Class 12 Commerce Applied Mathematics - Indefinite Integrals

Integrate the following function with respect to $x$ .
$x\sin { hx }$

Using integration by parts

$\int { \sin { hx } } dx=x\cos { hx } -\int { \cos { hx } } dx=x\cos { hx } -\sin { hx } +constant\quad$

$\Rightarrow \int xsinhx=xcoshx-\int coshxdx$

$\Rightarrow \int x sinhx =xcoshx-sinhx+constant$

Solve :

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

We have,

$I=\int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}$

$........(1)$

Apply property,

$I=\int\limits_a^b f(x) dx=\int\limits_a^bf(a+b-x)dx$

Therefore,

$I=\int\limits_0^a {\dfrac{{\sqrt {a-x} }}{{\sqrt {a-x} + \sqrt {a-a + x} }}dx}$

$I=\int\limits_0^a {\dfrac{{\sqrt {a-x} }}{{\sqrt {a-x} + \sqrt { x} }}dx}$

$.......(2)$

On adding both equations, we get

$2I=\int\limits_0^a {\dfrac{{\sqrt {a-x} +\sqrt x}}{{\sqrt {a-x} + \sqrt {x} }}dx}$

$2I=\int\limits_0^a 1dx$

$2I=[x]_0^a$

$2I=a-0$

$2I=a$

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

Hence, this is the answer.

The value of $\left ( \int x. e ^{-x} dx \right)$ is,

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

$\Rightarrow$ Using

$uv$ rule

consider (LATE)

$u=x, v=e^{-x}$

$\displaystyle\int uv\ dx$

$=u\displaystyle\int v\ dx-\int u^{3}\left(\int v\ dx\right)dx$

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

$=-xe^{-x}-(-(-e^{x}))$

$=-x^{e^{-x}}-e^{-x}$

$=e^{-x}(-x-1)$

The integration of $y=\dfrac{x^2}{2}+c$ . What does $c$ represent?

Given

$\int y=\left[ \dfrac { { x }^{ 2 } }{ 2 } \right] +c$

Here,

$c$ represents the constant of integration.

When we differentiate the constant function it becomes

$0$ .

So in integration, we assume

$c$ as that constant function.

$I= \displaystyle\int{e^x}\left(\frac{2}{\cos^2x}+\frac{2\sin x\cos x}{\cos^2x}\right)dx$

$= \displaystyle\int{e^x}\left(2\sec^2x+2\tan x\right)dx$

$= \displaystyle\int{e^x}2\tan xdx+$

$2 \displaystyle\int{e^x}\sec^2xdx$

$=2e^x\tan x -2\displaystyle\int{e^x}\sec^2 xdx+$

$2 \displaystyle\int{e^x}\sec^2xdx+c$ [

$c$ being integrating constant] [Using integration by parts]

$=2e^x\tan x+c$ .

Integrate the function $x \sin x$

$\displaystyle \int x \sin x dx$
Using by parts,

$\displaystyle =x\int \sin x dx-\int \left \{\left (\frac {d}{dx}x\right )\int \sin x dx\right \}dx$

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

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

Evaluate $\displaystyle \int x^{2}e^{x}dx$ .

$\Rightarrow$ Let

$u=x^2$ and

$v=e^x$ , then

$du=2xdx$

Now i ntegration by parts states that

$\int u(x)\,v'(x)dx=u(x)v(x)-\int v(x)u'(x)dx$

Hence,

$\int x^2 e^xdx=x^2 e^x-\int e^x\times 2xdx$

$\int x^2 e^xdx=x^2e^x-2\int xe^xdx+C$ ------- ( 1 )

Now, we set

$u=x,$ then

$du=dx$

and

$\int xe^x dx=xe^x-\int e^x\times 1\times dx$ or

$\int xe^xdx=xe^x-\int e^xdx=xe^x-e^x$

Putting this in ( 1 ), we get

$\int x^2 e^x dx=x^2 e^x-2(xe^x-e^x)+C$

$\int x^2 e^x dx=e^x(x^2-2x+2)+C$

Object: $\displaystyle\int cosec^{-1}x, dx, x > 1$ .

$I=\displaystyle\int cosec^{-1}xdx, x > 1$

$\rightarrow$ Now taking

$u=cosec^{-1}x$ and

$v=1\therefore \dfrac{du}{dx}=\dfrac{-1}{x\sqrt{x^2-1}}$ and

$\displaystyle\int v dx=\displaystyle \int 1 dx=x$

$\therefore I=u\displaystyle\int v dx-\displaystyle\int \left(\dfrac{du}{dx}\cdot \displaystyle\int v dx\right)dx$
(Rule of integration by parts)

$=x(cosec^{-1}x)-\displaystyle\int \dfrac{-1}{x\sqrt{x^2-1}}\cdot xdx$

$=xcosec^{-1}x+\displaystyle\int \dfrac{1}{\sqrt{x^2-1}}\cdot dx$

$\therefore I=x cosec^{-1}x+log\left|x+\sqrt{x^2-1}\right|+c$ .

The acceleration of a particle varies with time $t$ seconds according to the relation $a = 6t + 6\ ms^{-2}$ . Find velocity and position as functions of time. It is given that the particle starts from origin at $t = 0$ with velocity $2\ ms^{-1}$ .

We know that

$\int dv= \int a\,dt \Rightarrow \int_2^v dv=\int_0^t (6t+6)dt$

$\Rightarrow v-2=\left[\dfrac{6t^2}{2}+6t\right]_0^t \Rightarrow v=3t^2+6t+2$

Now

$\int dx=\int vdt \Rightarrow \int_0^x dx=\int_0^t(3t^2+6t+2)dt$

$\Rightarrow x=t^3+3t^2+2t$

Object: $\displaystyle\int (ex)^x(2+log x)dx$ .

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

$=\displaystyle\int e^x\cdot x^x(2+log x)dx$

$=\displaystyle\int e^x x^x[1+(1+log x)]dx$

$=\displaystyle\int e^x[x^x+x^x(1+log x)]dx$
Now taking

$f(x)=x^x$

$\therefore f'(x)=x^x(1+log x)+c$

$\therefore I=\displaystyle\int e^x(f(x)+f'(x))dx$

$=e^xf(x)+v$

$=e^x\cdot x^2+c$

$I=(e^x)^x+c$ .

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

Now,

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

$=\displaystyle\int e^x\dfrac{1}{x}\ dx-$

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

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

$\displaystyle\int e^x\dfrac{1}{x^2}\ dx$ [ Using method of by parts]

$=\dfrac{e^x}{x}+c$ [ Where

$c$ is integrating constant]

Evaluate:
$\displaystyle\int x^n\log_ex\ dx$

Now,

$\displaystyle\int x^n\log_ex\ dx$

$=\log_ex .\dfrac{x^{n+1}}{n+1}$

$-\displaystyle\int \dfrac{1}{x}\dfrac{x^{n+1}}{n+1}\ dx$

$=\log_ex .\dfrac{x^{n+1}}{n+1}$

$-\dfrac{x^{n+1}}{(n+1)^2}\ +c$ [ Where

$c$ being integrating constant]

Integrate with respect to $x$ :
$x \sin^{2}x$

$\int {\mathop x\limits_{(i)} } \,\,\mathop {{{\sin }^2}x}\limits_{\left( {ii} \right)} \,\,dx$

$= x\int {{{\sin }^2}x\,dx} - \int {1\left\{ {\int {{{\sin }^2}x\,dx} } \right\}} \,dx$

Now,

$\int {{{\sin }^2}x\,dx}$

$= \cfrac{1}{2}\int {2{{\sin }^2}x\,dx}$

$= \cfrac{1}{2}\int {\left( {1 - \cos 2x} \right)dx}$

$= \cfrac{1}{2}\left[ {x - \cfrac{{\sin 2x}}{2}} \right]$

$= \cfrac{x}{2} - \cfrac{{\sin 2x}}{4}$

$\int {x.{{\sin }^2}x\,dx} = x \times \left( {\cfrac{x}{2} - \cfrac{{\sin 2x}}{4}} \right) - \int {\left( {\cfrac{x}{2} - \cfrac{{\sin 2x}}{4}} \right)} \,dx$

$= \cfrac{{{x^2}}}{2} - \cfrac{{x\sin 2x}}{4} - \left[ {\cfrac{{{x^2}}}{4} + \cfrac{{\cos 2x}}{8}} \right] + c$

$= \cfrac{{{x^2}}}{2} - \cfrac{{x\sin 2x}}{4} - \cfrac{{{x^2}}}{4} - \cfrac{{\cos 2x}}{8} + c$

$= \cfrac{{{x^2}}}{4} - \cfrac{{x\sin 2x}}{4} - \cfrac{{\cos 2x}}{8} + c$

$\int {{e^x}.\,{a^x}} dx =$

$\int {{e^x}{a^x}dx}$

${e^x}{a^x} = u$

$dx = \dfrac{1}{{1n\left( a \right){e^x}{a^x} + {e^{x{a^x}}}}}dx$

$\int {\dfrac{1}{{1n\left( a \right) + 1}}dx}$

$\int {\dfrac{1}{{1n\left( a \right) + 1}}\int {1dx} }$

$\dfrac{1}{{1n\left( a \right) + 1}}u + c$

$=\dfrac{1}{{1n\left( a \right) + 1}}\left( {{e^x}{a^x}} \right) + c$

Integrate the following
i) $\log{x}$
ii) ${\cos}^{-1}(\sqrt{x})$

(i)

$\int_{}^{} {\log xdx}$

$= \int_{}^{} {1.\log xdx}$

$= x\log x - \int_{}^{} {\cfrac{1}{x}.xdx}$

$= x\log x - x + c$

(ii)

$\int_{}^{} {{{\cos }^{ - 1}}\sqrt x dx}$

putting

$x = {t^2}$

$\Rightarrow dx = 2tdt$

$= 2\int_{}^{} {t{{\cos }^{ - 1}}tdt}$

$= 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \int_{}^{} {\cfrac{{ - 1}}{{\sqrt {1 - {t^2}} }} \times \cfrac{{{t^2}}}{2}dt} } \right]$

$= 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \cfrac{1}{2}\int_{}^{} {\cfrac{{1 - {t^2} + 1}}{{\sqrt {1 - {t^2}} }}dt} } \right]$

$= 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \cfrac{1}{2}\int_{}^{} {\left( {\sqrt {1 - {t^2}} + \cfrac{1}{{\sqrt {1 - {t^2}} }}} \right)dt} } \right]$

$= {t^2}{\cos ^{ - 1}}t - \int_{}^{} {\sqrt {1 - {t^2}} dt - \int_{}^{} {\cfrac{1}{{\sqrt {1 - {t^2}} }}dt} }$

$= {t^2}{\cos ^{ - 1}}t - \left[ {\cfrac{t}{2}\sqrt {1 - {t^2}} + \cfrac{1}{2}{{\sin }^{ - 1}}t} \right] - {\sin ^{ - 1}}t + c$

$= {t^2}{\cos ^{ - 1}}t - \cfrac{t}{2}\sqrt {1 - {t^2}} - \cfrac{3}{2}{\sin ^{ - 1}}t + c$

$= x{\cos ^{ - 1}}\sqrt x - \cfrac{{\sqrt x }}{2}\sqrt {1 - x} - \cfrac{3}{2}{\sin {^{ - 1}\sqrt x }} + c$

$= x{\cos ^{ - 1}}\sqrt x - \cfrac{{\sqrt {x - {x^2}} }}{2} - \cfrac{3}{2}{\sin {^{ - 1}\sqrt x }} + c$

Evaluate $\int {\dfrac{1}{{\sqrt 3 \sin x + \cos x}}dx}$

We have,

$I=\int{\left( \dfrac{1}{\sqrt{3}\sin x+\cos x} \right)}dx$

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

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

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

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

We know that

$\int{\sec xdx=\log \left| \sec x+\tan x \right|}+C$

Therefore,

$I=\dfrac{1}{2}\left[ \log \left| \sec \left( x-\dfrac{\pi }{3} \right)+\tan \left( x-\dfrac{\pi }{3} \right) \right| \right]+C$

Hence, the value of integral is $\dfrac{1}{2}\left[ \log \left| \sec \left( x-\dfrac{\pi }{3} \right)+\tan \left( x-\dfrac{\pi }{3} \right) \right| \right]+C$ .

Evaluate:
$\int \dfrac{e^x}{\sqrt{5-4e^x-e^{2x}}}dx$

$I=\int \dfrac{e^x}{\sqrt{5-4e^x-e^{2x}}}dx$

Now, put

$e^x=t \implies e^x \ dx=dt$

Therefore,

$I=\int \dfrac{dt}{\sqrt{5-4t-t^2}}-\int \dfrac{dt}{\sqrt{-(t^2+4t-5)}}$

$I=\int \dfrac{dt}{\sqrt{-[(t+2)^2-5-4]}}$

$I=\int \dfrac{dt}{\sqrt{9-(t+2)^2}}=\int \dfrac{dt}{\sqrt{(3)^2-(t+2)^2}}$

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

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

Solve: $\displaystyle \int {e^x}{x^2} - {e^{2x}}dx$

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

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

$=\displaystyle {x^2}{e^x} - \int {2x{e^x} \ dx - {{{e^{2x}}} \over 2} + C}$

$=\displaystyle {x^2}{e^x} - 2\left[ {x{e^x} - \int {1.{e^x}dx} } \right] - {1 \over 2}{e^{2x}} + C$

$=\displaystyle {x^2}{e^x} - 2x{e^x} + 2{e^x} - {{{e^{2x}}} \over 2} + C$

Integrate: $\displaystyle \int {e^x}\left[ {\tan x + \log \,\sec x} \right] dx$

$\int {{e^x}} \tan xdx + \int {e_{II}^x\log {{\sec }_I}xdx}$

$= \int {{e^x}} \tan xdx + {\log \sec x.{e^x} - \int {{{\sec x\tan {x_e}^xdx} \over {\sec x}}} }$

$= \int {{e^x}} \tan xdx + {e^x}\log \sec x - \int {{e^x}\tan xdx}$

$= {e^x}\log \sec x + c$

Solve $\int {x{{\sin }^2}xdx}$

$\int x\sin^2xdx\quad (\because \cos2x=1-2\sin^2x)$

$\int x\left(\dfrac{1-\cos 2x}{2} \right )dx$

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

$=\dfrac{x^2}{4}-\dfrac{1}{2}\left[\dfrac{x\sin 2x}{2}-\int\dfrac{\sin 2x}{2}dx \right ]$

$=\dfrac{x^2}{4}-\dfrac{x}{4}\sin 2x-\dfrac{\cos 2x}{8}+c$

Evaluate $\int {\dfrac{{dx}}{{\sqrt {x + 1} \, - \sqrt x }}}$

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

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

$\int {\sqrt {x + 1} } \;dx + \int {\sqrt x } \;dx$

$= \dfrac{3}{2}{\left( {x + 1} \right)^{\cfrac{3}{2}}} + \dfrac{3}{2}{x^{\cfrac{3}{2}}} + C$

$\int {\dfrac{{x\,{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx}$

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

Let

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

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

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

Substituting this value,

$I=\int \sin t \times t \times \dfrac{dx}{\sqrt{1-x^2}}$

$I=\int \sin t \times t \ dt$

Hence, we take

First function :

$f(x)=t$

Second function :

$g(x)=\sin t$

$I=t \int \sin t \ dt-\int \dfrac{d(t)}{dt} \int \sin t dt \ dt$

$I=-t\cos t +\int \cos t \ dt$

$I=-t\cos t + \sin t+C$

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

$y=\displaystyle\int \dfrac{\cos x}{1+\sin x}dx$ .

We have,

$y=\displaystyle\int \dfrac{\cos x}{1+\sin x}dx$

Let

$t=1+\sin x$

$dt=\cos x\ dx$

Therefore,

$y=\displaystyle\int \dfrac{dt}{t}$

$y=\ln t+C$

On putting the value of

$t$ , we get

$y=\ln (1+\sin x)+C$

Hence, this is the answer.

Using integration, find the area of the triangle $PQR$ , whose vertices are at $P (2 , 5), Q (4 , 7)$ and $R (6 , 2)$ .

$e{q^n}\,of\,line\,PQ:$

$y - 5 = {{7 - 5} \over {4 - 2}}\left( {x - 2} \right)$

${y_{PQ}} = x - 3$

$e{q^n}\,of\,line\,QR:$

$y - 7 = {{2 - 7} \over {6 - 4}}\left( {x - 4} \right)$

$y - 7 = {{ - 5} \over 2}\left( {x - 4} \right)$

$2y - 14 = - 5x + 20$

$2y = - 5x + 34$

${y_{QR}} = {{ - 5x} \over 2} + 17$

$e{q^n}\,of\,line\,PR:$

$y - 5 = {{2 - 5} \over {6 - 2}}\left( {x - 2} \right)$

$y - 5 = - x + 2$

${y_{PR}} = - x + 3$

$Now\,Area = \smallint _2^4{y_{PS\,}}dx + \smallint _4^6\,{y_{QX}}dn - \smallint _2^6\,{y_{PR}}\,dx$

$\smallint _2^4\left( {x + 3} \right)\,dx + \smallint _4^6\,\left( {{{ - 5x} \over 2} + 17} \right)\,dn - \smallint _2^6\left( { - x + 3} \right)\,dx$

$\left( {{{{x^2}} \over 2} + 3x} \right)_2^4 + \left( {{{ - 5{x^2}} \over 4} + 17x} \right)_4^6 - \left( {{{ - x} \over 2} + 3x} \right)_2^6$

$\left( {8 + 12 - 2 - 6} \right) + \left( { - 45 + 102 + 20 - 68} \right) - \left( { - 18 + 18 + 2 - 6} \right)$

$12 + 9 + 4$

$25\,sq\,units.$

Evaluate: $\displaystyle\int \log \left( {\log x} \right) + {\left( {\log x} \right)^{ - 2}}= ?$

$I = \displaystyle \int {\left[ {\log \left( {\log x} \right) + {1 \over {{{\left( {\log x} \right)}^2}}}} \right]dx}$

$= \displaystyle \int {\log \mathop {\left( {\log x} \right)}\limits_I } .\mathop {1dx}\limits_{II} + \displaystyle \int {{1 \over {{{\left( {\log x} \right)}^2}}}dx}$

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

$= \log \left( {\log x} \right) - \displaystyle \int {\mathop {{1 \over {\log x}}}\limits_I .} \mathop {1dx}\limits_{II} + \displaystyle \int {{1 \over {{{\left( {\log x} \right)}^2}}}dx}$

$= \log \left( {\log x} \right) - {1 \over {\log x}}.x + \displaystyle \int { - {1 \over {{{\left( {\log x} \right)}^2}}}.1dx + } \displaystyle \int {{1 \over {{{\left( {\log x} \right)}^2}}}dx}$

$= x\log \left( {\log x} \right) - {x \over {\log x}} + c$

Evaluate: $\displaystyle\int e^x(x^2+2x) \ dx$

Now,

$\displaystyle\int e^x(x^2+2x) \ dx$

$=\displaystyle\int e^x x^2 \ dx$

$+\displaystyle\int e^x(2x) \ dx$

$=e^x.x^2$

$-\displaystyle\int e^x(2x) \ dx$

$+\displaystyle\int e^x(2x) \ dx+c$ [ Using method of by parts] [ Where

$c$ is integrating constant]

