^1 {{{\cot }^{ - 1}}} (1 - x + {x^2})dx$ .

Let

$I=\displaystyle\int_{0}^{1}{{\cot}^{-1}{\left(1-x+{x}^{2}\right)}dx}$

$=\displaystyle\int_{0}^{1}{{\tan}^{-1}{\left(\dfrac{1}{1-x+{x}^{2}}\right)}dx}$

$=\displaystyle\int_{0}^{1}{{\tan}^{-1}{\left(\dfrac{1}{1-x\left(1-x\right)}\right)}dx}$

$=\displaystyle\int_{0}^{1}{{\tan}^{-1}{\left(\dfrac{1+\left(1-x\right)}{1-x\left(1-x\right)}\right)}dx}$

$=\displaystyle\int_{0}^{1}{\left\{{\tan}^{-1}{x}+{\tan}^{-1}{\left(1-x\right)}\right\}dx}$

$=\displaystyle\int_{0}^{1}{{\tan}^{-1}{x}dx}+\displaystyle\int_{0}^{1}{{\tan}^{-1}{\left(1-x\right)}dx}$

$=\displaystyle\int_{0}^{1}{{\tan}^{-1}{x}dx}+\displaystyle\int_{0}^{1}{{\tan}^{-1}{\left(1-\left(1-x\right)\right)}dx}$ since

$\displaystyle\int_{0}^{a}{f\left(x\right)dx}=\displaystyle\int_{0}^{a}{f\left(a-x\right)dx}$

$=\displaystyle\int_{0}^{1}{{\tan}^{-1}{x}dx}+\displaystyle\int_{0}^{1}{{\tan}^{-1}{x}dx}$

$=2\displaystyle\int_{0}^{1}{{\tan}^{-1}{x}dx}$

$=2\displaystyle\int_{0}^{1}{{\tan}^{-1}{x}\,.1.\,dx}$

$=2\left[x{\tan}^{-1}{x}\right]_{0}^{1}-2\displaystyle\int_{0}^{1}{\dfrac{x\,dx}{1+{x}^{2}}}$

$=2\left[x{\tan}^{-1}{x}\right]_{0}^{1}-\displaystyle\int_{0}^{1}{\dfrac{2x\,dx}{1+{x}^{2}}}$

$=2\left[\dfrac{\pi}{4}-0\right]-\left[\log{\left(1+{x}^{2}\right)}\right]_{0}^{1}$

$=\dfrac{\pi}{2}-\left[\log{2}-\log{1}\right]$

$=\dfrac{\pi}{2}-\log{2}$

Evaluate $\ \int _{ 0 }^{ \pi }{ \dfrac { xdx }{ 1+\cos ^{ 2 }{ x } } }$

Solution: let

$I= \int _{ 0 }^{ \pi }{ \dfrac { xdx }{ 1+\cos ^{ 2 }{ x } } } =\int _{ 0 }^{ \pi }{ \dfrac { \left( \pi -x \right) dx }{ 1+\cos ^{ 2 }{ \left( \pi -x \right) } } = } \int _{ 0 }^{ \pi }{ \dfrac { xdx }{ 1+\cos ^{ 2 }{ x } } } -1$

$\Rightarrow 2I=\int _{ 0 }^{ \pi }{ \dfrac { xdx }{ 1+\cos ^{ 2 }{ x } } } =2\pi \int _{ 0 }^{ \pi /2 }{ \dfrac { dx }{ 1+\cos ^{ 2 }{ x } } } =2\pi \int _{ 0 }^{ \pi /2 }{ \dfrac { \sec ^{ 2 }{ x } dx }{ 2+\tan ^{ 2 }{ x } } }$ Let

$\tan{x}=t$ so that for

$x\rightarrow 0$ ,

$t\rightarrow 0$ and for

$x\rightarrow {\pi/2}$ ,

$t\rightarrow {\infty}$ .

Hence we can write.

$I=\pi \int _{ 0 }^{ \pi }{ \dfrac { dt }{ 2+{ t }^{ 2 } } } =\pi \dfrac { 1 }{ \sqrt { 2 } } { \left[ \tan ^{ -1 }{ \dfrac { t }{ \sqrt { 2 } } } \right] }_{ 0 }^{ \pi }=\dfrac { { \pi }^{ 2 } }{ 2\sqrt { 2 } }$

Evaluate:
$\displaystyle \int^{\frac{\pi}{4}}_{0}e^{\sin x}(1 + \tan x \sec x)dx$

given,

$\displaystyle \int^{\frac{\pi}{4}}_{0}e^{\sin x}(1 + \tan x \sec x)dx$

Let

$\displaystyle e^{\sin x} \sec x=z$

or,

$\displaystyle dz=(e^{\sin x}. \cos x.\sec x+e^{\sin x} \sec x.\tan x)dx$

or,

$dz=e^{\sin x}(1+\sec x.\tan x)dx$

So the given integration will take the form

$\Rightarrow$ $\displaystyle \int^{\frac{\pi}{4}}_{0}d\left(e^{\sin x}\sec x\right)$

$=\left[e^{\sin x}\sec x\right]_0^{\frac{\pi}{4}}$

$=\left(\sqrt{2}e^{\frac{1}{\sqrt{2}}}-1\right)$

$\int {x{{\sin }^{ - 1}}xdx}$

$\int x\,\sin^{-1}x\,dx$

Let

$x=\sin\,\theta$

$dx=\cos\theta\,d\theta$

Substituting values, we get

$\int x\,\sin^{-1}x\,dx=\int\sin\theta\,\sin^{-1}(\sin\theta)\cos\theta\,d\theta$

$=\int\sin\theta\,\theta\,\cos\theta\,d\theta$

[ Since

$\sin^{-1}(\sin\theta)=\theta$ ]

$=\dfrac{1}{2}\int\theta(2\sin\theta\cos\theta)d\theta$

$=\dfrac{1}{2}\int\theta \sin\,2\theta\,d\theta$ [ Since,

$2\sin x\,\cos x=\sin2x$ ]

$=\dfrac{1}{2}\left[\theta\int\sin 2\theta d\theta-\int\left(\dfrac{d\theta}{d\theta}\int \sin 2\theta\right)d\theta\right]$

$=\dfrac{1}{2}\left[\theta \dfrac{-(\cos\,2\theta)}{2}-\int 1.\dfrac{-(\cos\,2\theta)}{2}d\theta\right]$

$=\dfrac{1}{2}\left[\dfrac{-\theta}{2}\cos 2\theta+\dfrac{1}{2}\int \cos 2\theta d\theta\right]$

$=\dfrac{1}{2}\left[\dfrac{-\theta}{2}\cos\,2\theta+\dfrac{1}{2}\dfrac{\sin 2\theta}{2}\right]+c$

$=\dfrac{-\theta}{4}\cos\,2\theta+\dfrac{1}{8}\sin\,2\theta+c$

$=\dfrac{-\theta}{4}(1-2\sin^2\theta)+\dfrac{1}{8}\times 2\sin\theta\,\cos\theta+c$

$=\dfrac{-\theta}{4}(1-2\sin^2\theta)+\dfrac{1}{4}\sin\theta \cos\theta+c$

$=\dfrac{-\theta}{4}(1-2\sin^2\theta)+\dfrac{1}{4}\sin\theta \sqrt{1-\sin^2\theta}+c$