$=x^2e^x+c$

$\int { \left( x+1 \right) \log { x } } dx$

1317491_1063145_ans_73660ce955a74667996f2294195657c2.jpg

$\int {{e^{2x}}\left( {\sin \left( {ax + b} \right)} \right)} \,dx$

$1 = \int {e_{ii}^{2x}\left( {\sin ax + b} \right)dx}$

$= \sin (ax + b)\frac{{{e^{2x}}}}{2} - \frac{a}{2}\left[ {\cos (ax + b)\frac{{{e^{2x}}}}{2} + \int {a\sin (ax + b)\frac{{{e^{2x}}}}{2}dx} } \right]$

$= \frac{{{e^{2x}}}}{2}\sin (ax + b) - \frac{a}{4}{e^{2x}}\cos (ax + b) - \frac{{{a^2}}}{4}\int {{e^{2x}}\sin (ax + b)dx}$

$1 + \frac{{{a^2}}}{4}I = \frac{{{e^{2x}}}}{4}\left( {2\sin (ax + b) - a\cos (ax + 6)} \right)$

$\left( {\frac{{4t{a^2}}}{4}} \right)I = \frac{{{e^{2x}}}}{4}\left( {2\sin (ax + b) - a\cos (ax + b)} \right)$

$I = \frac{{{e^2}x}}{{{a^2} + 4}}\left( {2\sin (ax + b) - a\cos (ax + b) + c} \right)$

$\int { x\tan ^{ 2 }{ x } } dx$

1315420_1063140_ans_b1ef7f7b75f4460096be77ac60789fc1.jpg

$\int {\left( {\frac{1}{{\left( {\ell nx} \right)}} - \frac{1}{{{{(\ell nx)}^2}}}} \right)\,dx}$

$\int {\left( {{1 \over {inx}} - {1 \over {inx}}} \right)dn}$

$= \int {1.{1 \over {inx}}dx - \int {{1 \over {{{(inx)}^2}}}dn} }$

$= {x \over {inx}} - \int{ - {1 \over {i{n^2}}}dn - \int {{1 \over {i{n^2}}}dn} }$

$= {x \over {inx}} + c$

$\int { { x }^{ 2 } } \sin ^{ -1 }{ x } dx$

1315433_1063143_ans_51198b08148d4d809b4d6ed0ee73fa43.jpg

Solve $\displaystyle\int {x{{\tan }^{ - 1}}x\,dx}$

$\int {x{{\tan }^{ - 1}}x} dx$

${\tan ^{ - 1}}x.{{{x^2}} \over 2} - \int {{1 \over {1 + + {x^2}}} \times {{{x^2}} \over 2}} dx$

${{{x^2}} \over 2}{\tan ^{ - 1}}x - {1 \over 2}\int {{{{x^2} + 1 - 1} \over {1 + {x^2}}}dx}$

${{{x^2}} \over 2}{\tan ^{ - 1}}x - {1 \over 2}\int {dx} + {1 \over 2}\int {{{dx} \over {1 + {x^2}}}}$

${{{x^2}} \over 2}{\tan ^{ - 1}}x - {x \over 2} + {1 \over 2}{\tan ^{ - 1}}x + C$

$\int { \sin ^{ -1 }{ \left( 3x-4{ x }^{ 3 } \right) } } dx$

1315432_1063142_ans_104a655279a141649de7e148035e7042.PNG

Evaluate:
$\int_{} {x{e^{2x}}dx}$

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

$= \,\,x.\dfrac{{{e^{2x}}}}{2} - \,\,\int {\dfrac{{{e^{2x}}}}{2}} dx$

$= \,\,x.\dfrac{{{e^{2x}}}}{2} - \dfrac{1}{4}{e^{2x}} + c$

Find $\int \left(1-\frac{1}{x^2}\right)e^{(x+\frac{1}{x}) }dx$ on I where $I= (0, \infty ).$

$x+(\frac{1}{x})=t\\ then (1-(\frac{1}{x^2}))dx=dt \\\therefore I=\int e^t dt \\=e^t+c \\ =e^{x+(\frac{1}{x})}+c$

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

Integrating by parts we get,

$u = xe^x \space ,v = \cfrac{1}{(1+x)^2}$

$u^{'} = e^x+xe^x , \int v dx = \cfrac{-1}{1+x}$

$\int \cfrac{xe^x}{(1+x)^2} dx = -\cfrac{xe^x}{1+x} - \int \cfrac{-e^x(x+1)}{1+x}dx$

$\int \cfrac{xe^x}{(1+x)^2} dx= -\cfrac{xe^x}{1+x} + \int e^x dx$

$\int \cfrac{xe^x}{(1+x)^2} dx= -\cfrac{xe^x}{1+x} + e^x$

$\int \cfrac{xe^x}{(1+x)^2} dx= \cfrac{e^x}{1+x} + C$

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

Let

$x = tan \theta$

$dx = sec^2\theta d\theta$

$\cfrac{2tan\theta}{1+tan^2 \theta} = sin\cfrac{\theta}{2}$

$\int sin^{-1}(sin\cfrac{\theta}{2})sec^2 \theta d\theta$

$\int \cfrac{\theta}{2}sec^2 \theta d\theta$

Integrating by parts, we get

$u = \cfrac{\theta}{2} , v = sec^2 \theta$

$du = \cfrac{1}{2} , \int v d\theta = tan\theta$

$\cfrac{\theta}{2} tan\theta - \cfrac{1}{2} \int tan\theta d\theta$

$\cfrac{xtan^{-1} x}{2} + \cfrac{log|cosx|}{2}+C$

If $\quad \int { \cfrac { \sqrt { 4+{ x }^{ 2 } } }{ { x }^{ 6 } } } dx=\cfrac { { { \left( a+{ x }^{ 2 } \right) }^{ 3/2 } }.\left( { x }^{ 2 }-b \right) }{ 120{ x }^{ 5 } } +C$ then $a+b$ equals to

$\displaystyle\int \dfrac{\sqrt{4+x^{2}}}{x^{6}}dx$

put

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

$\displaystyle\int \dfrac{\sqrt{4+x^{2}}}{x^{6}}dx=-\displaystyle\int t^{3}\sqrt{4{t^{2}}+1}dt=-t^{2}\displaystyle\int t\sqrt{4{t^{2}}+1}dt+\displaystyle\int \dfrac{d(t^{2})}{dt}\bigg(\displaystyle\int t\sqrt{4{t^{2}}+1}dt\bigg)dt$

$=-\dfrac{1}{12}t^{2}(4{t^{2}}+1)^{3/2}+\dfrac{1}{6}\displaystyle\int t(4{t^{2}}+1)^{3/2}dt$

$=-\dfrac{1}{12}t^{2}(4{t^{2}}+1)^{3/2}+\dfrac{1}{120}(4{t^{2}}+1)^{5/2}+C=\dfrac{(4+x^{2})^{3/2}(x^{2}-6)}{120{x^{5}}}+C$

$a=4,b=-6\implies a+b=-2$

$\int {{x^3}\log xdx = }$

$\begin{array}{l}\int {{x^3}\log xdx} \\ = \log x\int {{x^3} - } \int {\left[ {\cfrac{d}{{dx}}\log x\int {{x^3}dx} } \right]} \\ = \cfrac{{\log x.{x^4}}}{4} - \int {\left[ {\cfrac{1}{x}.\cfrac{{{x^4}}}{4} + c} \right]} \\ = \cfrac{{\log x.{x^4}}}{4} - \cfrac{1}{4}\int {{x^3}dx} \\ = \cfrac{{\log x.{x^4}}}{4} - \cfrac{1}{{16}}{x^4} + c\\ = \cfrac{1}{4}{x^4}\left[ {\log x - \cfrac{1}{4}} \right] + c\end{array}$

Integrate with respect to $x$ :
$x\ ln\ x$

Consider the given integral.

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

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

Therefore,

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

$I=\dfrac{{{x}^{2}}}{2}\ln x-\dfrac{1}{2}\int{xdx}$

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

$I=\dfrac{{{x}^{2}}}{2}\ln x-\dfrac{{{x}^{2}}}{4}+C$

Hence, this is the answer.

$\displaystyle \int \dfrac{(t^{2}+1)^{2}}{t^6+1} dt$

Let,

$I=\int{\dfrac{{{\left( {{t}^{2}}+1 \right)}^{2}}}{{{t}^{6}}}}dt$

$=\int{\dfrac{{{t}^{4}}+1+2{{t}^{2}}}{{{t}^{6}}}dt=}\int{\left( \dfrac{1}{{{t}^{2}}}+\dfrac{1}{{{t}^{4}}}+\dfrac{2}{{{t}^{6}}} \right)}dt$

$=\int{{{t}^{-2}}dt+\int{{{t}^{-4}}dt+2\int{{{t}^{-6}}dt}}}$

$=\dfrac{{{t}^{-1}}}{-1}+\dfrac{{{t}^{-3}}}{-3}+\dfrac{2{{t}^{-5}}}{-5}+C$

$=\dfrac{1}{t}-\dfrac{1}{3{{t}^{3}}}-\dfrac{2}{5{{t}^{5}}}+C$

Hence, this is the answer.

Integrate the function w.r.t x : $x{\left( {\log x} \right)^2}$

Putting

$logx = m$

$\cfrac{dx}{x} = dm$

$dx = e^m dm$

$\int m^2 e^{2m}dm$

Integrating by parts we get,

$u = m^2 , v = e^{2m}$

$u^{'} =2m, \int v dm = \cfrac{1}{2} e^{2m}$

$\implies \cfrac{m^2 e^{2m}}{2} -\int me^{2m} dm$

Again integrating by parts

$u = m , v = e^{2m}$

$u^{'} = 1 , \int v dm = \cfrac{e^{2m}}{2}$

$\implies \cfrac{m^2 e^{2m}}{2} -\cfrac{me^{2m}}{2}+\cfrac{e^{2m}}{4}$

$\implies \cfrac{(logx)^2 x^2}{2} -\cfrac{x^2logx}{2}+\cfrac{x^2}{4}$

Solve: $\int\limits_0^{\pi /4} {\left( {x\cos x - \cos 3x} \right)dx}$

$\begin{array}{l}\int\limits_0^{\pi /4} {\left( {x\cos x - \cos 3x} \right)dx} \\ = \int\limits_0^{\pi /4} {x\cos xdx} - \int\limits_0^{\pi /4} {\cos 3xdx} \\ = \left[ {x\int {\cos xdx} - \int {\sin xdx} } \right]_0^{\pi /4} - \left[ {\frac{{\sin 3x}}{x}} \right]_0^{\pi /4}\\ = \left[ {x\sin x - \cos x} \right]_0^{\pi /4} - \left[ {\frac{{\sin 3\pi /4}}{3}} \right]\\ = \left[ {\frac{\pi }{4}.\frac{1}{2} + \frac{1}{2} - 1} \right] - \left[ {\frac{{1/\sqrt 2 }}{3}} \right]\\ = \frac{\pi }{{4\sqrt 2 }} - 1 - \frac{{\sqrt 2 }}{3}\end{array}$

Integrate with respect to $x$ :
$\ell n\ x$

Consider the given integral.

$I=\int{\ln xdx}$

$I=\int{\left( 1 \right)\ln xdx}$

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

Therefore,

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

$I=x\ln x-\int{1}dx$

$I=x\ln x-x+C$

Hence, this is the answer.

Simplify:-
$\int {x\ell n\sqrt x dx}$

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

$=\dfrac{1}{2}\left[\dfrac{x^2}{2}lnx \Rightarrow \dfrac{x^2}{4}lnx\dfrac{-x^2}{4}lnx \dfrac{-x^2}{8}+c \right.$

Integrate with respect to $x$ :
$x\sin^{2}x$

We have,

$\int{x{{\sin }^{2}}xdx}$

Using formula,

$\int{u.vdx=u\int{vdx-\left( \int{\dfrac{du}{dx}}\int{vdx} \right)dx}}$

Then,

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

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

$=\dfrac{{{x}^{2}}}{4}-\dfrac{1}{2}\left[ x\int{\cos 2xdx-\int{\left( \dfrac{dx}{dx}\cos 2xdx \right)}}dx \right]$

$=\dfrac{{{x}^{2}}}{4}-\dfrac{1}{2}\left[ x\dfrac{\sin 2x}{2}-\dfrac{\sin 2x}{2} \right]+C$

$=\dfrac{{{x}^{2}}}{4}-\dfrac{1}{2}\left[ x\dfrac{\sin 2x}{2}-\dfrac{1}{2}\int{\sin 2xdx} \right]+C$

$=\dfrac{{{x}^{2}}}{4}-\dfrac{1}{2}\left[ x\dfrac{\sin 2x}{2}+\dfrac{1}{4}\cos 2x \right]+C$

$=\dfrac{{{x}^{2}}}{4}-\dfrac{1}{4}x\sin 2x+\dfrac{1}{8}\cos 2x+C$

Hence, this is the answer.

Integrate with respect to x:
(i) $\dfrac{x}{nx}$
(ii) $x\sin^2x$

$(i)\quad \int { \dfrac { x }{ nx } dx } =\int { \dfrac { 1 }{ n } dx } \\ =\dfrac { x }{ n } +C\\ (ii)\quad \int { x{ \sin }^{ 2 }xdx } =\int { x.\dfrac { 1 }{ 2 } (1-\cos 2x)dx } \\ =\dfrac { 1 }{ 2 } \int { xdx } -\dfrac { 1 }{ 2 } \int { x\cos2xdx } \\ =\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { 1 }{ 2 } \left[ x\int { \cos2xdx } -\int { \left[ \dfrac { dx }{ dx } .\int { \cos2xdx } \right] } dx \right] \quad (integrating\quad by\quad part)\\ =\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { x\sin2x }{ 4 } -\dfrac { \cos2x }{ 4 } +C\\ =\dfrac { { x }^{ 2 }-x\sin2x-\cos2x }{ 4 } +C$

Solve : $\int \, 4 \, x^3 \sqrt{5 - x^2 } \, dx$

$I=\displaystyle \int 4x^{3}\sqrt{5-x^{2}}dx$

$5-x^{2}=t^{2}\Rightarrow x^{2}=(5-t^{2})$

$-2xdx=2tdt$

$I=-\displaystyle 2\int (-2x)x^{2}\sqrt{(5-x^{2})}dx$

$=\displaystyle -2\int (5-t^{2})\sqrt{t^{2}}(2t)(dt)$

$=\displaystyle 2\int (t^{2}-5)\times 2t^{2}dt$

$=\displaystyle 4\int (t^{4}-5t^{2})dt$

$=\displaystyle 4\times \left[\frac{t^{5}}{5}-\frac{5t^{3}}{3}\right]+C$

$\Rightarrow \displaystyle 4t^{3}\left[\frac{t^{2}}{5}-\frac{5}{3}\right]+C$

$\Rightarrow \displaystyle4(5-x^{2})^{\frac{3}{2}}\left[\frac{315-3x^{2}-25}{15}\right]+C$

$I=-\dfrac{4}{15}\times (5-x^{2})^{\frac{3}{2}}[10+3x^{2}]+C$

Solve : $\displaystyle \int \, \dfrac{dx}{x - \sqrt{x}}$

$I\displaystyle =\int \frac{dx}{x-\sqrt{x}}=\int \frac{dx}{\sqrt{x}(\sqrt{x-1})}$

$\sqrt{x}-1=t$

$\dfrac{1}{\sqrt{2}}dx=dt$

$\dfrac{dx}{\sqrt{x}}=2dt$

$I=\displaystyle \int \frac{(d\frac{x}{\sqrt{x}})}{\sqrt{x}-1}=\int \frac{2dt}{t}$

$I=2ln|t|+ln|c|$

$\Rightarrow I=ln|ct^{2}|$

$\Rightarrow I=ln|c(\sqrt{x-1})^{2}|=ln|x(x+1-2\sqrt{x})|$

$I=ln\left | c(x+1-2\sqrt{x}) \right |$

Solve : $\displaystyle \int \dfrac{2x}{x^2+5}dx$

Consider

$I=\displaystyle \int \dfrac{2x}{x^2+5}dx$

Let

$x^2+5=t$

$2x \ dx=dt$

Therefore,

$I=\displaystyle \int \dfrac{dt}{t}=\log t+C$

$I=\log|x^2+5|+C$

Solve: $\int 3^x.e^x dx =$ ?

$\displaystyle I = \int 3^{x}e^{x}dx$

$\displaystyle I = 3^x \int e^{x}dx -\int \left ( \dfrac{d}{dx}3^{x} \right )\int e^{x}dx$

$\displaystyle I = 3^{x} e^{x} -\int 3^{x}log 3 e^{x}dx$

$\displaystyle I = 3^{x}e^{x}- log 3 \int 3^{x}e^{x}dx$

$\displaystyle I = 3^{x} e^{x} - log 3 I$

$(1+log 3) I = 3^{x}e^{x}$

$I = \dfrac{1}{(1+log3)} 3^{x}e^{x}$

$\boxed{\therefore 3^x e^x dx = \frac{1}{(1+log3)}3^x e^x}$

1125820_1103675_ans_237fea83e4ed46648618807b302f084b.jpg

Solve $\int e^{ax}\sin bxdx$ .