$=\dfrac{-\sin^{-1}x}{4}(1-2x^2)+\dfrac{x}{4}\sqrt{1-x^2}+c$ [ Since,

$\theta=\sin^{-1}x$ ]

$=\dfrac{(2x^2-1)}{4}\sin^{-1}x+\dfrac{x}{4}\sqrt{1-x^2}+c$

If $f\left( x \right) =\int { \dfrac { ({ x }^{ 2 }+\sin ^{ 2 }{ x } ) }{ 1+{ x }^{ 2 } } } \sec ^{ 2 }{ x } dx$ and $f\left( 0 \right)=0$ , then $f\left( 1 \right)$ is equal to

1179134_1101420_ans_5c1e36455a8a419f9671c1b1c312ccfb.jpg

Evaluate:
$\displaystyle \int_0^1 \dfrac {2x}{x^2+1}dx$

$\displaystyle \int_0^1 \dfrac {2x}{x^2+1}dx$

Let

$t=x^2+1 \implies dt=2xdx$

$\implies x\to 0\to 1$ so

$t\to 1\to 2$

$\displaystyle \int_1^2 \dfrac 1tdt$

By using

$\int\dfrac 1 n =\log t+c$

$=\left.\log t \right|_1^2$

$\Rightarrow \log 2-\log 1=\log 2$

Find solution in terms of indefinite integration, using substitution
$\int_0^1 {x\cos \left( {{{\tan }^{ - 1}}x} \right)} \,dx$

$\int_0^1 {x\cos \left( {{{\tan }^{ - 1}}x} \right)} \,dx$

Let,

$\tan^{-1}{x} = t$

at,

$x = 0, t= 0 ; x= 1, t= \dfrac{\pi}{4}$

$dt = \dfrac {1}{1+x^2} dx$

$\therefore \int_0^1 {x\cos \left( {{{\tan }^{ - 1}}x} \right)} \,dx$

$\int_0^{\dfrac{\pi}{4}} \tan t \times \cos t \times (1+\tan^2 t) dt$

$\int_0^{\dfrac{\pi}{4}} \tan t \times \cos t \times (\sec^2 t) dt$ ......{

$\because 1+\tan^2 t = \sec^2 t$ }

$\int_0^{\dfrac{\pi}{4}} \dfrac{\sin t} {\cos^{2}t} dt$ ......{

$\because \dfrac {1}{\cos t} = \sec t \quad and \quad \tan t = \dfrac{\sin t}{\cos t}$ }

${[\dfrac{{(\cos t)}^{(-2+1)}}{-2+1}]}_{0}^{\dfrac{\pi}{4}}$

${[\dfrac{(\cos t)^{-1}}{-1}]}_{0}^{\dfrac{\pi}{4}}$

$[(-\sec \dfrac{\pi}{4})-(-\sec 0)]$

$-\sqrt {2} + 1$

Evaluate: $\displaystyle {\int {\left( {\dfrac{{1 - x}}{{1 + {x^2}}}} \right)} ^2}{e^x}\,dx$

Consider,

$I=\int e^x\dfrac{(1-x)^2}{(1+x^2)^2}dx$

$I=\int e^x\dfrac{(1+x^2-2x)}{(1+x^2)^2}dx$

$I=\int e^x\dfrac{1+x^2}{(1+x^2)^2}dx$ $-$ $\int e^x\dfrac{2x}{(1+x^2)^2}dx$

$I=\int e^x\dfrac{1}{(1+x^2)}dx$ $-$ $\int e^x\dfrac{2x}{(1+x^2)^2}dx$

By integration by parts :

$\int [f(x) g(x)]dx=f(x)\cdot \int g(x)dx-\int[f'(x) \int g(x) dx]dx$

By integration by parts of first integral

$I=$ $\dfrac { 1 }{ 1+{ x }^{ 2 } } \int { e } ^{ x }-\int { \{ (\dfrac { d(\dfrac { 1 }{ 1+{ x }^{ 2 } } ) }{ dx } } )\int { { e }^{ x } } dx\} dx$ $-$ $\int e^x\dfrac{2x}{(1+x^2)^2}dx$

$I= e^x\dfrac{1}{(1+x^2)}$ $+$ $\int e^x\dfrac{2x}{(1+x^2)^2}dx$ $-$ $\int e^x\dfrac{2x}{(1+x^2)^2}dx$

$I= e^x\dfrac{1}{(1+x^2)}$ $+$ $C$

$\int (\sec^{2} x+\csc^{2} x)dx$

=∫sec2x(1+cot2x)dx

Now recall that cotangent function is the reciprocal of the tangent function and the secant function is the reciprocal of the cosine function.

=∫1cos2x(1+cos2xsin2x)dx

=∫1cos2x+1sin2xdx

=∫sec2x+csc2xdx

=∫sec2x+∫csc2xdx

These are both widely known integrals. If you haven't already, I would recommend learning them by heart.

=tanx−cotx+C

Find the following integral.
$\displaystyle\int \dfrac{x^3-1}{x^2}dx$ .

$\int { \frac { { { x^{ 3 } }-1 } }{ { { x^{ 2 } } } } dx } \\ \Rightarrow \int { \frac { { { x^{ 3 } } } }{ { { x^{ 2 } } } } dx-\int { \frac { 1 }{ { { x^{ 2 } } } } dx } } \\ \Rightarrow \int { xdx-\int { \frac { 1 }{ { { x^{ 2 } } } } dx } } \\ \Rightarrow \frac { { { x^{ 2 } } } }{ 2 } -\int { { x^{ -2 } }dx } \\ \Rightarrow \frac { { { x^{ 2 } } } }{ 2 } -\frac { { { x^{ -1 } } } }{ { -1 } } +c \\ \, \, \, \, \, \frac { { { x^{ 2 } } } }{ 2 } +\frac { 1 }{ x } +c\, \,$

Find the following integral.
$\displaystyle\int (x^{\frac{1}{2}}+2e^x-\frac{1}{x})dx$ .

$\int { \left( { { x^{ \frac { 1 }{ 2 } } }+2{ e^{ x } }-\frac { 1 }{ x } } \right) dx } \\ \Rightarrow \int { { x^{ \frac { 1 }{ 2 } } }dx+2\int { { e^{ x } }dx-\int { \frac { 1 }{ x } dx } } } \\ \Rightarrow \frac { { { x^{ \frac { 3 }{ 2 } } } } }{ { \frac { 3 }{ 2 } } } +2{ e^{ x } }-\log x+c \\ \Rightarrow \frac { 2 }{ 3 } { x^{ \frac { 3 }{ 2 } } }+2{ e^{ x } }-\log x+c\, \, \,$

$\displaystyle\int^{\pi/2}_0\dfrac{\sin 8xlog (\cot x)dx}{\cos 2x}$ .

$\begin{matrix} I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { { \sin 8x\log \left( { co+x } \right) } }{ { \cos 2x } } dx\to (i) } \\ by\, \, applying\, \, property\, \, of\, \, { int }egration\, \, \\ I=\int _{ 0 }^{ a }{ faxdx=\int _{ 0 }^{ a }{ f\left( { a-x } \right) dx } } \\ I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { { \sin 8\left( { \dfrac { \pi }{ 2 } -x } \right) \log \left[ { \cot \left( { \dfrac { \pi }{ 2 } -x } \right) } \right] } }{ { \cos 2\left( { \dfrac { \pi }{ 2 } -x } \right) } } } \\ I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { { \sin \left( { 4x-8x } \right) \log \left( { \tan x } \right) } }{ { \cos \left( { \pi -2x } \right) } } dx } \\ I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { { -\sin 8x\log \left( { \dfrac { 1 }{ { \cot x } } } \right) } }{ { -\cos 2x } } } \, dx \\ I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { { { - }\sin 8x\log { { \left( { \cot x } \right) }^{ -1 } } } }{ { { - }\cos 2x } } } \\ I=\int _{ 0 }^{ \pi }{ -\dfrac { { \sin 8x\log \left( { \cot x } \right) } }{ { \cos 2x } } } \\ I=\int _{ 0 }^{ \pi }{ \dfrac { { \sin 8x\log \cot x } }{ { \cos 2x } } dx=-I } \\ I=-I \\ 2I=0 \\ \boxed { I=0 } \, \ \end{matrix}$

Find : $\int {\log x.dx}$

Sol

$\displaystyle \int (logx)dx$

$\displaystyle =\int (logx).1dx$

Now, we know that

$\displaystyle \int uv dx = u\int vdx -\int ({u}' \int v dx)dx ...(1)$

Here, u = logx

v = 1

putting values in eq (1), we get

$\displaystyle \int (logx).1dx = logx \int 1.dx-\int (\dfrac{d(logx)}{dx}.\int 1.dx)dx$

$= (logx)x-\int 1/x . 2 dx$

$= x log x- \int 1.dx$

$\boxed{\int logx dx = xlogx-x+c}$

1172132_1135572_ans_9f0ce84309464d2998c2a8134e671ed8.jpg

Evaluate: $\int\limits_0^{\pi /2} {\dfrac{{\left( {\sin x + \cos x} \right)dx}}{{9 + 16\,\sin 2x}}}$

$\displaystyle I= \int \frac{\sin x+\cos x}{9+16 \sin 2x}dx$

We have

$1- \sin 2x= \sin^{2}x+ \cos^{2}x-2 \sin x\, \cos x$

$=(\sin\, x- \cos\, x)^{2}$

$\sin2x = 1-(\sin x-\cos x)^{2}$

Now multiplying by

$16$

$16 \sin 2x= 16-16(\sin x- \cos x)^{2}$

Now

$9+16 \sin 2x= 25-16(\sin x- \cos x)^{2}$

$\displaystyle\therefore I= \int \dfrac{sin x+ \cos x}{25-16(\sin x- \cos x)^{2}}dx$

Let

$4(\sin x- \cos x)=t$

$\Rightarrow 4(\cos x+\sin x)dx= dt$

$\displaystyle \therefore I=\frac{1}{4}\int \frac{dt}{25-t^{2}}$

$=\displaystyle \frac{1}{4}\int \frac{dt}{5^{2}-t^{2}}$

$=\displaystyle \frac{1}{40}log \left | \frac{5+t}{5-t} \right |+c$

$=\displaystyle\frac{1}{40}log\left | \frac{5+4(\sin x- \cos x)}{5-4(\sin x-\cos x)} \right |+c$

Integrate $\dfrac{1}{x(x^n+1)}$ with respect to x.

2060198_1166714_ans_dc9b9074794f47bc80899bd964d0c4f5.JPG

Evaluate:
$\displaystyle\int_{\tfrac{-3\pi}{2}}^{\tfrac{-\pi}{2}}{\left[{\left(x+\pi\right)}^{3}+{\cos}^{2}{\left(x+3\pi\right)}\right]dx}$

1353904_1246775_ans_46df915134884bbfbaf1f1b624e6e60a.PNG

Evaluate

$\textbf{Step -1: Integrating separately.}$

$\int_1^3 (2x^2+5x)dx = \int_1^3 2x^2\ dx+ \int_1^3 5x\ dx$

$\text{W.K.T., } \int x^n dx = \dfrac{{x^{n+1}}}{{n+1}}$

$= 2 \left[\dfrac{{x^3}}{{3}} \right]_1^3 + 5 \left[\dfrac{{x^2}}{{2}} \right]_1^3$

$\textbf{Step -2: Substituting the integral values.}$

$= 2\left[\dfrac{{27}}{{3}} - \dfrac{{1}}{{3}} \right] +5 \left[ \dfrac{{9}}{{2}}-\dfrac{{1}}{{2}} \right]$

$= 2 \left[\dfrac{{26}}{{3}} \right] + 5 \left[ \dfrac{{8}}{{2}} \right]$

$= \dfrac{{52}}{{3}} + \dfrac{{40}}{{2}}$

$= \dfrac{{104 + 120}}{{6}}$

$= \dfrac{{224}}{{6}}$

$=\dfrac{{112}}{{3}}$

$\textbf{Hence, the value of } \mathbf{ \int_1^3 (2x^2+5x)dx = \dfrac{{112}}{{3}}.}$

Find $\displaystyle \int \dfrac{x^{4}dx}{(x-1)(x^{2}+1)}$

$\int { \dfrac { { x }^{ 4 }dx }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } }$

$=\int { \dfrac { \left( { x }^{ 4 }-1+1 \right) }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } } dx$

$=\int { \dfrac { \left( { x }^{ 4 }-1 \right) }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } } dx+\int { \dfrac { dx }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } } dx$

$=\int { \dfrac { \left( { x }^{ 2 }+1 \right) \left( x+1 \right) \left( x-1 \right) }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } } dx+\int { \dfrac { dx }{ \left( x-1 \right) \left( { x }^{ 2 }+1 \right) } }$

$=\int { \left( x+1 \right) } dx+\dfrac{1}{2}\int { \left( \dfrac { 1 }{ { x }^{ 2 }-1 } -\dfrac { x+1 }{ { x }^{ 2 }+1 } \right) } dx$

$=\int { \left( x+1 \right) } dx+\dfrac{1}{2}\int { \left( \dfrac { 1 }{ { x }-1 } -\dfrac { x }{ { x }^{ 2 }+1 } -\dfrac { 1 }{ { x }^{ 2 }+1 } \right) } dx$

$=\dfrac{x^2}{2}+x+\dfrac{1}{2}\int \dfrac{1}{x-1}dx-\dfrac{1}{2}\int \dfrac{x}{x^2+1}dx-\dfrac{1}{2}\int \dfrac{dx}{x^2+1}$

$=\dfrac{x^2}{2}+x+\dfrac{1}{2}ln\left| x-1\right| -\dfrac{1}{4}ln \left| x^2+1\right| -\dfrac{1}{2}\tan ^{-1}x +c$

Hence, solved.