Integrating using by part, we get,

$v = e^{ax} , u = \sin bx$

$\int v = \cfrac{1}{a} e^{ax} , u^{'} = b \cos bx$

$\cfrac{e^{ax} \sin bx}{a} - \int \cfrac{b e^{ax} \cos bx \space dx}{a}$

Again using by parts we get,

$\cfrac{e^{ax} \sin bx}{a} - \cfrac{b e^{ax} \cos bx \space}{a^2} - \int \cfrac{b^2 e^{ax} \sin bx \space dx}{a^2}$

$\cfrac{(a^2+b^2)I}{a^2} = \cfrac{e^{ax} \sin bx}{a} - \cfrac{b e^{ax} \cos bx \space}{a^2}$

$I = \cfrac{1}{a^2 + b^2}(a e^{ax} \sin bx - b e^{ax} \cos bx \space)$

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

Using by parts,we get

$\displaystyle \int cosec^3x dx$

$\displaystyle -cosec x \cot x - \int \cot^2 x \space cosec x dx$

$\displaystyle -cosec x \cot x - \int (cosec^2x-1) cosec x dx$

$\displaystyle -cosec x \cot x - \int cosec^3x + \int cosec x dx$

$\displaystyle \cfrac{-cosec x \cot x}{2} + \cfrac{\ln|cosec x - \cot x|}{2} + C$

Evaluate:
$\displaystyle\int \dfrac{1}{24+1}dx=?$

$\displaystyle \int \cfrac{1}{25} dx$

$\displaystyle \cfrac{x}{25} +C$

Solve: $\int tan (x)dx=?$

$\int \tan x dx=-\log|\cos x|+c$

$\displaystyle \int \dfrac {3ax}{ b^{2}+c^{2}x^{2}} dx$

$I=\int { \cfrac { 3ax }{ { b }^{ 2 }+{ c }^{ 2 }{ x }^{ 2 } } dx }$

Putting

${ b }^{ 2 }+{ c }^{ 2 }{ x }^{ 2 }=t\quad \quad \Rightarrow 2{ c }^{ 2 }xdx=dt$

$xdx=\cfrac { dt }{ 2{ c }^{ 2 } } \\ I=\int { \cfrac { 3a.\cfrac { dt }{ 2{ c }^{ 2 } } }{ t } } \\ =\cfrac { 3a }{ 2{ c }^{ 2 } } \int { \cfrac { dt }{ t } } =\cfrac { 3a }{ 2{ c }^{ 2 } } \log { t } \\ \therefore I=\cfrac { 3a }{ 2{ c }^{ 2 } } \log { ({ b }^{ 2 }+{ c }^{ 2 }{ x }^{ 2 }) }$

Solve: $\displaystyle\int\dfrac{(\ln x )^n}{x}dx$

$I=\displaystyle\int\dfrac{(\ln x )^n}{x}dx$
Let

$\ln x=t$ , then

$\dfrac{1}{x}dx=dt$

$\therefore I=\displaystyle\int t^ndt$

$\implies I = \dfrac{t^{n+1}}{n+1} + C$

$\implies I =\boxed{\dfrac{(\ln x)^{n+1}}{n+1}+C}$

$\int e^{\ 10x} dx$

$\int e^{10x}dx = e^{10x} \times 10 = 10e^{10x} + c$

Evaluate: $- 2 \int \cos 2 \theta \sin 2\theta d \theta$

$I = -2\int cos2\theta \, sin2\theta \,d\theta = - \int sin(4\theta )(d\theta )$

$\Rightarrow I=\dfrac{\cos 4\theta}4$

$\Rightarrow I = \dfrac{1}{4}\cos(4\theta )+c$

Solve: $\sqrt{\sin x}dx$

$I=\int { \sqrt { \sin { x } } dx } \\ \quad =\int { \sqrt { 2\cos ^{ 2 }{ \left( \dfrac { 2x-\pi }{ 4 } \right) -1 } } } dx\\ \quad =\int { \sqrt { 1-2\sin ^{ 2 }{ \left( \dfrac { 2x-\pi }{ 4 } \right) } dx } } \\ Let,\quad u=\dfrac { 2x-\pi }{ 4 } \Rightarrow dx=2du\\ \therefore I=2\int { \sqrt { 1-2\sin ^{ 2 }{ u } du } } \\ \quad \quad =2E\left( \dfrac { u }{ 2 } \right) \\ \quad \quad =2E\left( \dfrac { 2x-\pi }{ 8 } \right) +C\\$

Prove that
$\int e ^ { x } \left( \tan ^ { - 1 } x + \frac { 1 } { 1 + x ^ { 2 } } \right) d x$

Using

$\int (f(x)+f'(x)e^xdx=f(x)$

where

$f(x)=\tan ^{-1}x$

we get,

$\int e^x(\tan ^{-1}x+\cfrac{1}{1+x^2})dx=e^x\tan ^{-1}x+c$

I= $\int { \left( { 1-x } \right) \left( { 2+3x } \right) \left( { 5-4x } \right) dx }$

$\begin{array}{l} \int { \left( { 1-x } \right) \left( { 2+3x } \right) \left( { 5-4x } \right) dx } \\ =\int { \left( { 10-3x-19{ x^{ 2 } }+12{ x^{ 3 } } } \right) dx } \\ =10\int { dx } -3\int { xdx } -19\int { { x^{ 2 } }dx } +12\int { { x^{ 3 } }dx } \\ =10x-3.\frac { { { x^{ 2 } } } }{ 2 } -19.\frac { { { x^{ 3 } } } }{ 3 } +12.\frac { { { x^{ 4 } } } }{ 4 } +C \\ =10x-\frac { { 3{ x^{ 2 } } } }{ 2 } -\frac { { 19{ x^{ 3 } } } }{ 3 } +3{ x^{ 4 } }+C. \end{array}$

$\int {\tan x}dx$

$\int { tanx } dx=log\left| secx \right| +C$

$I = \int {\left( {3\sin x - 4\cos x + 5{{\sec }^2}x - 2\cos e{c^2}x} \right)dx}$

$\begin{array}{l} \int { \left( { 3\sin x-4\cos x+5{ { \sec }^{ 2 } }x-2\cos e{ c^{ 2 } }x } \right) dx } \\ =3\int { \sin xdx } -4\int { \cos x } dx+5\int { { { \sec }^{ 2 } }xdx } -2\int { \cos e{ c^{ 2 } }x } dx \\ =3\left( { -\cos x } \right) -4\sin x+5\tan x-2\left( { -\cot x } \right) +C \\ =\left( { -3\cos x-4\sin x+5\tan x+2\cot x+C } \right) \end{array}$

Evaluate: $\displaystyle \int{\dfrac{dx}{32-2x^{2}}}$

$int{\cfrac{dx}{32-2x^2}}\\=\cfrac{1}{2}\int{\cfrac{dx}{16-x^2}}\\=\cfrac{1}{2}\int{\cfrac{dx}{(4)^2-x^2}}\\=\cfrac{1}{2}.\cfrac{1}{2\times4}log|\cfrac{4+x}{4-x}|+C\\=\cfrac{1}{16}log|\cfrac{4+x}{4-x}|+C$

$\int {\cot xdx}$

$\int { cotx } dx=log\left| sinx \right| +C$

$\int {cosec\;x dx}$

$\int { cosecx } dx=log\left| cosecx-cotx \right| +C$

$\int {\sec xdx}$

$\int { secx } dx=log\left| secx+\ tanx \right| +C$

$I = \int {{x^9}dx}$

$\begin{array}{l} I=\int { { x^{ 9 } }dx } \\ =\frac { { { { \left( x \right) }^{ 9+1 } } } }{ { 9+1 } } +C \\ =\frac { { { x^{ 10 } } } }{ { 10 } } +C. \end{array}$

Evaluate: $\int {\frac{{\tan x}}{{{{\left( {\cos x} \right)}^2}}}dx}$

$\begin{array}{l} Let\, u=\tan x \\ du={ \sec ^{ 2 } }xdx \\ Now, \\ \int { \frac { { \tan x } }{ { { { \left( { \cos x } \right) }^{ 2 } } } } dx } \\ =\int { \tan xse{ x^{ 2 } }xdu } \\ =\int { udu } \\ =\frac { { { u^{ 2 } } } }{ 2 } \\ =\frac { 1 }{ 2 } { \tan ^{ 2 } }x+C \end{array}$

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

1346765_1261636_ans_af2a390e2c9b4ab29b3210d2015d005a.jpg

Find the value of -
$\int xdx$

$\int xdx$ =

$\dfrac{x^{2}}{2} +C$

$I = \int {\sqrt[3]{x}dx}$

$\begin{array}{l} I=\int { \sqrt [ 3 ]{ x } dx } \\ =\int { { x^{ \frac { 1 }{ 3 } } }dx } \\ =\frac { { { x^{ \left( { \frac { 1 }{ 3 } +1 } \right) } } } }{ { \frac { 1 }{ 3 } +1 } } +C \\ =\frac { 3 }{ 4 } { x^{ \frac { 4 }{ 3 } } }+C. \end{array}$

Solve:
$\displaystyle \int \dfrac{\log{x}}{x^{2}}\ dx$

$I=\displaystyle\int \dfrac{log x}{x^2}dx$

Let

$t=log x$

$\Rightarrow x=e^t$

$dt=\dfrac{1}{x}dx$

$I=\displaystyle\int t e^{-t}dt$

$I=t \dfrac{e^{-t}}{-1}-\displaystyle\int 1.\dfrac{e^{-t}}{-1}dt$

$I=-t e^{-t}+\dfrac{e^{-t}}{-1}+c$

$I=-\dfrac{t}{e^t}-e^{-t}+c$

$I=\dfrac{-log x}{x}-\dfrac{1}{x}+c$

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

1209519_1286896_ans_e4043d0cac094649bbe68796ad0a03f4.jpg

Solve
$\int \dfrac{\cos (x+a)}{\cos (x-a)} dx$

We have,

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

$Put\, \, \, x-\alpha =t \\ dx=dt \\ x=\left( { \alpha +t } \right)$

Therefore,

$I= \int { \dfrac { { \cos \left( { 2\alpha+t } \right) } }{ { \cos t } } } dx \\ =\int { \dfrac { { \cos 2\alpha .\cos t-\sin 2\alpha .\sin t } }{ { \cos t } } } dt \\ =\int { \left( { \cos 2\alpha -\sin 2\alpha .\tan t } \right) } dt \\ =t.\cos 2\alpha +\sin 2\alpha .\log \left| { \cos t } \right| +C \\$

On putting the value of

$t$ , we get

$I=\left( { x-\alpha } \right) \cos 2\alpha +\sin 2\alpha .\log \left| { \cos \left( { x-\alpha } \right) } \right| +C$

Hence, this is the answer.

Solve:
$\displaystyle \int \dfrac{x\tan^{-1}{x^{2}}}{(1+x^{4})}\ dx$

Consider the given integral.

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

Let

${ \tan ^{ -1 } }{ x^{ 2 } }=t$

$\dfrac { { dt } }{ { dx } } =\dfrac { { 2x } }{ { 1+{ x^{ 4 } } } }$

$\dfrac { 1 }{ 2 } dt=\frac { { xdx } }{ { 1+{ x^{ 4 } } } }$

Therefore,

$=\dfrac { 1 }{ 2 } \dfrac { { { t^{ 2 } } } }{ 2 } +C \\ =\dfrac { { { t^{ 2 } } } }{ 4 } +C \\ ={ \dfrac { { \left[ { { { \tan }^{ -1 } }\left( { { x^{ 2 } } } \right) } \right] } }{ 4 } ^{ 2 } }+C$

Hence, this is the answer.

Write the formula for integration by parts.

$\int f(x).g(x)dx = f(x)\int g(x)dx-\int f'(x).( \int g(x)dx )dx$

Find $\displaystyle \int \dfrac {1}{9x^{2}+6x+5}dx$

$\int \frac{dx}{ax^{2}+ 6x + 5 } = \frac{1}{9} \int \frac{dx}{x^{2} + \frac{2x}{3} + \frac{5}{9}}$

$= \frac{1}{9} \int \frac{dx}{x^{2} + \frac{2x}{3} + \frac{1}{9} - \frac{1}{9} + \frac{5}{9}}$

$= \frac{1}{9} \int \frac{dx}{(x+\frac{1}{3})^{2} + (\frac{2}{3})^{2}}$

$[\because \int \frac{dx}{x^{2}+a^{2}}$ =

$\frac{1}{a} tan^{-1} (\frac{x}{a}) + c ]$

$= \frac{1}{9} \times \frac{1}{(\frac{2}{3})} tan^{-1} (\frac{3x+1}{2}) + c$

$\frac{1}{6} tan^{-1} (\frac{3x+1}{2}) + c$

1233182_1301999_ans_6e23d6a0e9714fa9b66bfa314688b809.jpg

$\int {{x^9}dx}$

$\begin{array}{l} We\, have \\ \int { { x^{ 9 } }dx=\dfrac { { { x^{ \left( { 9+1 } \right) } } } }{ { \left( { 9+1 } \right) } } +C } \\ =\dfrac { { { x^{ 10 } } } }{ { 10 } } +C. \end{array}$

Evaluate $\displaystyle \int{\cos^{2}x\, dx}$ .

As we know that,

$\cos{2x} = 2 \cos^{2}{x} - 1$

$\Rightarrow \cos^{2}{x} = \cfrac{1 + \cos{2x}}{2}$

Therefore,

$\displaystyle \int{\cfrac{1 + \cos{2x}}{2} dx}$

$= \cfrac{x}{2} + \cfrac{\sin{2x}}{4} +C$

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

We have,

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

Let

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

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

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

Therefore,

$I=-\int t\ dt$

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

Put

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

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

Hence, this is the answer.

Evaluate:
$\displaystyle\int{\dfrac{1}{4+9x^{2}}}dx$

$\int \frac{dx}{4+9x^{2}} = \frac{1}{9}\int \frac{dx}{(\frac{2}{3})^{2}+x^{2}}$

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

$= \frac{1}{a}\times \frac{1}{(\frac{2}{3})} tan^{-}(\frac{3x}{2})+c$

$= \frac{1}{6}tan^{-1}(\frac{3x}{2})+c$

1233189_1303677_ans_d0724537f70d4b1ab2025caf1a28521d.jpg

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

$\int \frac{x^{3}dx}{x+2}$ put x+2=t x

x=t-2

dx=dt

$\int \frac{(t-2)^{3}}{t}dt$

$\int (\frac{t^{3}-6t^{2}+12t-8}{t})dt$

$\int (t^{2}-6t+12-\frac{8}{t})dt$

$=\frac{t^{3}}{3}-3t^{2}+12t-8ln(t)+c$

$=\frac{(x+2)^{3}}{3}-3(x+2)^{2}+12(x+2)-8ln(x+2)+c$

1233863_1302763_ans_c9c854954b34408e9f6349e9b603a13a.jpg

Integrate $\displaystyle \int \dfrac{\sqrt{a-x}}{x}\ dx$

Let

$x=a\sin^2\theta$

$\Rightarrow dx=2a\sin\theta \cos\theta d\theta$

$\Rightarrow \displaystyle\int \dfrac{\sqrt{a(1-\sin^2\theta)}\cdot 2a\sin\theta \cos\theta}{a\sin^2\theta}d\theta=\displaystyle\int \dfrac{\sqrt{a}(\cos\theta)\cdot 2\cos\theta}{\sin\theta}d\theta$

$=2\sqrt{a}\displaystyle\int \dfrac{1-\sin^2\theta}{\sin\theta}d\theta=2\sqrt{a}\left[\displaystyle\dfrac{1}{\sin \theta}d\theta -\displaystyle\int\sin \theta d\theta\right]$

$=2\sqrt{a}\left[\displaystyle\int cosec \theta d\theta -\displaystyle\int \sin\theta d\theta\right]$

$=2\sqrt{a}\left[ln|cosec \theta -\cot\theta|+\cos\theta\right]+c$

$\sin^2\theta =\dfrac{x}{a}$

$\Rightarrow \sin\theta =\sqrt{\dfrac{x}{a}}, \cos \theta =\sqrt{1-\dfrac{x}{a}}$

$\Rightarrow I=2\sqrt{a}\left[ln \left|\sqrt{\dfrac{a}{x}}-\dfrac{\sqrt{a-x}}{\sqrt{x}}\right|+\sqrt{1-\dfrac{x}{a}}\right]+c$ .

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Evaluate $\int \dfrac{sec^2x}{3+tan^2x}dx$

$I=\int \dfrac{sec^2x}{3+tan^2x}dx$

Put

$\tan x =t$

$sec^2xdx=dt$

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

$I=\int \dfrac{dt}{(\sqrt{3})^2+t^2}$

$I=\dfrac{1}{\sqrt{3}}tan^{-1} (\dfrac{t}{\sqrt{3}})+C$

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

Evaluate $\int \dfrac{log(\sin x)}{\tan x}dx$

$I=\int \dfrac{log(\sin x)}{\tan x}dx$

Put

$log(\sin x)=t$

$\cot x dx=dt$

$\dfrac{1}{\tan x}dx=dt$

Therefore,

$I=\int tdt$

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

$I=\dfrac{[log(sin x)]^2}{2}+C$

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

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

$I=\int 2 cosec x \cot x dx$

$I=-2 cosec x +C$

$\int \sqrt{\frac{x}{a-x}} d x=$

Put

$x=a\sin^2 \theta\implies d x=2{a}\sin \theta\cos \theta d\theta$

$\displaystyle\int \sqrt{\dfrac{x}{a-x}}d x=\int \sqrt{\dfrac{a\sin ^2 \theta}{a-a\sin^2 \theta}}(2 a\sin \theta\cos \theta)d\theta=2{a}\int \sin^2\theta d\theta=a\int (1-\cos 2\theta)d\theta=a\theta-\dfrac{a}{2}\sin 2\theta+c$

$=a\text{sin}^{-1}\bigg(\sqrt{\dfrac{x}{a}}\bigg)-a\sqrt{x/a}\times \sqrt{1-x/a}+c=a\text{sin}^{-1}\bigg(\sqrt{\dfrac{x}{a}}\bigg)-\sqrt{x-a x^2}+c$

Evaluate :
$\int 2^x dx$

$I=\int 2^xdx$

$I=\dfrac{2^x}{\log 2}+C$

Evaluate :
$\int (\dfrac{2}{1+2x}-\dfrac{1}{1-x})dx$

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

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

$I=log(1+2x)+log(1-x) +C$

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

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

$I=x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+C$

Prove that $\displaystyle \int x^{2}a^{x}dx=\frac{a^{x}}{\left ( \log a \right )^{3}}\left [ x^{2}\left ( \log a \right )^{2}-2x\left ( \log a \right )+2 \right ].$

Let

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

Applying integration by parts

$\displaystyle =x^{ 2 }\frac { a^{ x } }{ \log a } -\int 2x.\frac { a^{ x } }{ \log a } dx$
Again applying integration by parts

$\displaystyle =\frac { x^{ 2 }a^{ x } }{ \log a } -\frac { 2 }{ \log a } \left[ x.\frac { a^{ x } }{ \log a } -\int 1.\frac { a^{ x } }{ \log a } dx \right]$

$\displaystyle =x^{ 2 }\frac { a^{ x } }{ \log a } -\frac { 2xa^{ x } }{ \left( \log a \right) ^{ 2 } } +\frac { 2 }{ \left( \log a \right) ^{ 2 } } \int a^{ x }dx$

$\displaystyle =\frac { x^{ 2 }a^{ x } }{ \log a } -\frac { 2xa^{ x } }{ \left( \log a \right) ^{ 2 } } +\frac { 2a^{ x } }{ \left( \log a \right) ^{ 3 } }$

$\displaystyle =\frac { a^{ x } }{ \left( \log a \right) ^{ 3 } } \left[ x^{ 2 }\left( \log a \right) ^{ 2 }-2x\log a+2 \right]$

$\displaystyle\int x^{3}\left ( \log x \right )^{2}dx=\frac{x^{4}}{4}\left ( \log x \right )^{2}-\frac{1}{8}x^{4}\log x+\frac{1}{4k}x^{4}.$ Find the value of $k$

$\displaystyle I=\int x^{3}\left ( \log x \right )^{2}dx$

$\displaystyle =\frac{x^{4}}{4}\left ( \log x \right )^{2}-\int2\left ( \log x \right ).\frac{1}{x}.\frac{x^{4}}{4}dx$

$\displaystyle =\frac{x^{4}}{4}\left ( \log x \right )^{2}-\frac{1}{2}\int x^{3}\log dx$

$\displaystyle =\frac{x^{4}}{4}\left ( \log x \right )^{2}-\frac{1}{2}\left [ \frac{x^{4}}{4}\log x-\int \frac{1}{x}\frac{x^{4}}{4}dx \right ]$

$\displaystyle =\frac{x^{4}}{4}\left ( \log x \right )^{2}-\frac{1}{8}x^{4}\log x+\frac{1}{8}\int x^{3}dx$

$\displaystyle =\frac{x^{4}}{4}\left ( \log x \right )^{2}-\frac{1}{8}x^{4}\log x+\frac{1}{32}x^{4}dx$

$\displaystyle\int \frac{1}{x^{2}}\log \left ( x^{2}+a^{2} \right )dx=\frac{-1}{x}\log \left ( x^{2}+a^{2} \right )+k.\frac{1}{a}\tan^{-1}\frac{x}{a}.$ Find $k$ .