Let $f\left(y\right)={e}^{y},\,\,g\left(y\right)=y,\,\,y>0$ then $F\left(t\right)=\displaystyle\int_{0}^{t}{f\left(t-y\right)g\left(y\right)dy}=$

1353923_1258722_ans_0240259fd98a4efa82df84d0ccda8d21.PNG

Integrate:
$\int { \frac { dx }{ 3cosx+4sinx+6 } }$

1174670_1297399_ans_b1f46abfe3394f2cb39aec69b2925f44.jpg

$I=\displaystyle \int x^{3}\log{x}\ dx$

$\displaystyle\int x^3log x\cdot dx$

$\displaystyle\int udv=uv-\displaystyle\int v.du$

$u=log x$ ;

$dv=x^3dx$

$du=\dfrac{1}{x}dx$

$v=\dfrac{x^4}{4}$

$\displaystyle\int x^3 log x=\dfrac{x^4}{4}log(x)-\displaystyle\int \dfrac{x^4}{4}\cdot \dfrac{1}{x}dx$

$=\dfrac{x^4}{4}log x-\dfrac{1}{4}\displaystyle\int x^3dx$

$=\dfrac{x^4}{4}log x-\dfrac{1}{4}\dfrac{x^4}{4}$

$=\dfrac{x^4}{4}[log x-\dfrac{1}{4}]+c$ .

1201606_1280373_ans_4057b9d1d9f746e6a9b92b5381c70f9e.jpg

$\int\dfrac{dx}x({x^{3/2}+1})$ Solve it.

$\begin{array}{l} \int { \dfrac { { \left( { { x^{ \frac { 3 }{ 2 } } }+1 } \right) } }{ x } dx } \\ \Rightarrow \int { { x^{ \frac { 1 }{ 2 } } }dx } +\int { \dfrac { 1 }{ x } dx } \\ \Rightarrow \dfrac { { { x^{ \frac { 3 }{ 2 } } } } }{ { \frac { 3 }{ 2 } } } +\log \left( x \right) +c \\ \Rightarrow \dfrac { 2 }{ 3 } { x^{ \frac { 3 }{ 2 } } }+\log \left( x \right) +C. \end{array}$

Hence, this is the answer.

Evaluate $\int (\log x)^{2}dx$ .

$\begin{array}{l} \int { { { \left( { \log x } \right) }^{ 2 } } } \\ 4={ \left( { \log x } \right) ^{ 2 } } \\ du=\frac { { 2;ogx } }{ x } dx\, \, \, \left[ { \begin{array} { *{ 20 }{ c } }{ V=x } \\ { DV=dx } \end{array} } \right] \\ \int { { { \left( { \log x } \right) }^{ 2 } } } dx=\int { udv=uv-\int { udv } } \\ =x{ \left( { \log x } \right) ^{ 2 } }-2\left( { x\log x-x } \right) +c \\ =x{ \left( { \log x } \right) ^{ 2 } }2x\log x+2x+c \end{array}$

Evaluate:
$\displaystyle\int \dfrac {\sin x\ dx}{\sin^{3}x+\cos^{3}x}$

Consider,

$I=\displaystyle\int{\dfrac{\sin{x}}{{\sin}^{3}{x}+{\cos}^{3}{x}}dx}$

Dividing the numerator and denominator by

${\cos}^{3}{x}$

$I=\displaystyle\int{\dfrac{\dfrac{\sin{x}}{{\cos}^{3}{x}}}{{\tan}^{3}{x}+1}dx}$

Put

$t=\tan{x}\Rightarrow dt={\sec}^{2}{x}dx$

$\Rightarrow dx=\dfrac{dt}{{\sec}^{2}{x}}$

Substituting the value of

$dx$ above we get

$I=\displaystyle\int{\dfrac{t{\sec}^{2}{x}}{{t}^{3}+1}\dfrac{dt}{{\sec}^{2}{x}}}$

$=\displaystyle\int{\dfrac{t.\,dt}{{t}^{3}+1}}$

Using

${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$

$=\displaystyle\int{\dfrac{t.\,dt}{\left(t+1\right)\left({t}^{2}-t+1\right)}}$

Let

$\dfrac{t}{{t}^{2}-t+1}=\dfrac{A}{t+1}+\dfrac{Bt+C}{{t}^{2}-t+1}$

$\Rightarrow \dfrac{t}{{t}^{2}-t+1}=\dfrac{A\left({t}^{2}-t+1\right)+\left(t+1\right)\left(Bt+C\right)}{\left({t}^{2}-t+1\right)\left(t+1\right)}$

Simplifying we get :

$t={t}^{2}\left(A+B\right)+t\left(A+B+C\right)+A+C$

Comparing coefficients we get :

$A+B=0,\,\,A+B+C=1,\,\,A+C=0$

Solving the three equations simultaneously, we get

$A=\dfrac{−1}{3},\,\,B=C=\dfrac{1}{3}$

Putting the values of

$A,B$ and

$C$ our integral becomes,

$\dfrac{1}{3}\displaystyle\int{\dfrac{t+1}{\left({t}^{2}-t+1\right)}dt}-\dfrac{1}{3}\displaystyle\int{\dfrac{dt}{t+1}}$

$=\displaystyle\int{\dfrac{1}{6}\left(\dfrac{2t-1}{{t}^{2}-t+1}+\dfrac{3}{{t}^{2}-t+1}\right)}-\dfrac{\log{t}+1}{3}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{{t}^{2}-t+1}}+\dfrac{\log{{t}^{2}}-t+1}{6}-\dfrac{\log{t}+1}{3}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{{t}^{2}-2\times t\times \dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1}}+\dfrac{\log{{t}^{2}}-t+1}{6}-\dfrac{\log{t}+1}{3}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{{\left(t-\dfrac{1}{2}\right)}^{2}+{\left(\dfrac{\sqrt{3}}{2}\right)}^{2}}}+\dfrac{\log{{t}^{2}}-t+1}{6}-\dfrac{\log{t}+1}{3}$

$=\dfrac{{\tan}^{-1}{\left(\dfrac{2t-1}{\sqrt{3}}\right)}}{\sqrt{3}}+\dfrac{\log{{t}^{2}}-t+1}{6}-\dfrac{\log{t}+1}{3}+c$

$=\dfrac{{\tan}^{-1}{\left(\dfrac{2\tan{x}-1}{\sqrt{3}}\right)}}{\sqrt{3}}+\dfrac{\log{{\tan}^{2}{x}}-\tan{x}+1}{6}-\dfrac{\log{\tan{x}}+1}{3}+c$ where

$t=\tan{x}$

$\int _ { - 5 } ^ { 5 } | x + 2 | d x$

${\textbf{Step -1: Find limit of given function}}$

${\text{Let I}}$ $= \int_{ - 5}^5 {\mid x + 2\mid } \;dx$

${\text{we know}}$ $\left| {x + 2} \right| = \left \{\begin{gathered} - \left( {x + 2} \right),x \leqslant - 2 \hfill \\ x + 2;x > - 2 \hfill \\ \end{gathered} \right.$

${\textbf{Step -2: Calculation}}$

${\text{So I}} = \int_{ - 5}^{ - 2} {\mid x + 2\mid dx} + \int_{ - 2}^5 {\mid x + 2\mid dx}$

$= \int_{ - 5}^{ - 2} { - (x + 2)dx + \int_{ - 2}^5 {(x + 2)dx} }$

$= {\left[ { - \dfrac{{{x^2}}}{2} - 2x} \right]_{ - 5}}^{- 2} + \left[ {\dfrac{{{x^2}}}{2} + 2x} \right]_{ - 2}^5$

$= \left[ { - \dfrac{4}{2} + 4 + \dfrac{{25}}{2} - 10} \right] + \left[ {\dfrac{{25}}{2} + 10 - \dfrac{4}{2} + 4} \right]$

$= \left[ { - 2 + 4 + \dfrac{{25}}{2} - 10} \right] + \left[ {\dfrac{{25}}{2} + 14 - 2} \right]$

$= 4 + \dfrac{{25}}{2} - 12 + 12 + \dfrac{{25}}{2}$

$= \dfrac{{8 + 25 + 25}}{2}$

$= \dfrac{{58}}{2}$

$= 29$

${\textbf{So, the correct answer is 29}}{\text{.}}$

Prove that $\displaystyle \int^b_a f(x)dx = \int^b_a f(a + b - x) dx$ and evaluate $\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\dfrac{dx}{1+\sqrt{\tan x}}$

2070327_1501139_ans_4daf4ea3c4d24fe1a380b825ec68a0d5.jpg

Evaluate: $\displaystyle \int \sqrt{4-3x-2x^{2}} dx$

1237254_1502390_ans_be00db3a58284ddc84c0c44871a207f4.PNG

Evaluate the following:
$\displaystyle \int e^{-3x} \,cos^3 \,xdx$

Given

$I=\int { { e }^{ -3x }{ cos }^{ 3 }x } dx$

$\therefore I=\int { { e }^{ -3x }\left( \frac { cos3x+3cosx }{ 4 } \right) } dx$

$\therefore I=\frac { 1 }{ 4 } \int { { e }^{ -3x }\left( cos3x+3cosx \right) } dx$

$\therefore I=\frac { 1 }{ 4 } \left[ \int { { e }^{ -3x } } cos3xdx+3\int { { e }^{ -3x } } cosxdx \right]$ (1)

Let

${ I }_{ 1 }=\int { { e }^{ -3x } } cos3xdx$

$\therefore { I }_{ 1 }=cos3x\int { { e }^{ -3x } } dx-\int { \left[ \frac { d }{ dx } \left( cos3x \right) \int { { e }^{ -3x } } dx \right] dx }$

$\therefore { I }_{ 1 }=cos3x\frac { { e }^{ -3x } }{ -3 } -\int { \left[ -3sin3x\left( \frac { { e }^{ -3x } }{ -3 } \right) \right] dx }$

$\therefore { I }_{ 1 }=\frac { { e }^{ -3x }cos3x }{ -3 } -\int { \left[ { e }^{ -3x }sin3x \right] dx }$

$\therefore { I }_{ 1 }=\frac { { e }^{ -3x }cos3x }{ -3 } -\left\{ sin3x\int { { e }^{ -3x }dx } -\int { \frac { d }{ dx } \left( sin3x \right) \int { { e }^{ -3x }dx } } \right\}$

$\therefore { I }_{ 1 }=\frac { { e }^{ -3x }cos3x }{ -3 } -\left\{ sin3x\frac { { e }^{ -3x } }{ -3 } -\int { 3cos3x } \frac { { e }^{ -3x } }{ -3 }dx \right\}$

$\therefore { I }_{ 1 }=\frac { { e }^{ -3x }cos3x }{ -3 } -\left\{ \frac { { e }^{ -3x }sin3x }{ -3 } +\int { { e }^{ -3x }cos3xdx } \right\}$

$\therefore { I }_{ 1 }=\frac { { e }^{ -3x }cos3x }{ -3 } -\left\{ \frac { { e }^{ -3x }sin3x }{ -3 } +{ I }_{ 1 } \right\}$

$\therefore { I }_{ 1 }=\frac { { e }^{ -3x }cos3x }{ -3 } +\frac { { e }^{ -3x }sin3x }{ 3 } -{ I }_{ 1 }$

$\therefore 2{ I }_{ 1 }=\frac { { e }^{ -3x } }{ 3 } \left[ sin3x-cos3x \right]$

$\therefore { I }_{ 1 }=\frac { { e }^{ -3x } }{ 6 } \left[ sin3x-cos3x \right]$

Now, let

${ I }_{ 2 }=\int { { e }^{ -3x } } cosxdx$

$\therefore { I }_{ 2 }=cosx\int { { e }^{ -3x } } dx-\int { \left[ \frac { d }{ dx } \left( cosx \right) \int { { e }^{ -3x } } dx \right] dx }$

$\therefore { I }_{ 2 }=cosx\frac { { e }^{ -3x } }{ -3 } -\int { \left[ -sinx\left( \frac { { e }^{ -3x } }{ -3 } \right) \right] dx }$

$\therefore { I }_{ 2 }=\frac { { e }^{ -3x }cosx }{ -3 } -\frac { 1 }{ 3 } \int { { e }^{ -3x }sinxdx }$

$\therefore { I }_{ 2 }=\frac { { e }^{ -3x }cosx }{ -3 } -\frac { 1 }{ 3 } \left\{ sinx\int { { e }^{ -3x }dx } -\int { \left[ \frac { d }{ dx } \left( sinx \right) \int { { e }^{ -3x }dx } \right] dx } \right\}$

$\therefore { I }_{ 2 }=\frac { { e }^{ -3x }cosx }{ -3 } -\frac { 1 }{ 3 } \left\{ sinx\frac { { e }^{ -3x } }{ -3 } -\int { \left[ cosx\left( \frac { { e }^{ -3x } }{ -3 } \right) \right] } dx \right\}$

$\therefore { I }_{ 2 }=\frac { { e }^{ -3x }cosx }{ -3 } -\frac { 1 }{ 3 } \left\{ sinx\frac { { e }^{ -3x } }{ -3 } +\frac { 1 }{ 3 } \int { { e }^{ -3x }cosx } dx \right\}$

$\therefore { I }_{ 2 }=\frac { { e }^{ -3x }cosx }{ -3 } +\frac { 1 }{ 9 } { e }^{ -3x }sinx-\frac { 1 }{ 9 } { I }_{ 2 }$

$\therefore { I }_{ 2 }+\frac { 1 }{ 9 } { I }_{ 2 }=\frac { { e }^{ -3x }cosx }{ -3 } +\frac { 1 }{ 9 } { e }^{ -3x }sinx$

$\therefore \frac { 10 }{ 9 } { I }_{ 2 }=\frac { { e }^{ -3x }cosx }{ -3 } +\frac { 1 }{ 9 } { e }^{ -3x }sinx$

$\therefore { I }_{ 2 }=\frac { 9 }{ 10 } \left[ \frac { { e }^{ -3x }cosx }{ -3 } +\frac { 1 }{ 9 } { e }^{ -3x }sinx \right]$

$\therefore { I }_{ 2 }=\frac { -3 }{ 10 } { e }^{ -3x }cosx+\frac { 1 }{ 10 } { e }^{ -3x }sinx$

$\therefore { I }_{ 2 }=\frac { { e }^{ -3x } }{ 10 } \left[ sinx-3cosx \right]$

Thus, from (1),

$I=\frac { 1 }{ 4 } \left[ \frac { { e }^{ -3x } }{ 6 } \left[ sin3x-cos3x \right] +3\frac { { e }^{ -3x } }{ 10 } \left[ sinx-3cosx \right] \right]$

$\therefore I=\frac { { e }^{ -3x } }{ 24 } \left[ sin3x-cos3x \right] +\frac { 3{ e }^{ -3x } }{ 40 } \left[ sinx-3cosx \right]$