$\displaystyle \int \frac { 1 }{ x^{ 2 } } \log \left( x^{ 2 }+a^{ 2 } \right) dx$

$\displaystyle I=\frac{-1}{x}\log \left ( x^{2}+a^{2} \right )+\int \frac{1}{x}.\frac{1}{x^{2}+a^{2}}.2x\:dx$

$\displaystyle =\frac{-1}{x}\log \left ( x^{2}+a^{2} \right )+2\int \frac{1}{x^{2}+a^{2}}dx$

$\displaystyle =\frac{-1}{x}\log \left ( x^{2}+a^{2} \right )+2.\frac{1}{a}\tan^{-1}\frac{x}{a}.$

Solve $\displaystyle \int \frac{x}{1+\sin x}dx$

$\displaystyle I= \int \frac{x}{1+\sin x}dx$

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

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

$\displaystyle I= \int x\sec^{2}dx-\int x\sec x\tan x\:dx.$
Integrate each by parts

$\displaystyle I= x\tan { x } -\int { 1.\tan { x } dx } -x\sec { x } +\int { 1.\sec { x } dx }$

$\displaystyle I= \left [ x\tan x-\log \sec x \right ]-\left [ x\sec x-\log \left ( \sec x+\tan x \right ) \right ]+C$

$I=x\tan x-\log \sec x-x\sec x+\log \left( \sec x+\tan x \right) +C$

Show that $\displaystyle \int \left ( e^{\log x}+\sin x \right )\cos dx=x\sin x+\cos x+\frac{1}{2}\sin^{2}x.$

$\displaystyle I=\int { \left( { e }^{ \log { x } }+\sin { x } \right) \cos { x } dx }$

$\displaystyle =\int { \left( x+\sin { x } \right) \cos { x } dx }$

$=\displaystyle \int { x\cos { x } dx } +\int { \sin { x } \cos { x } dx }$

$={ { I } }_{ 1 }+{ I }_{ 2 }$

Where

$\displaystyle { I }_{ 1 }=\int { x\cos { x } dx }$

Applying integration by parts

$=x\sin { x } -\int { \sin { x } dx }$

$=x\sin { x } +\cos { x }$

${ I }_{ 2 }=\int { \sin { x } \cos { x } dx }$
Put

$t=\cos { x } \Rightarrow dt=-\sin { x } dx$

$\displaystyle { I }_{ 1 }=-\int { tdt }$

$\displaystyle I_1 =-\frac { { t }^{ 2 } }{ 2 } =-\frac { \cos ^{ 2 }{ x } }{ 2 }$

Hence

$\displaystyle I=x\sin { x } -\frac { 1 }{ 2 } \cos ^{ 2 }{ x } +\cos { x } +C$

$\displaystyle \int \left [ \frac{1}{\log x}-\frac{1}{\left ( \log x \right )^{2}} \right ]dx=x \left ( \log x \right )^{-1}.$ If this is true enter 1, else enter 0.

Let

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

$\displaystyle I=I_{1}-I_{2}$

$\displaystyle I_{1}=\int \frac{1}{\log x}dx=x.\frac{1}{\log x}-\int x.\left ( -\frac{1}{\log x} \right )^{2}.\frac{1}{x}dx$

$\displaystyle I_{1}=x. \frac{1}{\log x}+\int \left ( \frac{1}{\log x} \right )^{2}dx= x.\frac{1}{\log x}+I_{2}$

$\displaystyle I=I_{1}-I_{2}=x \left ( \log x \right )^{-1}.$

Show that $\int \displaystyle x\sin x\cos x = f(x)$ , taking const. of integration as zero. Find $f(\pi /4)$

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

$=\displaystyle \frac{1}{2}\int x\sin 2x\:dx$

Applying integration by parts

$\displaystyle=\frac{1}{2}\left [ x\left ( -\frac{\cos 2x}{2} \right )-\int 1.\left ( -\frac{\cos 2x}{2} \right )dx \right ]$

$\displaystyle=-\frac{1}{4}x\cos 2x+\frac{1}{4}.\frac{\sin 2x}{2}+C$

$\displaystyle=-\frac{1}{4}x\cos 2x+\frac{1}{8}\sin 2x$ (Given

$C=0$ )

Hence,

$\displaystyle f(x)=-\frac{1}{4}x\cos 2x+\frac{1}{8}\sin 2x$

Hence

$\displaystyle f\left( \frac { \pi }{ 4 } \right) =-\frac { 1 }{ 4 } \frac { \pi }{ 4 } \cos 2\frac { \pi }{ 4 } +\frac { 1 }{ 8 } \sin 2\frac { \pi }{ 4 } =\frac { 1 }{ 8 }$

$\displaystyle \int \sec ^{2}x\log \left ( 1+\sin^{2}x \right )dx=\tan x\log \left ( 1+\sin ^{2}x \right )-2x+\sqrt{k}\tan^{-1}\sqrt{k}\tan x$ . Find the value of $k$ .

Integrating by parts, we get

$\displaystyle I=\tan x\log \left ( 1+\sin^{2}x \right )-\int \frac{\tan x}{1+\sin ^{2}x}.2\sin x\cos x\:dx$ or

$\displaystyle I=\tan x \log \left ( 1+\sin^{2}x \right )-2I_{1}$ ...(A)

where

$\displaystyle I_{1}=\int \frac{\sin^{2}x}{1+\sin ^{2}x}dx=\int \left ( 1-\frac{1}{1+\sin ^{2}x} \right )dx$

$\displaystyle =x-\int \frac{\sec ^{2}x\:dx}{\left ( \tan ^{2}x+1 \right )+\tan^{2}x}$

$\displaystyle =x-\int \frac{dt}{\left ( 2t^{2}+1\right)}=x-\frac{1}{\sqrt{2}}\tan ^{-1}t\sqrt{2}$

Putting the value of

$\displaystyle I_{1}$ in (A), we get

$\displaystyle I=\tan x\log \left ( 1+\sin ^{2}x \right )-2x+\sqrt{2}\tan^{-1}\sqrt{2}\tan x$

Find the value of $\int { x\sin { x } dx }$ .

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

Using integration by parts,

$I = \displaystyle -x\cos x +\int\cos x dx$

$I = -x\cos x +\sin x +c$

Evaluate: $\displaystyle\int \log_ex dx$ .

$\displaystyle \int \log _ex dx$

$=\displaystyle \int \log _ex\cdot 1dx$

$=\log _e x\displaystyle \int 1dx-\displaystyle \int\left[\displaystyle\frac{d}{dx}\log _ex\displaystyle \int 1 dx\right] dx$

$=x\log _ex-\displaystyle \int \displaystyle\frac{1}{x}\cdot xdx$

$=x\log _ex-\displaystyle \int 1 dx$

$=x \log _ex-x+c$ , where

$C$ is integration constant.

Integration of $\dfrac{\cos x}{\cos 3x}$

$\displaystyle \int \frac{\cos x}{\cos 3x}=I=?$

$\cos 3x=4 \cos^{3}x-3 \cos x$ .. Identity

$\cos 3x= \cos x(4 \cos^{2}x-3)$

$\therefore \displaystyle I=\int \frac{\cos x dx}{\cos x (4 \cos^{2}x-3)}$

$=\displaystyle \int \frac{1 dx}{4 \cos^{2}x}-\int \frac{1}{3}dx$

$=\displaystyle \frac{1}{4}\int \sec^{2}xdx -\frac{1}{3}\int dx$

$\displaystyle I=\frac{1}{4} \tan x -\frac{1}{3}x+c$

$\therefore \displaystyle \int \frac{\cos x}{\cos 3x}=\frac{1}{4} \tan x-\frac{1}{3}x+C$

Evaluate: $\displaystyle\int\displaystyle\frac{x^3+5x^2+4x+1}{x^2}dx$ .

$I=\displaystyle \int \dfrac{x^3+5x^2+4x+1}{x^2}dx$

$I=\displaystyle \int \left (x+5+\dfrac{4}{x}+\dfrac{1}{x^2} \right )dx$

Now integral of

$\displaystyle \int x^a dx=\dfrac{x^{a+1}}{a+1}$ and of

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

Applying this in

$I$ we get,

$I=\dfrac{x^2}{2}+5x+4\ln x-\dfrac{1}{x}$

The value of $\int { \cos ^{ 6 }{ x } } dx$ is

1450185_837459_ans_c8514e223ef6471b9b046695068ed516.jpeg

Evaluate:
$\displaystyle\int\dfrac{2^{x+1}-5^{x-1}}{10^x}dx$

$\int_{}^{} {\dfrac{{{2^{x + 1}} - {5^{x - 1}}}}{{{{10}^x}}}} dx$

$= \int_{}^{} {\dfrac{{2 \times {2^x} - \dfrac{{{5^x}}}{5}}}{{{2^x} \times {5^x}}}dx}$

$= 2\int_{}^{} {{5^{ - x}}dx} - \dfrac{1}{5}\int_{}^{} {{2^{ - x}}dx}$

$= - 2 \times {5^{ - x}}\log 5 + \dfrac{1}{5}{2^{ - x}}\log 2 + c{\text{ }}$

$\left( {\because \int_{}^{} {{a^{bx}}dx = \dfrac{{{a^{bx}}\log a}}{b}} } \right)$

$= \dfrac{{ - 2\log 5}}{{{5^x}}} + \dfrac{{\log 2}}{{5 \times {2^x}}} + c$

Find $\displaystyle \int e^{x}\sin x\ dx$ .

Let

$I = \int e^{x} . \sin x dx$ (Integrating by parts)

$= \sin x . \int e^{x} dx - \int e^{x} dx . \dfrac {d}{dx}(\sin x). dx$

$= \sin x . e^{x} - \int e^{x} . \cos x dx$

$= \sin x . e^{x} - \left [\cos x \int e^{x} dx - \left [\int e^{x} dx . \dfrac {d}{dx} (\cos x)dx\right ]\right ]$

$= \sin x . e^{x} - \cos x e^{x} + \int e^{x} (-\sin x)dx$

$= e^{x} \sin x - \cos x e^{x} - \int e^{x} \sin x\ dx$

$2I = e^{x}(\sin x - \cos x) \Rightarrow I = \dfrac {e^{x}}{2} (\sin x - \cos x) + c$ .

$\int { { e }^{ 3x } } \cdot { x }^{ 2 }dx$

1945161_1029567_ans_ccf2be6b895544538bceec6a4d67de80.jpeg

Integrate the following function: $e^x\left(\dfrac{1+ sinx}{1+ cosx}\right)$

$\\ \int e^x(\frac{1+sinx}{1+cosx})dx\\=\int e^x(\frac{1+2sin(\frac{x}{2})cos(\frac{x}{2})}{1+2cos^2(\frac{x}{2})})dx\\=\int e^x[(\frac{1}{2})sec^2(\frac{x}{2})+tan(\frac{x}{2})]dx\\as f(x) =tan(\frac{x}{2}) \\ thenf'(x)=sec^2(\frac{x}{2})\cdot (\frac{1}{2})dx\\\therefore\> integration\> is\> of\> the\> form \\\int e^x[f(x)+f'(x)]dx=e^xf(x)\\\therefore I=e^xtan(\frac{x}{2})+c$

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

Given:

$\int { \log x } dx$

$u=\log x; v=1$

$\Rightarrow du=\dfrac { 1 }{ x } ; \int { vdx } =x$

$\Rightarrow x\log x-\int { dx }$

$\Rightarrow x\log x-x+c$

Solve $\displaystyle \int_0^{\pi/4} ln (1+tan\,\,x)dx$

$I = \int_{0}^{\pi /4}ln(1+tanx)dx$

$I = \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$

$I = \int_{0}^{\pi /4}ln(1+tan(\frac{\pi }{4}-x))dx$

$I = \displaystyle\int_{0}^{\pi /4}ln\left(1+\dfrac{tan\dfrac{\pi }{4}-tanx}{1+tan\dfrac{\pi }{4}.tanx}\right)dx$

$I = \displaystyle\int_{0}^{\pi /4}ln\left(1+\dfrac{1-tanx}{1+tanx}\right)dx$

$I = \displaystyle \int_{0}^{\pi /4}ln\left(\dfrac{2}{1+tanx}\right)dx$

$I = \int_{0}^{\pi /4}ln(2)dx-\int_{0}^{\pi /4}ln(1+tan(x))dx$

$I = \int_{0}^{\pi /4}ln(2)dx-I$

$2I = \int_{0}^{\pi /4}ln(2)dx$

$I = \dfrac{ln(2)}{2}(x)_{0}^{\pi /4}$

$I = \dfrac{\pi }{8}ln(2)$

$\displaystyle \int {\sin x\sqrt {1 - \cos 2x} } \;dx$

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

$\displaystyle \Rightarrow \sqrt{2}\int sin\,x(sin\,x)dx$

$\displaystyle \Rightarrow \sqrt{2}\int sin^{2}x\,dx \Rightarrow \frac{+\sqrt{2}}{2}\int 1-cos\,2x\,dx$

$\displaystyle \frac{1}{\sqrt{2}}(x)-\frac{1}{2\sqrt{2}}sin\,2x$

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Integrate $\displaystyle \int {\sqrt {1 + {x^2}} } dx$ is equal to:

$= \mathop \smallint \nolimits^ {\sec ^2}\left( u \right)\sqrt {{{\tan }^2}\left( u \right) + 1} {\mkern 1mu} {\rm{d}}u$

$= \mathop \smallint \nolimits^ {\sec ^3}\left( u \right){\mkern 1mu} {\rm{d}}u$

$=\dfrac { { \sec \left( u \right) \tan \left( u \right) } }{ 2 } +\dfrac { 1 }{ 2 } \int { sec\left( u \right) } { { d } }u$

$\int {\sec \left( u \right)du}$

$= \ln \left( {\tan \left( u \right) + \sec \left( u \right)} \right)$

$\dfrac{{\sec \left( u \right)\tan \left( u \right)}}{2} + \dfrac{1}{2}\int {\sec \left( u \right)du}$

$= \dfrac{{\ln \left( {\tan \left( u \right) + \sec \left( u \right)} \right)}}{2} + \dfrac{{\sec \left( u \right)\tan \left( u \right)}}{2}$

$\mathop \smallint \nolimits^ \sqrt {{x^2} + 1} {\mkern 1mu} {\rm{d}}x$

$= \dfrac{{\ln \left( {\left| {\sqrt {{x^2} + 1} + x} \right|} \right)}}{2} + \dfrac{{x\sqrt {{x^2} + 1} }}{2} + C$

$\dfrac{{ar\sinh \left( x \right)}}{2} + \dfrac{{x\sqrt {{x^2} + 1} }}{2} + C$

Find $\displaystyle \int_{}^{} {x{e^x}dx.}$

$\begin{array}{l}\int_{}^{} {x{e^x}dx} = x{e^x} - \int_{}^{} {{e^x}} dx\\ = x{e^x} - {e^x} + c\end{array}$

$\int \frac{x^{-2/3}}{\sqrt(x^{1/3})^{2} - 4} dx$ =

1311768_1031270_ans_8a75afbda6a6497ab31d09a0ee87c4c2.jpeg

Number of real solution of the given equation for $x$ , $\int x^{2}\ e^{x}dx=0$

$\int_{}^{} {{x^2}{e^x}dx = 0}$

$\Rightarrow {x^2}{e^x} - \int_{}^{} {2x{e^x}dx} = 0$

$\Rightarrow {x^2}{e^x} - 2\left[ {x{e^x} - \int_{}^{} {{e^x}dx} } \right] = 0$

$\Rightarrow {x^2}{e^x} - 2x{e^x} + 2{e^x} = 0$

$\Rightarrow {e^x}\left( {{x^2} - 2x + 2} \right) = 0$

$\Rightarrow \left( {{x^2} - 2x + 2} \right) = 0$

Now,

$D = {\left( { - 2} \right)^2} - 4 \times 1 \times 2 = 4 - 8 = - 4 < 0$

So, it has no real roots

$\displaystyle \int \dfrac {1-x}{1+x}$

$\int {\cfrac{{1 - x}}{{1 + x}}dx}$

$\int {\left( {\cfrac{1}{{1 + x}} - \cfrac{x}{{1 + x}}} \right)dx}$

$\int {\left( {\cfrac{1}{{1 + x}} - \left( {\cfrac{{1 + x - 1}}{{1 + x}}} \right)} \right)dx}$

$\int {\left( {\cfrac{1}{{1 + x}} - 1 + \cfrac{1}{{1 + x}}} \right)dx}$

$\int {\cfrac{2}{{1 + x}}dx - \int {1} dx}$

$=2 \log (x+1) -x+c$

If $\int \dfrac{1}{1+ \sin x} dx=\tan x-m\sec x$ .Find $m$

Solution:-

$\cfrac{1}{1 + \sin{x}} = \cfrac{1}{1 + \sin{x}} \times \cfrac{1 - \sin{x}}{1 - \sin{x}}$

$\Rightarrow \; = \cfrac{1 - \sin{x}}{1 - \sin^{2}{x}}$

$\Rightarrow \; = \cfrac{1 - \sin{x}}{\cos^{2}{x}} \; \left\{ \because \sin^{2}{x} + \cos^{2}{x} = 1 \right\}$

$\Rightarrow \; = \cfrac{1}{\cos^{2}{x}} - \cfrac{\sin{x}}{\cos^{2}{x}}$

$\Rightarrow \; = \sec^{2}{x} - \sec{x} \tan{x}$

$R.H.S.-$

$\tan{x} - m \sec{x}$

$L.H.S.-$

$\therefore \; \int{\cfrac{1}{1 + \sin{x}} dx} = \int{\left( \sec^{2}{x} - \sec{x} \tan{x} \right) dx}$

$\Rightarrow \; \int{\sec^{2}{x} \; dx} - \int{ \sec{x} \tan{x} \; dx}$

$\Rightarrow \; = \tan{x} - \sec{x}$

On comparing the values of L.H.S. and R.H.S., we have

$m = 1$

Hence, the correct answer is 1.

Solve:
$\int { \tan { x } \tan { 2x } \tan { 3xdx } }$

$\displaystyle\int \tan x\tan 2x\tan 3x dx$ .

$\tan 3x=\tan (x+2x)=\dfrac{\tan +\tan 2x}{1-\tan x\tan 2x}$

$\tan 3x(1-\tan x\tan 2x)=\tan x+\tan 2x$

$\tan 3x-\tan x-\tan 2x=\tan x\tan 2x\tan 3x$

$\displaystyle\int \tan 3x-\tan x-\tan 2x dx$

$\Rightarrow \dfrac{1}{3}log|\sec 3x|-\dfrac{1}{2}log|\sec 2x|-log|\sec x|+C$ .

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

$I=\displaystyle\int { \left( \log(\log x)+\dfrac { 1 }{ { (\log x) }^{ 2 } } \right) dx } \\ \quad =\displaystyle\int { \log(\log x)dx } +\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ Let\quad { I }_{ 1 }=\displaystyle\int { \log(\log x)dx } \\ \quad \quad \quad\quad =\displaystyle\int { \log(\log x).1dx } \\ \text{continue with integration by part}:\\ \displaystyle\int { f(x)g(x)dx } =f(x)\displaystyle\int { g(x)dx } -\int { \left[ f'(x)\int { g(x)dx } \right] } dx\\ now,\\ { { I }_{ 1 } }=\log(\log x)\displaystyle\int { 1 } dx\quad -\quad \int { \left[ \dfrac { d }{ dx } (\log(\log x)).\int { 1dx } \right] } dx\\ \quad \quad =x\log(\log x)\quad -\quad \displaystyle\int { \left[ \dfrac { 1 }{ \log x } .\dfrac { 1 }{ x } .x \right] } dx\\ \quad \quad =x\log(\log x)\quad -\quad\displaystyle \int { \dfrac { 1 }{ \log x } } dx\\ Let\quad { I }_{ 2 }=\displaystyle\int { \dfrac { 1 }{ \log x } } dx\\ \quad \quad \quad \quad \quad =\displaystyle\int { \dfrac { 1 }{ \log x } .1dx } \\ \quad \quad \quad \quad \quad =\dfrac { 1 }{ \log x } .\displaystyle\int { 1dx } \quad -\quad \int { \left[ \dfrac { d }{ dx } \left( \dfrac { 1 }{ \log x } \right) .\displaystyle\int { 1dx } \right] } dx\\ \quad \quad \quad \quad \quad =\quad \dfrac { x }{ \log x } \quad -\quad \displaystyle\int { \dfrac { -1 }{ { (\log x) }^{ 2 } } .\dfrac { 1 }{ x } .xdx } \\ \quad \quad \quad \quad \quad =\dfrac { x }{ \log x } +\displaystyle\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ putting\quad the\quad value\quad of\quad { I }_{ 2 }\quad in\quad { I }_{ 1 }\\ { I }_{ 1 }=x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad -\quad \displaystyle\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ now\quad putting\quad the\quad value\quad of\quad { I }_{ 1 }\quad in\quad I\\ I\quad =\quad x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad -\quad\displaystyle \int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \quad +\quad \int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ \quad \quad =\quad x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad +\quad C,\quad where\quad C\quad =\quad integration\quad constant\quad$

Solve $\displaystyle\int^{\pi/2}_0e^x\cdot\sin \left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)dx$ .

Given: $\int _0^{\dfrac{\pi }{2}}e^x\sin \left(\dfrac{\pi }{4}+\dfrac{x}{2}\right)dx$

Apply u-substitution, $u=\dfrac{x}{2}$

$= \int _0^{\dfrac{\pi} {2}} \:e^{2u}\sin \left(\dfrac{\pi }{4}+u\right)\cdot \:2du$

$= 2\cdot \int _0^{\dfrac{\pi} {2}} \:e^{2u}\sin \left(\dfrac{\pi }{4}+u\right)du$

Apply integration by parts,

$u=e^{2u},\:v'=\sin \left(\dfrac{\pi }{4}+u\right)$

$= 2\left(-e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)-\int \:-2e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)du\right)$

$=2\left(-e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)-\left(-2\cdot \int \:e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)du\right)\right)$

Apply integration by parts,

$u=e^{2u},\:v'=\cos \left(\dfrac{\pi }{4}+u\right)$

$=2\left(-e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)-\left(-2\left(e^{2u}\sin \left(\dfrac{\pi }{4}+u\right)-\int \:e^{2u}\cdot \:2\sin \left(\dfrac{\pi }{4}+u\right)du\right)\right)\right)$

$=-\dfrac{2}{5}e^{2u}\left(\cos \left(u+\dfrac{\pi }{4}\right)-2\sin \left(u+\dfrac{\pi }{4}\right)\right)$

Substitute back $u=\dfrac{x}{2}.$

$=-\dfrac{2}{5}e^{2\cdot \dfrac{x}{2}}\left(\cos \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)-2\sin \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)\right)$

$=-\dfrac{2}{5}e^x\left(\cos \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)-2\sin \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)\right)$

$= \left [-\dfrac{2}{5}e^x\left(\cos \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)-2\sin \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)\right) \right ] _0^{\dfrac{\pi }{2}}$

$= \dfrac{4e^{\dfrac{\pi }{2}}}{5}-\dfrac{\sqrt{2}}{5}$

$= \dfrac{4e^{\dfrac{\pi }{2}}-\sqrt{2}}{5}$

Solve $\displaystyle\int {\frac{{{e^x}\left( {{x^2} + 5x + 7} \right)}}{{{{\left( {x + 3} \right)}^2}}}} dx$

$\displaystyle\int {\dfrac{{{e^x}\left( {{x^2} + 5x + 7} \right)}}{{{{\left( {x + 3} \right)}^2}}}} dx$

let

${u^2} = {\left( {x + 3} \right)^2}\\$

$\implies {x^2} + 5x + 7 = {u^2} - u + 1\\$

$\displaystyle \int {\dfrac{{{e^{u - 3}}\left( {{u^2} - u + 1} \right)}}{{{{\left( u \right)}^2}}}} du\\$

$\displaystyle={e^{-3}}\int {{e^u}\left( {\dfrac{1}{{{u^2}}} - \dfrac{1}{u} + 1} \right)du\,} \,$

$\displaystyle= {e^{ - 3}}\int {{e^u}du} \, + {e^{ - 3}}\int {\left( {\dfrac{1}{u} - \dfrac{1}{{{u^2}}}} \right){e^u}du} \,\\$

$\displaystyle= {e^{ - 3}}{e^u} + \dfrac{{{e^{ - 3}}}}{u}$

$= {e^{u - 3}} + \dfrac{{{e^{ - 3}}}}{{u }}$

Putting the value of u, we get

$= {e^x} + \dfrac{{{e^{ - 3}}}}{{x + 3}} + c$

Integrate: $\displaystyle \int e^x\sin x\ dx$

Consider the given integral.

$I=\int{{{e}^{x}}\sin xdx}$

We will use integration by parts method.

We know that,

$\int{udv=uv-\int{vdu}}$

Let,

$u={{e}^{x}}\Rightarrow du={{e}^{x}}dx$

$dv=\sin x\Rightarrow v=-\cos x$

Then,

$I={{e}^{x}}\left( -\cos x \right)-\int{{{e}^{x}}\left( -\cos x \right)dx}$

$I=-{{e}^{x}}\cos x+\int{{{e}^{x}}\cos xdx}$

Again, we will use integration by parts.

Let,

$p={{e}^{x}}\Rightarrow dp={{e}^{x}}dx$

$dq=\cos x\Rightarrow q=\sin x$

Then,

$I=-{{e}^{x}}\cos x+{{e}^{x}}\sin x-\int{{{e}^{x}}\sin xdx}+C$

$I=-{{e}^{x}}\cos x+{{e}^{x}}\sin x-I+C$

$2I={{e}^{x}}\left( \sin x-\cos x \right)+C$

$I=\dfrac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)+C$

Hence, this is the required value of the integral.

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

Given: $\displaystyle\int \dfrac{\sqrt{1+x^2}}{x^2}dx$

Apply the integration by parts,

$u = \sqrt{1 + x^{2}}, v = \dfrac{1}{x^{2}}$

$\displaystyle\int \dfrac{\sqrt{1+x^2}}{x^2}dx = -\dfrac{\sqrt{1+x^2}}{x}-\int \:-\dfrac{1}{\sqrt{1+x^2}}dx$

$= -\dfrac{\sqrt{1+x^2}}{x}-\left(-\ln \left|\sqrt{x^2+1}+x\right|\right)$

$= -\dfrac{\sqrt{1+x^2}}{x}+\ln \left|\sqrt{x^2+1}+x\right|+C$

Hence, the required result is found.

Solve $\dfrac{1}{4}\displaystyle\int \left(\dfrac{1+\cos 2x}{2}\right)^2dx$ .

Given: $\dfrac{1}{4}\cdot \int \:\left(\dfrac{1+\cos \left(2x\right)}{2}\right)^2dx$

Simplify the integral.

$\dfrac{1}{4}\cdot \displaystyle\int \:\left(\dfrac{1+\cos \left(2x\right)}{2}\right)^2dx$

$=\dfrac{1}{4}\cdot \int \:\cos ^4\left(x\right)dx$

$=\dfrac{1}{4}\cdot \int \:\cos ^3\left(x\right)\cos \left(x\right)dx$

Apply integration by parts,

$u=\cos ^3\left(x\right),\:v'=\cos \left(x\right)$

$=\dfrac{1}{4}\cdot \left(\cos ^3\left(x\right)\sin \left(x\right)-\int \:-3\cos ^2\left(x\right)\sin ^2\left(x\right)dx\right)$

$=\dfrac{1}{4}\cdot \left(\cos ^3\left(x\right)\sin \left(x\right)-\left(-\dfrac{3}{8}\left(x-\dfrac{1}{4}\sin \left(4x\right)\right)\right)\right)$

$=\dfrac{1}{4}\cdot (\cos ^3\left(x\right)\sin \left(x\right)+\dfrac{3}{8}\left(x-\dfrac{1}{4}\sin \left(4x\right)\right))$

$=\dfrac{1}{4}\left(\cos ^3\left(x\right)\sin \left(x\right)+\dfrac{3}{8}\left(x-\dfrac{1}{4}\sin \left(4x\right)\right)\right)+C$

Integrate $\int x.\cos^{-1}x$

$\int {x{{\cos }^{ - 1}}xdx}$

$putting\,x = \cos \theta$

$dx = i\sin \theta d\theta$

$= \int {\cos \theta \times {{\cos }^{ - 1}}\left( {\cos \theta } \right) \times - \sin \theta d\theta }$

$= - \int {\sin \theta \cos \theta \times \theta d\theta }$

$= - \dfrac{1}{2}\int {2\sin \theta cos\theta .\theta d\theta }$

$= - \dfrac{1}{2}\int {\theta .\sin 2\theta d\theta }$

$= - \dfrac{1}{2}\left[ {\theta \times \dfrac{{ - \cos 2\theta }}{2} - \int {1 \times \frac{{ - \cos 2\theta }}{2}d\theta } } \right]$

$= - \dfrac{1}{2}\left[ { - \theta \dfrac{{\cos 2\theta }}{2} + \frac{1}{2} \times \frac{{\sin 2\theta }}{2}} \right] + c$

$= \dfrac{{\theta \cos 2\theta }}{4} - \dfrac{1}{8}\sin 2\theta + c$

$= \dfrac{{\theta \left( {2{{\cos }^2}\theta - 1} \right)}}{4} - \dfrac{1}{8}\sin 2\theta + c$

$= \dfrac{{\theta \left( {2{{\cos }^2}\theta - 1} \right)}}{4} - \dfrac{1}{4}\cos \theta \sqrt {1 - {{\cos }^2}\theta } + c$

$= \dfrac{{\left( {2{x^2} - 1} \right){{\cos }^{ - 1}}x}}{4} - \dfrac{x}{4}\sqrt {1 - {x^2}} + c$

Evaluate $\displaystyle\int a^{x} e^{x}\ dx$ .

$\int{a^xe^x}dx$

$\Rightarrow$ Let

$I=\int a^xe^xdx$

Here we use intergration by parts,

$\int{udv}=uv-\int{vdu}$

$\Rightarrow I=a^xe^x-\int{a^x\space (\ln{a})\space e^x}dx$

$\Rightarrow I=a^xe^x-\ln{a}\int{a^x\space \space e^x}dx$

This can be written as

$\Rightarrow I=a^xe^x-(\ln{a})I$

$\Rightarrow (1+\ln{a})I=a^xe^x$

$\Rightarrow I=\dfrac{a^xe^x}{1+\ln{a}}$

Therefore,

$\Rightarrow \int{a^xe^x}dx=\dfrac{a^xe^x}{1+\ln{a}}$

$\displaystyle \int {\dfrac{7^{2x+3}\sin^2 2x+ \cos^22x}{\sin^22x}}=\dfrac{7^{2x+3}}{2\log 7}-\dfrac{(\cot x+x)}{b}$ .Find $b$

$\\\int \left[7^{2x+3}+cot^2(2x)\right]dx\\=7^3\int 7^{2x}dx+\int [cosec^2(2x)-1]dx\\=7^3\int 49^x dx-(\frac{cot(2x)}{2})-x+C\\=7^3 (\frac{49^x}{log 49})-(\frac{cot(2x)}{2})-x+C\\=(\frac{7^{2x+3}}{2log7})-(\frac{[cot(2x)+2x]}{2})+C\\\therefore\>b=2$

$\displaystyle\int x^3\tan^{-1}xdx$ .

When the integer and can be expressed as a product of two Function. If

$f(x)=$ first Function (that can be differentiated) and

$g(x)=$ second Function (that can be integrated ), then preference of

$f(x)$ & decided by

$ILATE$

So,

$f(x) =\tan^{-1}x$ &

$g(x)=x^3$

$\displaystyle \int f(x) g(x) dx=f(x) \displaystyle \int g(x) dx= \displaystyle \int \left\{\dfrac {d}{dx} f(x) \displaystyle \int g(x) \right\}dx$

so,

$\displaystyle \int x^3 \tan^{-1} x\ dx=\tan ^{-1}x \displaystyle \int x^3 dx-\displaystyle \int \left\{\dfrac {d}{dx} (\tan^{-1}x \displaystyle \int x^3 dx\right\}$

$=\tan^{-1}x \dfrac {x^4}{4}-\displaystyle \int \dfrac {1}{(x^2 +1)}\times \dfrac {x^4}{4}dx$

$=\dfrac {x^4}{4}. \tan^{-1}dx-\dfrac {1}{4}\displaystyle \int \dfrac {x^4}{x^2+1}dx$

$=\dfrac {x^4}{4}\tan^{-1} x-\dfrac {1}{4} \displaystyle \int \dfrac {x^4 -1+1}{x^2 +1}dx$

$=\dfrac {x^4}{4}.\tan^{-1}x-\dfrac {1}{4} \left [\displaystyle \int \dfrac {x^4 -1}{x^2 +1} dx \displaystyle \int \dfrac {1}{1+x^2} dx \right]$

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

$=\dfrac {x^4}{4}\times \tan^{-1} x-\dfrac {1}{4} [\displaystyle \int (x^2 -1)dx +\tan^{-1}x]$

$=\dfrac {x^4}{4}.\tan^{-1}x \dfrac {-1}{4} \left [\dfrac {x^3}{3}-x+\tan^{-1}x\right]+C$

$=\left(\dfrac {x^4}{4}-\dfrac {1}{4}\right) \tan^{-1}x -\dfrac {1}{4}x^3 +\dfrac {1}{4}x+C$

$=\dfrac {(x^4 -1)}{4}\tan^{-1}x-\dfrac {x^3}{4}+\dfrac {x}{4}+C$

Integrate with respect to $x$ .:
$e^{x}\sin x$

$\int e^x \sin x dx$

This can be solved by applying the concept of integration by parts

$\int{udv}=uv-\int{vdu}$

Take

$\sin x$ as

$u$ and

$e^x$ as

$dv$ , we get

$\Rightarrow \int{e^x \sin x}\space dx=I$ (let) ....

$(1)$

$\Rightarrow I=\sin x (e^x)-\int{\cos x \space e^x}dx$

Again apply integration by parts

$\Rightarrow I=\sin x \space e^x-(\cos x \space e^x-\int{(-\sin x)e^x}dx)$

$\Rightarrow I=\sin x \space e^x-(\cos x \space e^x+\int{(\sin x)e^x}dx)$

$\Rightarrow I=\sin x \space e^x-(\cos x \space e^x+I)$ (from

$(1)$ )

$\Rightarrow I=\sin x \space e^x-\cos x \space e^x-I$

$\Rightarrow 2I=\sin x \space e^x-\cos x \space e^x$

$\Rightarrow 2I=e^x(\sin x -\cos x)$

$\Rightarrow I=\dfrac{e^x(\sin x -\cos x)}{2}$

Therefore,

$\Rightarrow \int e^x \space \sin x\space dx=\dfrac{e^x(\sin x -\cos x)}{2}$

Evaluate $\int {{e^x}\left( {\tan x - \log \,\cos \,x} \right)\,dx}$

It is just an observation to make perfect differential;

it is in the form of

$\displaystyle \int (e^{x}(f(x))+e^{x}(\dfrac{d}{dx}f(x)))dx=e^{x}(f(x))$

$\dfrac{d}{dx}\log(\cos x)=-\tan x$

so here

$f(x)=\log(\cos x)$

so,

$\displaystyle\int e^{x}(\tan x-\log \cos x)dx=e^{x}(-\log\cos x)$

Evaluate $\int {x{\mathop{\rm sintx}\nolimits} \,dx}$

$\displaystyle\int{x\sin{tx}dx}$

Let

$u=x\Rightarrow\,du=dx$

$dv=\sin{tx}dx\Rightarrow\,v=-\dfrac{\cos{tx}}{t}$

$=-\dfrac{x\cos{tx}}{t}-\displaystyle\int{-\dfrac{\cos{tx}}{t}dx}$

$=-\dfrac{x\cos{tx}}{t}+\dfrac{1}{t}\displaystyle\int{\cos{tx}dx}$

$=-\dfrac{x\cos{tx}}{t}+\dfrac{1}{t}\dfrac{\sin{tx}}{t}+c$

$=-\dfrac{x\cos{tx}}{t}+\dfrac{\sin{tx}}{{t}^{2}}+c$

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

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

Integrating by parts, we get

Let

$u={x}^{2}\Rightarrow\,du=2x\,dx$

$dv={e}^{x}dx\Rightarrow\,v={e}^{x}$

$I={x}^{2}{e}^{x}-\displaystyle\int{2x{e}^{x}dx}$

$I={x}^{2}{e}^{x}-2\displaystyle\int{x{e}^{x}dx}$

Let

$u=x\Rightarrow\,du=dx$

$dv={e}^{x}dx\Rightarrow\,v={e}^{x}$

$I={x}^{2}{e}^{x}-2\left[x{e}^{x}-\displaystyle\int{{e}^{x}dx}\right]$

$I={x}^{2}{e}^{x}-2\left[x{e}^{x}-{e}^{x}\right]+c$

$\therefore\,I={x}^{2}{e}^{x}-2x{e}^{x}+2{e}^{x}+c$

$\int { \cos ^{ -1 }{ \left( 4{ x }^{ 3 }-3x \right) dx } }$

1323449_1063144_ans_6f01176fb1ca4695861eae6c99bac84b.jpg

$\int \sin x \log (\cos x)dx$

Let

$t=\cos{x}\Rightarrow dt=-\sin{x}dx$

$\displaystyle\int{\sin{x}\log{\cos{x}}dx}$

$=-\displaystyle\int{\log{t}dt}$

Integrating by parts,

Let

$u=\log{t}\Rightarrow du=\dfrac{1}{t}dt$

$dv=dt\Rightarrow v=t$

$=-\left[t\log{t}-\displaystyle\int{t\times\dfrac{dt}{t}}\right]$

$=-t\log{t}+\displaystyle\int{dt}$

$=-t\log{t}+t+c$ where

$c$ is the constant of integration

$=-\cos{x}\log{\left(\cos{x}\right)}+\cos{x}+c$ where

$t=\cos{x}$

$\int x^{3}\cos x^{2}dx$

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

$=\displaystyle\int{\left({x}^{2}\cos{{x}^{2}}\right)xdx}$

Let

$t={x}^{2}\Rightarrow dt=2xdx$

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

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

Let

$u=t\Rightarrow du=dt$

and

$dv=\cos{t}dt \Rightarrow v=\sin{t}$

$=\dfrac{1}{2}\left[t\sin{t}-\displaystyle\int{\sin{t}dt}\right]$

$=\dfrac{1}{2}\left[t\sin{t}+\cos{t}\right]+c$ where

$t={x}^{2}$

$=\dfrac{1}{2}\left[{x}^{2}\sin{{x}^{2}}+\cos{{x}^{2}}\right]+c$ where

$c$ is the constant of integration.