Evaluate the following integrals:
$\int { \cfrac { 1 }{ 5+4\cos { x } } } dx$

Let

$\text{I}=\displaystyle\int \dfrac{1}{5+4\cos x} d x$

$=\displaystyle\int \dfrac{1}{5+4\bigg(\dfrac{1-\tan^2 (x/2)}{1+\tan^2 (x/2)}\bigg)}d x$

$\bigg(\because \cos 2 A=\dfrac{1-\tan^2A}{1+\tan^2 A}\bigg)$

$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{5\bigg(1+\tan^2 \dfrac{x}{2}\bigg)+4\bigg(1-\tan^2 \dfrac{x}{2}\bigg)}d x$

$=\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{9+\tan^2 \dfrac{x}{2}}d x$

$(\because \text{sec}^2 A-\tan^2 A=1)$

Let

$t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$

$I=\displaystyle\int \dfrac{2 d t}{t^2+9}$

$=2\displaystyle\int \dfrac{d t}{t^2+3^2}$

$=\dfrac{2}{3}\text{tan}^{-1} \dfrac{t}{3}+c$

$\bigg(\because \displaystyle\int \dfrac{d x}{x^2+a^2}=\dfrac{1}{a}\text{tan}^{-1} \dfrac{x}{a}+c\ \bigg)$

$=\dfrac{2}{3}\text{tan}^{-1} \bigg(\dfrac{1}{3}\tan \dfrac{x}{2}\bigg)+c$

$\bigg(\because t=\tan \dfrac{x}{2}\bigg)$

If f(x) is a function satisfying $f\left( \dfrac{1}{x}\right)+x^2f(x)=0$ for all non-zero x, then $\displaystyle \int_{sin\Theta}^{cosec\Theta}f(x)dx$ is equal to

1715102_1775675_ans_8c0f87d8e2de46a4a467a80603135415.jpg

Evaluate : $\int_0^{\pi/2}{\cfrac{x}{\sin{x}+\cos{x}}}dx$

2042808_1534813_ans_fad32f65af8d404695381482d7bd2261.jpg

Integrals - Class 12 Commerce Maths - Extra Questions

Evaluate :∫√1−cos2x1+cos2xdx\displaystyle\int \sqrt{\dfrac{1-\cos 2x}{1+\cos 2x}}dx

Evaluate : ∫0−1(e3x+7x−5)dx\int_{-1}^{0} (e^{3x}+7x-5)dx.

∫10xex(1+x)2dx\int_{0}^{1} \dfrac {xe^{x}}{(1 + x)^{2}}dx.

Find ∫sin6xcos8x dx\displaystyle\int \dfrac{\sin^6x}{\cos^8x} \ dx

Evaluate the integrals∫201(x2+1)dx\int _{ 0 }^{ 2 }{ \cfrac { 1 }{ \left( { x }^{ 2 }+1 \right) } dx }

Evaluate : ∫102x+35x2+1dx\displaystyle \int_{0}^{1}\dfrac {2x + 3}{5x^{2} + 1} dx

Evaluate :\int \dfrac{2\cos x}{3 \sin^2x} \ dx\int \dfrac{2\cos x}{3 \sin^2x} \ dx

Evaluate :\displaystyle\int \dfrac{x^3-x^2+x-1}{x-1}dx\displaystyle\int \dfrac{x^3-x^2+x-1}{x-1}dx

\int { \dfrac { 3x }{ 3x-1 } dx } \int { \dfrac { 3x }{ 3x-1 } dx }

Evaluate :\int \dfrac{dx}{x(x^5+3)}\int \dfrac{dx}{x(x^5+3)}

The acceleration of a motorcycle is given by a_{x}(t) = At - Br^{2}a_{x}(t) = At - Br^{2}, where A = 1.50\ ms^{-2}A = 1.50\ ms^{-2} and B = 0.120\ ms^{-4}B = 0.120\ ms^{-4}. The motorcycle is at rest at the origin at time t = 0t = 0.Find its position and velocity as functions of time.

\displaystyle _{0}^{5}\int \dfrac{4\sqrt{x+4}}{4\sqrt{x+4}+4\sqrt{9-x}}\displaystyle _{0}^{5}\int \dfrac{4\sqrt{x+4}}{4\sqrt{x+4}+4\sqrt{9-x}}

Evaluate \int {x\ln \sqrt {xdx} } \int {x\ln \sqrt {xdx} }

Solve: \displaystyle\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin^5 x.\cos^{100} x \,dx \displaystyle\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin^5 x.\cos^{100} x \,dx

\int_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}} \int_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}}

\int \dfrac {1}{\sqrt {3x+5}-\sqrt {3x+2}}dx\int \dfrac {1}{\sqrt {3x+5}-\sqrt {3x+2}}dx

Solve \displaystyle \int { \dfrac { 1 }{ \log _{ x }{ e } } dx }\displaystyle \int { \dfrac { 1 }{ \log _{ x }{ e } } dx }

\displaystyle \int_{0}^\cfrac{\pi}{4} \sin 3x \sin 2x\ dx\displaystyle \int_{0}^\cfrac{\pi}{4} \sin 3x \sin 2x\ dx

Evaluate \int {\dfrac {\sin x}{\sin (x+a)}}dx\int {\dfrac {\sin x}{\sin (x+a)}}dx

\displaystyle\int e^{log x}dx\displaystyle\int e^{log x}dx.

Integrate with respect to xx:\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx

Solve \displaystyle\int\limits_0^x {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {8 - x} }}dx} \displaystyle\int\limits_0^x {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {8 - x} }}dx}

Solve : \int \dfrac{dx}{x(n + 1) (n + 2)}\int \dfrac{dx}{x(n + 1) (n + 2)}

Solve:\int {\dfrac{1}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right|} \int {\dfrac{1}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right|}

Evaluate \int \dfrac {1}{a^{x}b^{x}}dx\int \dfrac {1}{a^{x}b^{x}}dx

Solve\displaystyle \int \dfrac {1-x^{2}}{x(1-2x)} = \displaystyle \int \dfrac {1-x^{2}}{x(1-2x)} =

Solve \displaystyle \overset{4}{\underset{2}{\int}} 2^x dx\displaystyle \overset{4}{\underset{2}{\int}} 2^x dx.

The value of \int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { dx }{ 1+\tan ^{ 3 }{ x } } } \int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { dx }{ 1+\tan ^{ 3 }{ x } } } is

\int {x^2}{e^x} dx\int {x^2}{e^x} dx =

Solve:\frac{1}{2}\int {\frac{{2x}}{{{{\left( {1 + x} \right)}^2}}}dx} \frac{1}{2}\int {\frac{{2x}}{{{{\left( {1 + x} \right)}^2}}}dx}

\int{\dfrac{{x}^{2}dx}{{(1+{x}^{2})}^{2}}}\int{\dfrac{{x}^{2}dx}{{(1+{x}^{2})}^{2}}}.