Find the following integral.
$\displaystyle\int e^x (\sec^2 x + \tan x).dx$

Now,

$\displaystyle\int e^x (\sec^2 x + \tan x).dx$

$=\displaystyle\int e^x (\sec^2 x ).dx+$

$\displaystyle\int e^x ( \tan x).dx$

$=e^x\displaystyle\int\sec^2 x.dx-\displaystyle\int e^x ( \tan x).dx$

$+\displaystyle\int e^x ( \tan x).dx$ [Using method of by parts]

$=e^x.\tan x-\displaystyle\int e^x ( \tan x).dx$

$+\displaystyle\int e^x ( \tan x).dx$

Evaluate:
$\displaystyle\int{{x}^{2}{(3-x)}^{5}dx}$

$\displaystyle =\int x^2 (3-x)^5dx$

$I$

$II$

$\displaystyle =x^2 \int(3-x)^5dx - \int\left(\dfrac{d(x^2)}{dx} \int (3-x)^5dx\right)dx$

$\displaystyle = x^2 \dfrac{(3-x)^6}{6} (-1) - \int 2x \dfrac{(3-x)^6}{6} (-1)dy$

$\displaystyle = \dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3} \int x(3-x)^6dy$

$I$

$II$

$\displaystyle = \dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3} \left[ x\int (3-x)^6 dx - \int\left(\dfrac{dx}{dx}\int (3-x)^6 dx\right)dx\right]$

$\displaystyle =\int \dfrac{-x^2}{6}(3-x)^6 + \dfrac{1}{3} \left[x \dfrac{(3-x)^7}{7} (-1)-\int \dfrac{(3-x)^7}{7}(-1)dx\right]$

$\displaystyle = \dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3} \left[-\dfrac{x}{7}(3-x)^7 + \dfrac{1}{7}\int (3-x)^7dx\right]$

$\displaystyle =-\dfrac{x^2}{6} (3-x)^6 + \dfrac{1}{3} \left[-\dfrac{x}{7} (3-x)^7 + \dfrac{1}{7} \dfrac{(3-x)^8}{8} (-1) \right]+C$

$\displaystyle =\dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3}\left[ -\dfrac{x}{7} (3-x)^7 - \dfrac{1}{56} (3-x)^8 \right]+ C$

$\int { { x }^{ 3 }\tan ^{ -1 }{ x } dx }$

1323450_1063155_ans_3ab873ed4037402192c2cf7ab90b1827.jpg

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

1323235_1063152_ans_31061d94577d4c78b4a36da558174c1b.jpg

$\int { \tan ^{ -1 }{ \left( \sqrt { x } \right) } } dx$

1323234_1063146_ans_25cd9250e1154e4c86d8b5753f46131f.jpg

Evaluate $\int\tan^{-1}x dx$

Let $I = \int {{{\tan }^{ - 1}}xdx}$

Applying by parts,

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

$= x{\tan ^{ - 1}}x - \int {\cfrac{x}{{1 + {x^2}}}dx}$

$= x{\tan ^{ - 1}}x - {I_1}$

Let ${I_1} = \int {\cfrac{x}{{1 + {x^2}}}dx}$ , then, put $1 + {x^2} = t$ and $2xdx = dt$ .

${I_1} = \cfrac{1}{2}\int {\cfrac{{dt}}{t}}$

$= \cfrac{1}{2}\log t + c$

$= \cfrac{1}{2}\log \left( {1 + {x^2}} \right) + c$

Therefore,

$I = x{\tan ^{ - 1}}x - \cfrac{1}{2}\log \left( {1 + {x^2}} \right) + c$

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

1317517_1064099_ans_468e8da83b764e20a46b929c1c257866.jpg

$\int {{e^x}\left( {\cos x - \sin x} \right)dx}$

$\int e^x(cosx-sinx) dx=\int d(e^xcosx)=e^xcosx+c$

$\int {\ell n\left( {{x^2} + 1} \right)dx}$

$\int ln(x^{2}+1)dx= xln(x^{2}+1)-\int\frac{x}{x^{2}+1}2xdx$

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

$=xln(x^{2}+1)-2x+2tan^{-1}x+ C$

$\int {{e^x}\left( {{{\sec }^2}x + \tan x} \right)} dx$

$\int e^{x}(sec^{2} x+tan x)dx=\int d(e^{x}tanx )$

$= e^{x}tan x +c$

Solve $\int { \cfrac { 1 }{ 9{ x }^{ 2 }+6x+10 } dx }$

$\int { \cfrac { 1 }{ 9{ x }^{ 2 }+6x+10 } dx } =\int { \cfrac { 1 }{ 9{ x }^{ 2 }+6x+1+9 } dx }$

$=\int { \cfrac { 1 }{ { (3x+1) }^{ 2 }+9 } dx }$

$=\cfrac { 1 }{ 9 } \int { \cfrac { 1 }{ { \left( x+\cfrac { 1 }{ 3 } \right) }^{ 2 }+1 } dx }$

$=\cfrac { 1 }{ 9 } \int { \cfrac { 1 }{ { 1+\left( x+\cfrac { 1 }{ 3 } \right) }^{ 2 } } d\left( x+\cfrac { 1 }{ 3 } \right) }$

$=\cfrac { 1 }{ 9 } \tan ^{ -1 }{ \left( x+\cfrac { 1 }{ 3 } \right) } +C$

Evaluate $\int{\sin ^{ - 1}}xdx$

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

Integrating by parts, we get

Let

$u={\sin}^{-1}{x}\Rightarrow\,du=\dfrac{dx}{\sqrt{1-{x}^{2}}}$

$dv=dx\Rightarrow\,v=x$

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

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

Let

$t=1-{x}^{2}\Rightarrow\,dt=-2x\,dx$

$=x{\sin}^{-1}{x}+\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{\sqrt{t}}}$

$=x{\sin}^{-1}{x}+\dfrac{1}{2}\displaystyle\int{{t}^{-\frac{1}{2}}dt}$

$=x{\sin}^{-1}{x}+\dfrac{1}{2}\left[\dfrac{{t}^{-\frac{1}{2}+1}}{\dfrac{-1}{2}+1}\right]+c$

$=x{\sin}^{-1}{x}+\dfrac{1}{2}\times\dfrac{2}{1}\sqrt{t}+c$

$=x{\sin}^{-1}{x}+\sqrt{1-{x}^{2}}+c$ where

$t=1-{x}^{2}$

Evaluate: $\displaystyle \int { \dfrac { \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) }{ \left( x-4 \right) \left( x-5 \right) \left( x-6 \right) } dx }$

Consider,

$I=\displaystyle \int { \dfrac { \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) }{ \left( x-4 \right) \left( x-5 \right) \left( x-6 \right) } dx }$

$I=\displaystyle \int {1+\dfrac{9x^{2}-63x+114}{(x-4)(x-5)(x-6)}{dx}}$

$let \ \dfrac{9x^{2}-63x+114}{(x-4)(x-5)(x-6)}=\dfrac{A}{x-4}+\dfrac{B}{x-5}+\dfrac{C}{x-6}$

$On \ solving \ the \ above \ equation \ we \ get \ A =3 ,B =-24, C=30$

$Now \ the \ integration \ reduces \ to;$

$\Rightarrow$

$\displaystyle \int1{dx}+\int\dfrac{3}{x-4}{dx}+\int\dfrac{-24}{x-5}{dx}+\int\dfrac{30}{x-6}{dx}$

$= x+3\ln (x-4)-24\ln (x-5)+30\ln (x-6)$

Evaluate $\displaystyle \int {\frac{{{x^2}\,{e^x}}}{{{{\left( {x + \,2} \right)}^2}}}\,dx}$

Let

$t=x+2\Rightarrow\,dt=dx$

$=\displaystyle\int{\dfrac{{\left(t-2\right)}^{2}{e}^{t-2}}{{t}^{2}}dt}$

$=\displaystyle\int{\dfrac{\left({t}^{2}-4t+4\right){e}^{t-2}}{{t}^{2}}dt}$

$=\displaystyle\int{\dfrac{{t}^{2}{e}^{t-2}}{{t}^{2}}dt}-4\displaystyle\int{\dfrac{t{e}^{t-2}}{{t}^{2}}dt}+4\displaystyle\int{\dfrac{{e}^{t-2}}{{t}^{2}}dt}$

$=\displaystyle\int{{e}^{t-2}dt}-4\displaystyle\int{\dfrac{{e}^{t-2}}{t}dt}+4\displaystyle\int{\dfrac{{e}^{t-2}}{{t}^{2}}dt}$

$={e}^{t-2}-4\displaystyle\int{\dfrac{{e}^{t-2}}{t}dt}+4\displaystyle\int{\dfrac{{e}^{t-2}}{{t}^{2}}dt}$

Let

$u={e}^{t-2}\Rightarrow\,du={e}^{t-2}dt$

$dv=\dfrac{dt}{{t}^{2}}\Rightarrow\,v=-\dfrac{1}{t}$

$={e}^{t-2}-4\displaystyle\int{\dfrac{{e}^{t-2}}{t}dt}+4\left[-\dfrac{1}{t}{e}^{t-2}+\displaystyle\int{\dfrac{{e}^{t-2}dt}{t}dt}\right]+c$

$={e}^{t-2}-4\displaystyle\int{\dfrac{{e}^{t-2}}{t}dt}-4\dfrac{1}{t}{e}^{t-2}+4\displaystyle\int{\dfrac{{e}^{t-2}dt}{t}dt}+c$

$={e}^{t-2}-4\dfrac{1}{t}{e}^{t-2}+c$

$=\left(1-\dfrac{4}{t}\right){e}^{t-2}+c$

$=\left(1-\dfrac{4}{x+2}\right){e}^{x+2-2}+c$

$=\dfrac{x+2-4}{x+2}{e}^{x}+c$

$=\dfrac{x-2}{x+2}{e}^{x}+c$

Evaluate:

$\displaystyle \int \dfrac{\sin (x)}{\sqrt{1+ \sin (x)}} dx$

1340205_1082965_ans_857dcd5773bd4f4ba1b10e54858b8340.jpg

Evaluate: $\displaystyle\int \dfrac{\cos 2x+2\sin^2x}{\sin^2x}dx$ .

1332339_1083069_ans_2c6e2ab273d14790bcf63c2a20e39400.jpg

$\displaystyle\int \dfrac{1}{\cos^4x+\sin^4x}dx$ .

1374406_1080952_ans_d4752319c87e41f99391fc8282389787.jpg

$I = \displaystyle \int{\dfrac{x\cos x+\sin x}{x\sin x}}dx$ Then $I = ?$

Consider the given integral.

$I=\int{\dfrac{x\cos x+\sin x}{x\sin x}}dx$

Let $t=x\sin x$

$\dfrac{dt}{dx}=x\cos x+\sin x\times 1$

$dt=\left( x\cos x+\sin x \right)dx$

Therefore,

$I=\int{\dfrac{1}{t}}dt$

$I=\ln \left( t \right)+C$

Thus,

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

Hence, this is the answer.

Evaluate:
$\displaystyle\int \dfrac{\sin^3xdx}{(\cos^4x+3\cos^2x+1)\tan^{-1}(\sec x+\cos x)}$ .

1329061_1084096_ans_babdf7d8c1e2428b989a79d197c80a28.jpg

Evaluate the following :
$\displaystyle \int{\left(\dfrac{\tfrac{x}{x+1}-ln(x+1) }{x(ln(x+1)) }\right)}dx$

1329062_1084095_ans_65c90cde328247efb5de418ee98dc541.jpg

Integrate:
$\displaystyle\int{\left( x+1 \right)\sqrt{3-2x-{{x}^{2}}}}dx =$

Consider the given integral.

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

Let $t=3-2x-{{x}^{2}}$

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

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

Therefore,

$I=-\dfrac{1}{2}\int{\sqrt{t}}dt$

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

$I=-\dfrac{{{t}^{3/2}}}{3}+C$

On putting the value of $t$ , we have

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

Hence, this is the answer.

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

Consider the given integral.

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

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

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

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

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

$I=\dfrac{1}{4}\left[ \ln \left( {{x}^{2}}+1 \right) \right]+\dfrac{1}{2}\left( {{\tan }^{-1}}x \right)-\dfrac{1}{2}\ln \left( x+1 \right)+C$

Hence, this is the answer.

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

Consider the following integral.

$\int\limits_{{}}^{{}}{\dfrac{1}{\sin x{{\cos }^{2}}x}}dx$

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

$=\int\limits_{{}}^{{}}{(\sec x\tan x dx +cscxdx} )$

$=\int \sec x \tan xdx+\int \csc xdx$

$= \sec x- \ln(\csc x+\cot x)$

Hence, this is the correct answer.

$\int \frac{1}{x(x^n+1)}dx$ .

$\int(\frac{1}{x(x^n+1)})dx\\=\int(\frac{x^{n-1}}{x^n(x^n+1)})dx\\let\>x^n=t\>\\nx^{n-1}dx=dt\>\\\therefore\>x^{n-1}dx=(\frac{dt}{n})\\\therefore\>I=(\frac{1}{n})\int\>(\frac{1}{t(t+1)}dt)\\=(\frac{1}{n})[\int(\frac{1}{t})dt-\int\>(\frac{1}{t+1})dt]\\=(\frac{1}{n})[logt-log(t+1)]+c\\=(\frac{1}{n})log((\frac{1}{n}))+c\\=\>(\frac{1}{n})log((\frac{x^n}{x^n+1}))+c$

Simplify: $\int {\dfrac{{{e^x}\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^3}}}dx}$

$I=\int e^x\frac{(x-1)}{(x+1)^3}dx$

$I=\int e^x\frac{(x+1-2)}{(x+1)^3}dx$

$I=\int e^x\frac{1+x}{(x+1)^3}dx$

$-$

$\int e^x\frac{2}{(1+x)^3}dx$

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

$-$

$\int e^x\frac{2}{(x+1)^3}dx$

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

$I=$

$\frac { 1 }{ (1+ x )^{ 2 } } \int { e } ^{ x }-\int { \{ (\frac { d(\frac { 1 }{ (1+ x )^{ 2 } } ) }{ dx } } )\int { { e }^{ x } } dx\} dx$

$-$

$\int e^x\frac{2x}{(1+x^2)^2}dx$

$I= e^x\frac{1}{(1+x)^2}$

$+$

$\int e^x\frac{2}{(1+x)^3}dx$

$-$

$\int e^x\frac{2}{(1+x)^3}dx$

$I= e^x\frac{1}{(1+x)^2}$

$+$

$C$

$\int\frac{dx}{x^2-16}$

$I=\int (\frac{dx}{x^2-16})\\=\int (\frac{1}{x^2-4^2})dx \>(in\>the\>form\>of\>(\frac{1}{x^2-a^2}))\\= (\frac{1}{2\times 4})log \left | { (\frac{x-4}{x+4})} \right |+c \\ =(\frac{1}{8})log \left | { (\frac{x-4}{x+4})} \right |+c$

Integrate with respect to $x$ :
$\dfrac { { x }^{ 2 }\tan ^{ -1 }{ x } }{ 1+{ x }^{ 2 } }$

Consider the given integral.

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

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

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

Therefore,

$I=\int{\left( {{\tan }^{2}}t \right)tdt}$

$I=\int{\left( {{\sec }^{2}}t-1 \right)tdt}$

$I=\int{t{{\sec }^{2}}tdt}-\int{tdt}$

$I=t\tan t-\int{1\tan tdt}-\dfrac{{{t}^{2}}}{2}$

$I=t\tan t-\left( -\log \left( \cos t \right) \right)-\dfrac{{{t}^{2}}}{2}+C$

$I=t\tan t+\log \left( \cos t \right)-\dfrac{{{t}^{2}}}{2}+C$

On putting the value of $t$ , we get

$I={{\tan }^{-1}}x\tan \left( {{\tan }^{-1}}x \right)+\log \left( \cos \left( ta{{n}^{-1}}x \right) \right)-\dfrac{{{\left( {{\tan }^{-1}}x \right)}^{2}}}{2}+C$

$I=x{{\tan }^{-1}}x+\log \left( \cos \left( ta{{n}^{-1}}x \right) \right)-\dfrac{{{\left( {{\tan }^{-1}}x \right)}^{2}}}{2}+C$

Hence, this is the answer.

Integrate with respect to $x$ :
$e^{x}\sin x$

Consider the following integral.

$I=\int\limits_{{}}^{{}}{{{e}^{x}}\sin xdx}$

Using a ilate rule in a given integral.

$I ={{e}^{x}}\int\limits_{{}}^{{}}{\sin x}dx-\int\limits_{{}}^{{}}{\left( \dfrac{d\left( {{e}^{x}} \right)}{dx}\int\limits_{{}}^{{}}{\left( \sin x \right)} \right)} dx$

$I =-{{e}^{x}}\cos x+\int\limits_{{}}^{{}}{{{e}^{x}}\cos xdx}$

$I =-{{e}^{x}}\cos x+{{e}^{x}}\sin x-\int\limits_{{}}^{{}}{{{e}^{x}}\sin xdx}$

$I =-{{e}^{x}}\cos x+{{e}^{x}}\sin x-I$

$2I={{e}^{x}}(\sin x-\cos x)+C$

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

Hence, this is the required answer.

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

Consider the given integral.

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

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

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

Therefore,

$I=\int{t\sin t}dt$

$I=t\left( -\cos t \right)-\int{1\left( -\cos t \right)dt}$

$I=-t\cos t+\int{\cos tdt}$

$I=-t\cos t+\sin t+C$

On putting the value of $t$ , we get

$I=-{{\sin }^{-1}}x\cos \left( {{\sin }^{-1}}x \right)+\sin \left( {{\sin }^{-1}}x \right)+C$

$I=-{{\sin }^{-1}}x\cos \left( {{\sin }^{-1}}x \right)+x+C$

$I=x-{{\sin }^{-1}}x\cos \left( {{\sin }^{-1}}x \right)+C$

Hence, this is the answer.