Integerate:\int {x\,\cos \,x\,dx} \int {x\,\cos \,x\,dx}

\int \dfrac {\cos 2x}{\cos^{2}x.\sin^{2}x}dx\int \dfrac {\cos 2x}{\cos^{2}x.\sin^{2}x}dx

Solve \displaystyle\int_{}^{} {\dfrac{{\left( {3{{\tan }^2}x + 2} \right){{\sec }^2}x}}{{{{\left( {{{\tan }^3}x + 2\tan x + 5} \right)}^2}}}dx} \displaystyle\int_{}^{} {\dfrac{{\left( {3{{\tan }^2}x + 2} \right){{\sec }^2}x}}{{{{\left( {{{\tan }^3}x + 2\tan x + 5} \right)}^2}}}dx}

Solve \displaystyle\int\limits_b^a {\frac{x}{{\sqrt {{a^2} + {x^2}} }}dx} \displaystyle\int\limits_b^a {\frac{x}{{\sqrt {{a^2} + {x^2}} }}dx}

\int x{\sin ^{ - 1}}x\int x{\sin ^{ - 1}}x

Evalute \int\limits_0^a x {\left( {{a^2} - {x^2}} \right)^{\frac{7}{2}}}dx\int\limits_0^a x {\left( {{a^2} - {x^2}} \right)^{\frac{7}{2}}}dx

Evaluate: \int \,lnx \cdot \dfrac{1}{(x + 1)^2}dx\int \,lnx \cdot \dfrac{1}{(x + 1)^2}dx equals

The value of \displaystyle\int_1^{37\pi } {\dfrac{{\pi \sin \left( {\pi \log x} \right)}}{x}} \,dx\displaystyle\int_1^{37\pi } {\dfrac{{\pi \sin \left( {\pi \log x} \right)}}{x}} \,dx

\int \cot^{2}xdx = \int \cot^{2}xdx =

\displaystyle\int 2x\sqrt{x} dx\displaystyle\int 2x\sqrt{x} dx.

\displaystyle\int \left(\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2dx\displaystyle\int \left(\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2dx.

\displaystyle\int \dfrac{1}{1+\cot}dx\displaystyle\int \dfrac{1}{1+\cot}dx.

Solve : \displaystyle \int \dfrac{x^2}{(4 + x)^{3/2}} dx\displaystyle \int \dfrac{x^2}{(4 + x)^{3/2}} dx

Evaluate :\displaystyle \int x.(\log x)^2 \ dx\displaystyle \int x.(\log x)^2 \ dx

Solve : \displaystyle \int\limits_{-\pi/2}^{\pi/2} \sin^7x \ dx\displaystyle \int\limits_{-\pi/2}^{\pi/2} \sin^7x \ dx

Evaluate :\displaystyle \int\limits_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\displaystyle \int\limits_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx

Evaluate : \displaystyle \int\limits_{0}^{\pi/4} \log(1+\tan x)dx\displaystyle \int\limits_{0}^{\pi/4} \log(1+\tan x)dx

Evaluate :\displaystyle \int x \ \sin x \ dx\displaystyle \int x \ \sin x \ dx

Solve :-I = \displaystyle \int_0^{\infty}\dfrac{dx}{e^x+e^{-x}}I = \displaystyle \int_0^{\infty}\dfrac{dx}{e^x+e^{-x}}

\displaystyle\int x\left(\dfrac{\sec 2x-1}{\sec 2x+1}\right)dx\displaystyle\int x\left(\dfrac{\sec 2x-1}{\sec 2x+1}\right)dx.

Evaluate:\int x.\sin{2x}dx\int x.\sin{2x}dx

Evaluate :\displaystyle\int_{0}^{\pi} \dfrac{1}{1+\sin x} dx\displaystyle\int_{0}^{\pi} \dfrac{1}{1+\sin x} dx

\displaystyle\int e^x\left(\dfrac{x-1}{x^2}\right)dx\displaystyle\int e^x\left(\dfrac{x-1}{x^2}\right)dx.

\displaystyle\int (x+1)e^xlog (xe^x)dx\displaystyle\int (x+1)e^xlog (xe^x)dx.

\displaystyle\int x \sin^3x dx\displaystyle\int x \sin^3x dx.

Solve: \int_0^1 x^3 \sqrt{1 +3x^4 } dx \int_0^1 x^3 \sqrt{1 +3x^4 } dx

Integrate:\int \sin^4xdx\int \sin^4xdx

\int \frac{2x^{2}}{3x^{4}2x} dx\int \frac{2x^{2}}{3x^{4}2x} dx

\int {\dfrac{{{e^x}dx}}{{\left( {1 + {e^{2x}}} \right)}}} \int {\dfrac{{{e^x}dx}}{{\left( {1 + {e^{2x}}} \right)}}}

Evaluate :\displaystyle \int e^{ax} \cos(bx+c) \ dx\displaystyle \int e^{ax} \cos(bx+c) \ dx

Evaluate; \int {\left( {\frac{{3{x^4} - 5{x^3} + 4{x^2} - x + 2}}{{{x^3}}}} \right)dx} \int {\left( {\frac{{3{x^4} - 5{x^3} + 4{x^2} - x + 2}}{{{x^3}}}} \right)dx}

I = \int\limits_0^1 {\frac{{\left( {1 - x} \right)dx}}{{\left( {1 + x} \right)}}} I = \int\limits_0^1 {\frac{{\left( {1 - x} \right)dx}}{{\left( {1 + x} \right)}}}

Calculate \int \frac{x}{x^{2}+1}dx\int \frac{x}{x^{2}+1}dx

Solve: \displaystyle \int \dfrac{x^3}{2x + 1}dx\displaystyle \int \dfrac{x^3}{2x + 1}dx.

Evaluate: \displaystyle\int \dfrac { \dfrac{- 1} 6 } { ( x + 3 ) } d x + \int \dfrac { \dfrac1 6 } { ( x - 3 ) } d x \displaystyle\int \dfrac { \dfrac{- 1} 6 } { ( x + 3 ) } d x + \int \dfrac { \dfrac1 6 } { ( x - 3 ) } d x

Find the following integrals:\int (ax^{2} + bx + c) dx\int (ax^{2} + bx + c) dx.

Evaluate \int \sqrt{1+t^{3}} dt\int \sqrt{1+t^{3}} dt

Solve:\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)dx} \int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)dx}

Evaluate \displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx

Evaluate \int_0^n { xdx} \int_0^n { xdx} .

\displaystyle \int { \dfrac { 1 }{ \sqrt { 9-25{ x }^{ 2 } } } } dx\displaystyle \int { \dfrac { 1 }{ \sqrt { 9-25{ x }^{ 2 } } } } dx

Evaluate:\int \dfrac{\sqrt{x^{2}-a^{2}}}{x}dx\int \dfrac{\sqrt{x^{2}-a^{2}}}{x}dx

Prove that:\displaystyle \int x^5(x+7)dx\displaystyle \int x^5(x+7)dx

Evaluate \displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt

Evaluate : \int _{ 0 }^{ \pi /2 }{ sinxdx } \int _{ 0 }^{ \pi /2 }{ sinxdx } .

Solve \int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } x d x\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } x d x

Solve it 2I=\int _{ O }^{ Q }{ dx } 2I=\int _{ O }^{ Q }{ dx }

Evaluate:\displaystyle\int\limits_{0}^{1} \dfrac{dx}{1+x^2}\displaystyle\int\limits_{0}^{1} \dfrac{dx}{1+x^2}

Evaluate :
$\displaystyle\int \sqrt{\dfrac{1-\cos 2x}{1+\cos 2x}}dx$

Evaluate : $\int_{-1}^{0} (e^{3x}+7x-5)dx$ .

$\int_{0}^{1} \dfrac {xe^{x}}{(1 + x)^{2}}dx$ .