Integrate with respect to $x$ :
$2x^{2}e^{x^{2}}$

Consider the given integral.

$I=\int{2{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$

$I=2\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$

Let $t={{x}^{3}}$

$\dfrac{dt}{dx}=3{{x}^{2}}$

$\dfrac{dt}{3}={{x}^{2}}dx$

Therefore,

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

$I=\dfrac{2}{3}\left( {{e}^{t}} \right)+C$

Thus,

$I=\dfrac{2}{3}\left( {{e}^{{{x}^{3}}}} \right)+C$

Hence, this is the answer.

$\displaystyle \int a^{x}e^{x}dx$

Consider the following Integral.

$I=\int\limits_{{}}^{{}}{{{a}^{x}}}{{e}^{x}}dx$

Using integration by part method.

$={{a}^{x}}\int\limits_{{}}^{{}}{{{e}^{x}}}dx-\int\limits_{{}}^{{}}{\dfrac{d}{dx}\left( {{a}^{x}} \right)}\int\limits_{{}}^{{}}{{{e}^{x}}}dx$

$I={{a}^{x}}{{e}^{x}}-\ln a\int\limits_{{}}^{{}}{{{e}^{x}}{{a}^{x}}}dx$

$I={{a}^{x}}{{e}^{x}}-I\ln a$

$I+I\ln a={{a}^{x}}{{e}^{x}}$

$I\left( 1+\ln a \right)={{a}^{x}}{{e}^{x}}$

$I=\dfrac{1}{2}\dfrac{{{a}^{x}}{{e}^{x}}}{\left( 1+\ln a \right)}+\dfrac{C}{2}$

Hence, this is the correct answer .

Find :
$\int {\log xdx}$

$\int \log x dx$

Now,

we can also write

$\int \log x\times 1 dx$

it is in from of partial fraction

$\int u\times v=u\int v dx-\int[\dfrac{du}{dx}\times \int v dx]dx$

$=>\log x\int 1dx-\int[\dfrac{d\log x}{dx}\times \int 1 dx]dx$

$=>\log x\times x-\int [\dfrac{1}{x}\times x]dx$

$=>x\log x-\int1dx$

$=>x\log x-x+c$

Solve:-
$\int {\dfrac{{{e^x}}}{x}} \left( {1 + x.\ln x} \right)dx$

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

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

We know that

$\displaystyle\int e^{x}(f(x)+f(x))\ dx=e^{x}f(x)+c$

$=\dfrac{e^{x}}{x}+c$ because

$\dfrac{d(\ln x)}{dx}=\dfrac{1}{x}$

Simplify:
$\int {x{{\tan }^{ - 1}}xdx}$

Consider the given integral.

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

We know that

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

Therefore,

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

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

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

$I=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{1}{2}\left[ \int{1}dx-\int{\dfrac{1}{1+{{x}^{2}}}}dx \right]$

$I=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{1}{2}\left[ x-{{\tan }^{-1}}x \right]+C$

Hence, this is the answer.

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

Consider the following question.

$I=\int_{{}}^{{}}{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx$

$=\int_{{}}^{{}}{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx$

$=\int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\cos }^{2}}x}}dx$

$=\int_{{}}^{{}}{{{\sec }^{2}}xdx}$

$=\tan x+C$

Hence, this is the required answer.

Solve : $\int (1 - x) \sqrt{x} \, dx$

$\int (1-x)\sqrt{x}dx$

$\int \sqrt{x}-x^{\frac{3}{2}}dx$

$=\dfrac{2}{3}x^{\frac{3}{2}}-\dfrac{2}{5}x^{\frac{5}{2}}+C$

Integrate:
$\int 2x^3 \,e^{x^2}.dx$

$\int { 2{ x }^{ 3 }{ e }^{ { x }^{ 2 } }dx } =\int { 2x.{ x }^{ 2 }{ e }^{ { x }^{ 2 } } } \\ Let,\quad { x }^{ 2 }=t\\ \Rightarrow 2xdx=dt\\ \therefore \int { 2{ x }^{ 3 }{ e }^{ { x }^{ 2 } }dx } =\int { t.{ e }^{ t }dt } \\ =t\int { { e }^{ t }dt } +\int { \left[ \dfrac { dt }{ dt } .\int { { e }^{ t }dt } \right] dt } \\ =t.{ e }^{ t }+{ e }^{ t }+C\\ ={ x }^{ 2 }{ e }^{ { x }^{ 2 } }+{ e }^{ { x }^{ 2 } }+C$

Integrate:
$\int e^x\sin x.dx$

$I=\int { { e }^{ x }\sin xdx } \\ ={ e }^{ x }\int { \sin xdx } +\int { \left[ \dfrac { d }{ dx } { e }^{ x }.\int { \sin xdx } \right] dx } \\ =-{ e }^{ x }\cos x-\int { { e }^{ x }\cos xdx } \\ =-{ e }^{ x }\cos x-\left[ { e }^{ x }\int { \cos xdx } +\int { \left[ \dfrac { d }{ dx } { e }^{ x }.\int { \cos xdx } \right] dx } \right] \\ =-{ e }^{ x }\cos x-{ e }^{ x }\sin x-\int { { e }^{ x }\sin xdx } \\ I=-{ e }^{ x }\cos x-{ e }^{ x }\sin x-I\\ 2I=-{ e }^{ x }(\cos x+\sin x)\\ I=-\dfrac { { e }^{ x } }{ 2 } (\cos x+\sin x)$

Integrate : $\int x\sin^2x$

$\int { x{ \sin }^{ 2 }xdx } \\ =\int { x\left[ \dfrac { 1 }{ 2 } (1-\cos2x) \right] dx } \quad (\because \cos2x=1-2{ \sin }^{ 2 }x)\\ =\dfrac { 1 }{ 2 } \int { xdx } -\dfrac { 1 }{ 2 } \int { x\cos2xdx } \\ =\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { 1 }{ 2 } \left[ x\int { \cos2xdx } -\int { \left( \dfrac { dx }{ dx } .\int { \cos2xdx } \right) } \right] dx\\ =\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { x\sin2x }{ 4 } -\dfrac { \cos2x }{ 8 } +C\\ =\dfrac { { 2x }^{ 2 }-2x\sin2x-\cos2x }{ 8 } +C$

Evaluate:
$\displaystyle \int x+5 dx$

Given

$\displaystyle \int x+5 dx$

$=\displaystyle \int xdx +\int 5 dx$

$=\dfrac {x^2}2+5x+c$

Evaluate
$\int { { e }^{ x }(\tan { x } +1)secx\quad dx}$

$\begin{array}{c}I = \int {{e^x}\left( {\tan x + 1} \right)\sec xdx} \\ = \int {{e^x}\tan x\sec x} dx + \int {{e^x}\sec x} dx\end{array}$

According to the integration by parts,

$\begin{array}{l}\int {uvdx} = u\int {vdx} - \int {\left( {\frac{{du}}{{dx}} \times \int {vdx} } \right)} dx\\Here,\\u = \sec x\\v = {e^x}\\So,\\\int {\sec x{e^x}dx} = \sec x\int {{e^x}dx} - \int {\left( {\frac{{d\sec x}}{{dx}} \times \int {{e^x}dx} } \right)} dx\\ = {e^x}\sec x - \int {\left( {\sec x\tan x{e^x}} \right)} dx\\So,\\I = \int {\left( {\sec x\tan x{e^x}} \right)} dx + {e^x}\sec x - \int {\left( {\sec x\tan x{e^x}} \right)} dx\\ = {e^x}\sec x\end{array}$

Find $\int {{e^x}\left( {\frac{{x - 1}}{{{x^2}}}} \right)} dx$

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

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

$=\displaystyle\int \dfrac{d}{dx}\left\{\dfrac{e^x}{x}\right\}dx$

$=\dfrac{e^x}{x}+c$ .

Solve:
$\int {\sin ^4}x.{\cos ^2}xdx$

$\displaystyle\int \sin^4x\cdot \cos^2xdx =\displaystyle\int (\sin^2x)(\sin^2x\cdot\cos^2x)dx$

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

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

$\sin^2x=\dfrac{1-\cos 2x}{2}$

$\sin 2x= 2\sin x\cos x$

The integral becomes

$=\displaystyle\int \dfrac{1}{2}(1-\cos 2x)\cdot\left(\dfrac{\sin 2x}{2}\right)^2dx$

$=\dfrac{1}{8}\left(\displaystyle\int (1-\cos 2x)\cdot \sin (2x)dx)\right)$

$=\dfrac{1}{8}\left(\displaystyle\int (\sin^22x-\cos 2x\cdot \sin^2(2x))dx\right)$

$=\dfrac{1}{8}\displaystyle\int \dfrac{1}{2}\cdot 2\sin^2(2x)dx-\displaystyle\int\dfrac{1}{2}\cdot 2\cos 2x\sin^22xdx$

$=\dfrac{1}{8}\left(\displaystyle\int \dfrac{1}{2}\cdot (1-\cos (4x))dx-\displaystyle\int\dfrac{1}{2}\sin^22xd\sin 2x\right)$

$=\dfrac{1}{16}\left(\displaystyle\int (1-\cos (4x)dx-\displaystyle\int \sin^2(2x)d\sin 2x\right)$

$\dfrac{d\sin 2x}{dx}=2\cos 2x$

$d\\sin (2x)=2\cos 2x\cdot dx$

$=\dfrac{1}{16}\left(x-\dfrac{\sin 4x}{4}-\dfrac{\sin^3(2x)}{3}\right)+C$ .

Integrate $\displaystyle \int e^{\log (\sec x+\tan x)}\sqrt{1+\tan^2 x}\ dx$ .

We have,

$I=\int e^{\log (\sec x+\tan x)}\sqrt{1+\tan^2 x}\ dx$

$I=\int e^{\log (\sec x+\tan x)}\sqrt{\sec^2 x}\ dx$

$I=\int e^{\log (\sec x+\tan x)}\sec x\ dx$

Let

$t=\log(\sec x+\tan x)$

$\dfrac{dt}{dx}=\dfrac{1}{\sec x+\tan x}\times (\sec x\tan x+\sec^2 x)$

$\dfrac{dt}{dx}=\dfrac{\sec x}{\sec x+\tan x}\times (\tan x+\sec x)$

$dt=\sec x\ dx$

Therefore,

$I=\int e^t\ dt$

$I=e^t+C$

On putting the value of

$t$ , we get

$I=e^{\log(\sec x+\tan x)}+C$

Hence, this is the answer.

The value of $\displaystyle\int{{e}^{\ln{\sqrt{x}}}dx}$ is

$\displaystyle\int{{e}^{\ln{\sqrt{x}}}dx}$

$=\displaystyle\int{\sqrt{x}dx}$

$=\displaystyle\int{{x}^{\frac{1}{2}}dx}$

$=\dfrac{{x}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+c$ ....where

$c$ is the constant of integration

$=\dfrac{{x}^{\frac{1+2}{2}}}{\dfrac{1+2}{2}}+c$

$=\dfrac{2}{3}{x}^{\frac{3}{2}}+c$

$=\dfrac{2}{3}x\sqrt{x}+c$

Find the integrals of the functions.
i) $sin^2 (2x + 5)$
ii) $sin \, 3x \, cos \, 4x$
iii) $cos \, 2x \, cos \, 4x \, cos \, 6x$
iv) $sin^3 (2x + 1)$

(i)

$\displaystyle\int \sin^2(2x+5)dx=I$

$I=\displaystyle\int \left[\dfrac{1-\cos 2(2x+5)}{2}\right]dx=\displaystyle\int\dfrac{1}{2}dx-\displaystyle\int \dfrac{\cos (4x+10)}{2}dx$

$I=\dfrac{x}{2}-\dfrac{\sin (4x+10)}{8}+C$ .

(ii)

$I=\displaystyle\int \sin 3x\cos 4x dx$

$\sin C\cos D=\dfrac{1}{2}\left[\sin \dfrac{(C+D)}{2}+\sin\left(\dfrac{C-D}{2}\right)\right]$

$=\dfrac{1}{2}\displaystyle\int \left[\sin (3x+4)+\sin (3x-4x)\right]dx$

$=\dfrac{1}{2}\displaystyle\int \left[\displaystyle\int \sin (7x)dx+\displaystyle\int \sin (-x)dx\right]$

$=\dfrac{-1}{2}\dfrac{\cos 7x}{7}+\dfrac{1}{2}\cos x+C$

$I=\dfrac{1}{2}\left[\dfrac{7\cos x-\cos 7x}{7}\right]+C$

(iii)

$I=\displaystyle\int \cos 2x\cos 4x\cos 6x dx$

$I=\displaystyle\int \dfrac{1}{2}\left[\cos (6x+2x)+\cos (6x-2x)\right]\cos 4xdx$

$=\displaystyle\int \dfrac{1}{2}\left[\cos 8x\right]\cos 4x dx+\dfrac{1}{2}\displaystyle\int \cos^24cdx$

$=\displaystyle\int \left[\dfrac{1}{4}\cos (12x)+\dfrac{1}{4}\cos (4x)\right]dx+\dfrac{1}{4}\displaystyle\int (4\cos 8x)dx$

$=\displaystyle \dfrac{1}{4.12}\sin (12x)+\dfrac{1}{4.4}\sin (4x)+\dfrac{x}{4}+\dfrac{1}{4.8}\sin (8x)+C$

$I\Rightarrow \dfrac{x}{4}+\dfrac{1}{16}\sin 4x+\dfrac{\sin 8x}{32}+\dfrac{\sin 12x}{48}+C$

(iv)

$I=\displaystyle\int \sin^3(2x+1)=\displaystyle\int\dfrac{1}{4}\left[3\sin(2x+1)-\sin (6x+3)\right]dx$

$(\sin 3x=3\sin x-4\sin^3x)$

$I=\dfrac{-3}{4.2}\cos (2x+1)+\dfrac{1}{6.4}\cos (6x+3)+C$

$I=\dfrac{1}{24}\cos (6x+3)-\dfrac{3}{8}\cos (2x+1)+C$ .

Find the integrals of the functions.
i) $sin^3 \, x \, cos^3 \, x$
ii) $sin \, x \, sin \, 2x \, sin \, 3x$
iii) $sin \, 4x \, sin \, 8x$
iv) $\dfrac{1 - cos \, x}{1 + cos \, x}$
v) $\dfrac{cos \, x}{1 + cos \, x}$

(i)

$I=\displaystyle\int \sin^3x\cos^3xdx=\dfrac{1}{8}\displaystyle\int(\sin 2x)^3dx\Rightarrow \dfrac{1}{32}\displaystyle\int (3\sin 3x-4\sin(6x))dx$

$(\sin 3x=3\sin x-4\sin^3x)$

$I=\dfrac{1}{32}\displaystyle\int 3\sin 2xdx-\dfrac{1}{32}4\sin (6x)dx$

$=\dfrac{3}{32.2}\times -(\cos 2x)+\dfrac{4}{6.32}\cos (6x)+C$

$I=\dfrac{\cos (6x)}{48}-\dfrac{3\cos (2x)}{64}+C$ .

(ii)

$\displaystyle\int \sin x\sin 2x\sin 3xdx$

$\Rightarrow \displaystyle\int\dfrac{1}{2}\left[\cos(3x-x)-\cos (3x+x)\right]\sin 2xdx$

$\Rightarrow \displaystyle\int \dfrac{1}{2}\sin 2x\cos 2xdx-\displaystyle\dfrac{1}{2}\sin 2x\cos 4xdx$

$\Rightarrow \dfrac{1}{4}\displaystyle\int \sin 4xdx-\dfrac{1}{4}\displaystyle\int (\sin (2x+4x)+\sin (2x-4x)dx)$

$\Rightarrow -\dfrac{1}{16}\cos 4x-\dfrac{1}{4}\displaystyle\int \sin 6xdx+\dfrac{1}{4}\displaystyle\int \sin 2xdx$

$I\Rightarrow \dfrac{1}{24}\cos 6x-\dfrac{1}{16}\cos 4x-\dfrac{1}{8}\cos 2x+C$

(iii)

$\displaystyle\int \sin 4x\sin 8xdx$

$\Rightarrow \displaystyle\int \dfrac{1}{2}\left[\cos (8x-4x)-\cos (8x+4x)\right]dx$

$\Rightarrow \displaystyle\int \dfrac{1}{2}(\cos 4x-\cos 12x)dx$

$\Rightarrow \dfrac{1}{8}\sin 4x-\dfrac{1}{24}\sin 12x+C$ .

(iv)

$\displaystyle\int \dfrac{(1-\cos x)}{(1+\cos x)}dx$

$\Rightarrow \displaystyle\int \dfrac{2\sin^2x/_{2}}{2\cos^2 x/_{2}}dx$

$\Rightarrow \displaystyle\int \dfrac{1-\cos^2x/_{2}}{\cos^2x/_{2}}dx$

(iv)

$\displaystyle\int \dfrac{(1-\cos x)}{(1+\cos x)}dx$

$\Rightarrow \displaystyle\int \dfrac{2\sin^2x/_{2}][2\cos^2x/_{2}}dx$

$\Rightarrow \displaystyle\int \dfrac{1-\cos^2x/_{2}}{\cos^2x/_{2}}dx$

$\Rightarrow \displaystyle\int \sin^2x/_{2}dx=\displaystyle\int dx$

$\Rightarrow 2\tan x/_{2}-x+C$

(vi)

$\displaystyle\int \dfrac{\cos x}{1+\cos x}dx$

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

$\Rightarrow \displaystyle\int dx-\displaystyle\int \dfrac{1}{1+\cos x}dx$

$\Rightarrow x-\dfrac{1}{2}\displaystyle\int \sec^2x/_{2}dx$

$\Rightarrow x-\dfrac{1}{2}\times 2\tan x/_{2}+C$

$\Rightarrow x-\tan x/_{2}+C$ .

Evaluate:
$\displaystyle\int x \cos^3x dx$ .

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

Integrating by parts, we get

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

$\displaystyle \int cos^3x dx = \int (\cos x - \sin^2x \cos x) dx$

$\sin x - \cfrac{\sin^3 x}{3}$

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

$\displaystyle x \sin x - \cfrac{x \sin^3x}{3} - \cfrac{\cos^3x}{9}+\cfrac{8\cos x}{9}$

Find :
$\int {\frac{e^x(x-1)}{(x+1)^3}dx}$

Find

$\displaystyle\int { { e }^{ x } } \dfrac { \left( x-1 \right) }{ { \left( x-1 \right) }^{ 3 } } dx$

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

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

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

$\Rightarrow \dfrac { { e }^{ x }{ \left( x+1 \right) }^{ 2 }-{ e }^{ x }2\left( x+1 \right) dx }{ { \left( x+1 \right) }^{ 4 } } =dz$

$\left\{ \dfrac { { e }^{ x } }{ { \left( x+1 \right) }^{ 2 } } -\dfrac { 2{ e }^{ x } }{ { \left( x+1 \right) }^{ 3 } } \right\} dx=dz$

$\displaystyle\int { dz=z+c }$

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

Ans

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

$\displaystyle\int e^{\tan x}\cdot \tan x\cdot \sec^2x \ dx$ .