Find $\displaystyle\int \dfrac{\sin^6x}{\cos^8x} \ dx$

Evaluate the integrals
$\int _{ 0 }^{ 2 }{ \cfrac { 1 }{ \left( { x }^{ 2 }+1 \right) } dx }$

Evaluate : $\displaystyle \int_{0}^{1}\dfrac {2x + 3}{5x^{2} + 1} dx$

Evaluate :
$\int \dfrac{2\cos x}{3 \sin^2x} \ dx$

Evaluate :
$\displaystyle\int \dfrac{x^3-x^2+x-1}{x-1}dx$

$\int { \dfrac { 3x }{ 3x-1 } dx }$

Evaluate :
$\int \dfrac{dx}{x(x^5+3)}$

The acceleration of a motorcycle is given by $a_{x}(t) = At - Br^{2}$ , where $A = 1.50\ ms^{-2}$ and $B = 0.120\ ms^{-4}$ . The motorcycle is at rest at the origin at time $t = 0$ .
Find its position and velocity as functions of time.

$\displaystyle _{0}^{5}\int \dfrac{4\sqrt{x+4}}{4\sqrt{x+4}+4\sqrt{9-x}}$

Evaluate $\int {x\ln \sqrt {xdx} }$

Solve: $\displaystyle\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin^5 x.\cos^{100} x \,dx$

$\int_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}}$

$\int \dfrac {1}{\sqrt {3x+5}-\sqrt {3x+2}}dx$

Solve $\displaystyle \int { \dfrac { 1 }{ \log _{ x }{ e } } dx }$

$\displaystyle \int_{0}^\cfrac{\pi}{4} \sin 3x \sin 2x\ dx$

Evaluate $\int {\dfrac {\sin x}{\sin (x+a)}}dx$

$\displaystyle\int e^{log x}dx$ .

Integrate with respect to $x$ :
$\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx$

Solve $\displaystyle\int\limits_0^x {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {8 - x} }}dx}$

Solve : $\int \dfrac{dx}{x(n + 1) (n + 2)}$

Solve:
$\int {\dfrac{1}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right|}$

Evaluate $\int \dfrac {1}{a^{x}b^{x}}dx$

Solve
$\displaystyle \int \dfrac {1-x^{2}}{x(1-2x)} =$

Solve $\displaystyle \overset{4}{\underset{2}{\int}} 2^x dx$ .

The value of $\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { dx }{ 1+\tan ^{ 3 }{ x } } }$ is

$\int {x^2}{e^x} dx$ =

Solve:
$\frac{1}{2}\int {\frac{{2x}}{{{{\left( {1 + x} \right)}^2}}}dx}$

$\int{\dfrac{{x}^{2}dx}{{(1+{x}^{2})}^{2}}}$ .

Integerate:
$\int {x\,\cos \,x\,dx}$

$\int \dfrac {\cos 2x}{\cos^{2}x.\sin^{2}x}dx$

Solve $\displaystyle\int_{}^{} {\dfrac{{\left( {3{{\tan }^2}x + 2} \right){{\sec }^2}x}}{{{{\left( {{{\tan }^3}x + 2\tan x + 5} \right)}^2}}}dx}$

Solve $\displaystyle\int\limits_b^a {\frac{x}{{\sqrt {{a^2} + {x^2}} }}dx}$

$\int x{\sin ^{ - 1}}x$

Evalute $\int\limits_0^a x {\left( {{a^2} - {x^2}} \right)^{\frac{7}{2}}}dx$

Evaluate: $\int \,lnx \cdot \dfrac{1}{(x + 1)^2}dx$ equals

The value of $\displaystyle\int_1^{37\pi } {\dfrac{{\pi \sin \left( {\pi \log x} \right)}}{x}} \,dx$

$\int \cot^{2}xdx =$

$\displaystyle\int 2x\sqrt{x} dx$ .

$\displaystyle\int \left(\dfrac{e^x+e^{-x}}{e^x-e^{-x}}\right)^2dx$ .

$\displaystyle\int \dfrac{1}{1+\cot}dx$ .

Solve : $\displaystyle \int \dfrac{x^2}{(4 + x)^{3/2}} dx$

Evaluate :
$\displaystyle \int x.(\log x)^2 \ dx$

Solve : $\displaystyle \int\limits_{-\pi/2}^{\pi/2} \sin^7x \ dx$

Evaluate :
$\displaystyle \int\limits_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

Evaluate : $\displaystyle \int\limits_{0}^{\pi/4} \log(1+\tan x)dx$

Evaluate :
$\displaystyle \int x \ \sin x \ dx$

Solve :-
$I = \displaystyle \int_0^{\infty}\dfrac{dx}{e^x+e^{-x}}$

$\displaystyle\int x\left(\dfrac{\sec 2x-1}{\sec 2x+1}\right)dx$ .

Evaluate:

$\int x.\sin{2x}dx$

Evaluate :

$\displaystyle\int_{0}^{\pi} \dfrac{1}{1+\sin x} dx$

$\displaystyle\int e^x\left(\dfrac{x-1}{x^2}\right)dx$ .

$\displaystyle\int (x+1)e^xlog (xe^x)dx$ .

$\displaystyle\int x \sin^3x dx$ .

Solve:
$\int_0^1 x^3 \sqrt{1 +3x^4 } dx$

Integrate:
$\int \sin^4xdx$

$\int \frac{2x^{2}}{3x^{4}2x} dx$

$\int {\dfrac{{{e^x}dx}}{{\left( {1 + {e^{2x}}} \right)}}}$

Evaluate :
$\displaystyle \int e^{ax} \cos(bx+c) \ dx$

Evaluate; $\int {\left( {\frac{{3{x^4} - 5{x^3} + 4{x^2} - x + 2}}{{{x^3}}}} \right)dx}$

$I = \int\limits_0^1 {\frac{{\left( {1 - x} \right)dx}}{{\left( {1 + x} \right)}}}$

Calculate $\int \frac{x}{x^{2}+1}dx$

Solve: $\displaystyle \int \dfrac{x^3}{2x + 1}dx$ .

Evaluate: $\displaystyle\int \dfrac { \dfrac{- 1} 6 } { ( x + 3 ) } d x + \int \dfrac { \dfrac1 6 } { ( x - 3 ) } d x$

Find the following integrals:
$\int (ax^{2} + bx + c) dx$ .

Evaluate
$\int \sqrt{1+t^{3}} dt$

Solve:
$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)dx}$

Evaluate $\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{2}x}{1+3\sin^{2}x}\ dx$

Evaluate $\int_0^n { xdx}$ .

$\displaystyle \int { \dfrac { 1 }{ \sqrt { 9-25{ x }^{ 2 } } } } dx$

Evaluate:
$\int \dfrac{\sqrt{x^{2}-a^{2}}}{x}dx$

Prove that:
$\displaystyle \int x^5(x+7)dx$

Evaluate $\displaystyle\int_{0}^{\pi/4}\sin^{3}2t\cos 2t\ dt$

Evaluate : $\int _{ 0 }^{ \pi /2 }{ sinxdx }$ .

Solve $\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } x d x$

Solve it
$2I=\int _{ O }^{ Q }{ dx }$

Evaluate:
$\displaystyle\int\limits_{0}^{1} \dfrac{dx}{1+x^2}$

Prove that:
$\displaystyle \int \dfrac {1}{4x^{2}-1}dx$