Putting

$\tan x = m$

$\sec^2x dx = dm$

$\displaystyle \int e^m m dm$

Integrating by parts, we get

$\displaystyle me^m - e^m+C$

$\tan x e^{\tan x} - e^{\tan x} + C$

$\displaystyle\int \sqrt{1-x^2}dx$ .

$\begin{array}{l} \int { \sqrt { 1-{ x^{ 2 } } } } dx \\ \int { \sqrt { { 1^{ 2 } }-{ x^{ 2 } } } } dx \\ \dfrac { 1 }{ 2 } x\sqrt { 1-{ x^{ 2 } } } +\dfrac { 1 }{ 2 } { 1^{ 1 } }{ \sin ^{ -1 } }\left( { \frac { x }{ 1 } } \right) +C \\ \dfrac { { x\sqrt { 1-{ x^{ 2 } } } } }{ 2 } +\dfrac { 1 }{ 2 } { \sin ^{ -1 } }\left( { \dfrac { x }{ 1 } } \right) +C \end{array}$

Solve:
$\int x^2e^x\sin xdx$

$I=\displaystyle\int x^{2}e^{x}\sin xdx$

Integrating by parts

$u=x^{2}. V=e^{x}\sin x$

$u^{1}=2x, \displaystyle\int vdx=\dfrac{1}{2}e^{x} (\sin x-\cos x)$ [Integrated by parts]

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

$=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)-\int xe^{x}(\sin x-\cos x)dx$

Now,

$\displaystyle\int e^{x}(\cos x-\sin x)dx=e^{x}\cos x [\because \int e^{x}f(x)+f'(x)dx=e^{x}f(x)]$

$I=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)+\left[x e^{x}\cos x-\int (1) e^{x}\cos xdx\right]$ [Using by parts]

$=\dfrac{x^{2}e^{x}}{2}(\sin x-\cos x)+xe^{x}\cos x-\dfrac{e^{x}}{2} (\sin x+\cos x)+C$ [Using by parts]

$=\dfrac{e^{x}}{2}[\sin (x)(x^{2}-1)+\cos x(-x^{2}+2x-1)]+C$

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

Evaluate: $\displaystyle \int { { x }^{ 2 }{ sin }^{ -1 } } x\ dx$

1356554_1128520_ans_e3d1918916cd4a64bc8e0cf4f96f0784.jpg

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

$\begin{array}{l} \int { { e^{ x } } } \left( { \log x+\frac { 1 }{ { { x^{ 2 } } } } } \right) dx \\ \int { { e^{ x } }\left( { \log x+\frac { 1 }{ x } -\frac { 1 }{ x } +\frac { 1 }{ { { x^{ 2 } } } } } \right) } dx \\ \int{{e^{x}(f(x)-f'(x))\ dx}}= e^x{f(x)}\\ ={ e^{ x } }\left( { \log x-\frac { 1 }{ x } } \right) +C \end{array}$

Hence,

Solved.

Solve $\int { x\sqrt { x+2 } } dx$

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

$\displaystyle \int \left( m-2 \right) \sqrt { 3 }\ dm \quad \left(let x=2 =m, dx=dm \right)$

$\displaystyle \int m \sqrt{m}\ dm -2 \int\sqrt {m}\ dm$

$\displaystyle \int m^{3/2}\ dm \ -2\int m^{1/2}\ dm$

$\dfrac {3}{2} \sqrt {m}\ - 2\dfrac {1}{2\sqrt{m}}+C$

$\dfrac {3}{2} \sqrt {m}\ - \dfrac {1}{\sqrt{m}}+C$

Substituting back

$m=x+2$

$\dfrac {3}{2} \sqrt {\left(x+2\right)}\ - \dfrac {1}{\sqrt{\left(x+2\right)}}+C$

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

1354281_1129617_ans_412cbdd4cd3e43e2884f34497cf5d41d.jpg

Solve:
$I = \int {\dfrac{{dx}}{{18 - 4x - {x^2}}}}$

$\int \dfrac { d x } { 18 - 4 x - x ^ { 2 } }$

$= - 1 \int \dfrac { d x } { x ^ { 2 } + 4 x - 18 }$

$= - 1 \int \dfrac { d x } { ( x + 2 ) ^ { 2 } - ( \sqrt { 22 } ) ^ { 2 } }$

$= - 1 \dfrac { 1 } { 2 \cdot \sqrt { 22 } } \ln \left( \dfrac { x + 2 - \sqrt { 2 } 2 } { x + 2 + \sqrt { 2 } } \right) + C$

$= \dfrac { - 1 } { 2 \sqrt { 22 } } \ln ( \dfrac { x + 2 - \sqrt { 2 } 2 } { x + 2 + \sqrt { 2 } 2 } ) + C$

Evaluate: $\displaystyle\int{\dfrac{1+\cos x} { x+\sin x}dx }$ .

Now,

$\displaystyle\int{\dfrac{1+\cos x} { x+\sin x}dx }$

$\displaystyle\int{\dfrac{d(x+\sin x)} { x+\sin x} }$

$=\log|(x+\sin x)|+c$ [ Where

$c$ is integrating constant]

$I= \int \dfrac{1}{x(1+\log x)}. dx$

$\displaystyle I= \int \dfrac{1}{x(1+\log x)}. dx$

Let

$(1+log x) = m$

$\dfrac{1}{x}dx = dm$

$\displaystyle \int \frac{dm}{m}$

$\Rightarrow ln|m|+c$

$\Rightarrow ln|1+;n(x)|+c$

1124769_1140100_ans_04bfb5bc0eaf4b4ca59efc2ab34a00fd.jpg

Solve :
$I=\int \sin^{6}{x}dx$

$\displaystyle I = \int sin^{6}x\,dx$

$\displaystyle = \int (sin^{2}x)^{2}(sin^{2}x)dx$

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

$\displaystyle = \frac{1}{8}\int (1-2cos2x+cos^{2}x)(1-cos2x)dx$

$\displaystyle = \frac{1}{8}\int (1-2cos2x+\frac{1+cos4x}{2})(1-cos2x)dx$

$\displaystyle = \frac{1}{16}\int (2-4cos2x+cos4x)(1-cos2x)dx$

$\displaystyle = \frac{1}{16}\int (3-4cos2x+cos4x)(1-cos2x)dx$

$\displaystyle = \frac{1}{16}\int (3-4cos2x-3cos2x+cos4x+4cos^{2}2x-cos4xcos2x)dx$

$\displaystyle = \frac{1}{16}\int [3-7cos2x+cos4x+2\times 2cos^{2}2x-\frac{1}{2}.2cos^{4}xcos2x]dx$

$\displaystyle = \frac{1}{16}[3-7cos2x+cos^{4}x+2(1+cos^{4}x)-\frac{1}{2}(cos6x+cos2x)]dx$

$\displaystyle = \frac{1}{16}(3-7cos2x+cos^{4}x+2+2cos^{4}x-\frac{1}{2}(cos6x+cos2x))dx$

$\displaystyle = \frac{1}{32}\int [6-14cos2x+2cos4x+4+4cos4x-cos6x-cos2x]dx$

$\displaystyle = \frac{1}{32}[10-15cos2x+6cos4x-cos6x]dx$

$\displaystyle \frac{1}{32}[10-15cos2x+6cos4x-cos6x]dx$

$\displaystyle \frac{1}{32}[10x-\frac{15sin2x}{2}+\frac{6sin4x}{4}-\frac{sin6x}{6}]dx$

$\displaystyle \frac{1}{32}[10x-\frac{15sin2x}{2}+\frac{3}{2}sin4x]$

$\displaystyle \frac{1}{92}[60x-45sin2x+9sin4x-sin6x]+c$

1120203_1140220_ans_b5f33a1bf15848b79fe40688c9404ef3.jpg

Solve :
$\int x^{ \sin x} \left( \dfrac {sinx}{x} + \cos x . logx \right) dx$

$\begin{array}{l} \int { { x^{ \sin x } }\left( { \frac { { \sin x } }{ x } +\cos x\log x } \right) } dx \\ y={ x^{ \sin x } } \\ \log y=\sin x\log x \\ \frac { 1 }{ y } \frac { { dy } }{ { dx } } =\cos x\log x+\frac { { \sin x } }{ x } \\ \frac { { dy } }{ { dx } } ={ x^{ \sin x } }\left[ { \cos x{ { logx } }+\frac { { \sin x } }{ x } } \right] \\ \int { dy } =y+C \\ ={ x^{ \sin x } }+C \end{array}$

Hence,

Solved

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

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

$\displaystyle \Rightarrow \int (1-cos^2 x) sinx cos^3x dx$

$\displaystyle \int sin^3x cosx (1-sin^2 x)dx$

Let

$sin x = m$

Differentiating w.r.t.

$x$ , we get,

$cosx \,dx = dm$

$\displaystyle \Rightarrow \int m^3 (1-m^2)dm$

$\displaystyle \Rightarrow \int m^{3} dm - \int m^5 dm$

$\displaystyle \Rightarrow \frac{m^4}{4}-\frac{m^6}{6}+c$

Back substituting the value of

$m$

$\displaystyle \Rightarrow \frac{sin^4x}{4}-\frac{sin^6x}{6}+c$

$\int \sin^{3}x\cos^{2}x dx$ is equal to

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

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

$\mu =wsx$

$dv=-\sin x \ dx$

$\Rightarrow \ \displaystyle \int -\mu^2 (1-\mu^2)dv$

$\Rightarrow \ \displaystyle \int -\mu ^2 d\mu +\displaystyle \int \mu^4 du$

$=-\dfrac {\mu^3}{3}+\dfrac {\mu^5}{5}+C$

$=\dfrac {\sin ^5}{5}+\dfrac {\cos ^5 x}{5}-\dfrac {\cos ^3 x}{3}+1$

$\Rightarrow \ \dfrac {\cos^5 x}{5}-\dfrac {\cos^3 x}{3}+C$

Integrate the function:
$\dfrac{\sin(\tan^{-1}x)}{1+x^2}$

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

Let

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

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

$\therefore I= \displaystyle\int \sin(t) dt$

$=-\cos t$

$I=-\cos(\tan^{-1}x)$

Evaluate:
$\int x \cdot \sin^{2}x dx$

$\displaystyle \int { x\sin ^{ 2 }{ x } dx }$

Using LATE Rule,

$x\rightarrow 1^{st}$ part

$\sin^2x\rightarrow II^{nd}$ part

$\Rightarrow \displaystyle\int { x\sin ^{ 2 }{ x } dx } -\displaystyle\int { \left( \frac { d }{ dx } \left( x \right) .\displaystyle\int { \sin ^{ 2 }{ x } dx } \right) dx } \\ \quad \quad \quad \quad \quad \left( { I }_{ 1 } \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( { I }_{ 2 } \right)$

$x\Rightarrow \displaystyle \int{\dfrac{1}{2}(1-\cos 2x)dx}$

$\Rightarrow \dfrac{x}{2}\displaystyle \int{(1-\cos 2x)dx}\Rightarrow \dfrac{x}{2}\left(x-\sin \dfrac{2x}{2}\right)$

$I_2=\displaystyle \int{1.\dfrac{x}{4}(2x-\sin 2x)dx}$

$=\dfrac{1}{4}\left[\displaystyle \int{2x^2dx}\displaystyle \int{x\sin 2s}dx\right]$

$=\dfrac{1}{4}\left[\dfrac{2x^3}{3}-x\left(\dfrac{-\cos 2x}{2}\right)-\left(\dfrac{-\sin 2x}{4}\right)\right]$

$=\dfrac{1}{4}\left(\dfrac{2x^3}{3}+\dfrac{x\cos 2x}{2}-\dfrac{\sin 2x}{4}\right)$

$I\Rightarrow I_1-I_2\Rightarrow \dfrac{x}{4}(2x-\sin 2x)-\dfrac{1}{4}\left(\dfrac{2x^3}{3}+x\dfrac{\cos 2x}{2}-\dfrac{\sin 2x}{4}\right)$

Solve: $\displaystyle \int { \dfrac {dx }{ e^x+e^{-x} } }$

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

$\displaystyle \Rightarrow \int \frac{e^{x}}{e^{2x}+1}dx\Rightarrow \int \frac{d(e^{x})}{(e^{x})^{2}+1}$

$\displaystyle \Rightarrow \int \frac{dz}{z^{2}+1}$ [let z =

$e^{x}$ ]

$\displaystyle \Rightarrow tan^{-1}(z)+c$ [c is arbitrary constant]

$\displaystyle \Rightarrow tan^{-1}(e^{x})+c$

$\int \frac{ln(\frac{x-1}{x+1})}{x^{2}-1}dx$ is equal to

$\begin{array}{l} Given, \\ I=\int { \frac { { ln\left( { \frac { { x-1 } }{ { x+1 } } } \right) } }{ { { x^{ 2 } }-1 } } } dx \\ Integration\, of\, \, \, ln\, \, \left( { \frac { { x-1 } }{ { x+1 } } } \right) =t, \\ differentiation\, \, both\, \, side\, \, is,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left( { \therefore \, \, from\, question\, tool=\frac { u }{ v } =\frac { { { u^{ ' } }v-u{ v^{ ' } } } }{ { { v^{ 2 } } } } } \right) \\ \frac { 1 }{ { \frac { { x-1 } }{ { x+1 } } } } \left( { \frac { { 1(x+1)-(x-1) } }{ { { { (x+1) }^{ 2 } } } } } \right) \\ \left( { \frac { { x+1 } }{ { x-1 } } } \right) \left( { \frac { { 1(x+1)-(x-1) } }{ { { { (x+1) }^{ 2 } } } } } \right) dx=dt \\ \frac { 1 }{ { x-1 } } \left( { \frac { 2 }{ { (x+1) } } } \right) dx=dt \\ \frac { { dx } }{ { { x^{ 2 } }-1 } } =\frac { 1 }{ 2 } dt \\ Now, \\ I=\frac { 1 }{ 2 } \int { t\, \cdot dt } \\ \, \, \, =\frac { 1 }{ 2 } \left( { \frac { { { t^{ 2 } } } }{ 2 } } \right) +c \\ \therefore \, \, \, \frac { 1 }{ 4 } { \left( { ln\frac { { (x-1) } }{ { x+1 } } } \right) ^{ 2 } }+c \\ so\, that\, \, the\, \, corect\, \, Answeer\, is\, \frac { 1 }{ 4 } { \left( { ln\frac { { (x-1) } }{ { x+1 } } } \right) ^{ 2 } }+c \end{array}$

Find the integral part of the greatest root of equation $x^{3}-10x^{2}-11x-100=0$ .

Given equation is

$x^{3}-10 x^{2}-11 x-100=0$

Let

$f(x)=x^{3}-10 x^{2}-11 x-100$

$\Rightarrow f^{\prime}(x)=3 x^{2}-20 x-11$

For

$3 x^{2}-20 x-11=0,$

we have

$x=\dfrac{20\pm\sqrt{400+132}}{6}=\dfrac{10\pm\sqrt{133}}{3}$

Hence, graph of

$y=f(x)$ is plotted

Now,

$(10+\sqrt{133}) / 3 \cong 7.16$

$f(8)=8^{3}-10(8)^{2}-11(8)-100<0$

$f(9)=9^{3}-10(9)^{2}-11(9)-100<0$

$f(10)=10^{3}-10(10)^{2}-11(10)-100<0$

$f(11)=11^{3}-10(11)^{2}-11(11)-100<0$

$f(12)=12^{3}-10(12)^{2}-11(12)-100>0$

$\Rightarrow \quad y \in(11,12)$

$\Rightarrow \quad[y] \quad=11$

1582657_1142567_ans_487982f8b44644d38b4b06c252ac3982.png

Solve :
$\int 2^x . e^x dx$

$\begin{array}{l} I=\int { { 2^{ x } }.{ e^{ x } }dx } \\ ={ 2^{ x } }.{ e^{ x } }-\int { \left( { { 2^{ x } }.\log { e^{ 2 } } .{ e^{ x } }dx } \right) } \\ ={ 2^{ x } }-{ e^{ x } }-\log _{ e } ^{ 2 }\int { { 2^{ x } }.{ e^{ x } }dx } \\ ={ 2^{ x } }.{ e^{ x } }-\log _{ e } ^{ 2 }.I+C \\ \left( { 1+\log _{ e } ^{ 2 } } \right) .I={ 2^{ x } }.{ e^{ x } }+C \\ I=\dfrac { { { 2^{ x } }.{ e^{ x } } } }{ { \left( { 1+\log _{ e } ^{ 2 } } \right) } } +C \end{array}$

Find the following integrals:
$\int \left (\sqrt {x} - \dfrac {1}{\sqrt {x}}\right )^{2}dx$ .

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

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

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

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

$=\dfrac{x^2}{2}+ln |x|-2x+c$ .

1177543_1156568_ans_8f0b730f274144888c989b626c75a972.jpg

Solve : $\int e^x. \sin \,3xdx$

$\int e^x\sin 3xdx$

Using Integration by Parts

Considering

$u=e^x$ and

$v=\sin 3x$

$=e^x\left ( -\dfrac{1}{3}\cos 3x \right )-\int e^x\left ( -\dfrac{1}{3}\cos 3x \right )dx$

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

$=e^x\left ( -\dfrac{1}{3}\cos 3x \right )+\dfrac{1}{3}\left ( e^x \dfrac{1}{3}\sin 3x-\dfrac{1}{3}\int e^x \sin 3xdx\right )$

$=-\dfrac{1}{10}e^x(3\cos 3x-\sin 3x)+C$

Solve:-
$\int \dfrac{{\cos \,x}}{{1 + \cos \,x}}d x$

$\displaystyle\int \dfrac{\cos x}{1+\cos x}d x=\int \dfrac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}d x=\int \dfrac{\cos x-\cos^2 x}{1-\cos ^2 x}d x=\int \dfrac{\cos x-\cos^2 x}{\sin^2 x}d x$

$=\displaystyle\int \text{cosec}x\cot x d x-\int \cot^2 x d x=-\text{cosec} x-\int (\text{cosec}^2 x-1)d x\\\\-\text{cosec} x-\int \text{cosec}^2 x d x+\int d x=-\text{cosec} x+\cot x+x+C$

Solve $\dfrac {1}{\sqrt {3}}\int \dfrac {dx}{\sin \left (x + \dfrac {\pi}{4}\right ) + 1}$ .

Let

$I=\cfrac { 1 }{ \sqrt { 3 } } \int { \cfrac { dx }{ \sin { \left( x+\cfrac { \pi }{ 4 } \right) } +1 } }$

Let

$\left( x+\cfrac { \pi }{ 4 } \right) =t\Rightarrow dx=dt$

$\therefore I=\cfrac { 1 }{ \sqrt { 3 } } \int { \cfrac { dt }{ 1+\sin { t } } }$

$=\cfrac { 1 }{ \sqrt { 3 } } \int { \cfrac { 1-\sin { t } }{ \cos ^{ 2 }{ t } } dt } =\cfrac { 1 }{ \sqrt { 3 } } \int { \left( \sec ^{ 2 }{ t } -\sec { t } \tan { t } \right) dt }$

$=\cfrac { 1 }{ \sqrt { 3 } } \left[ \tan { t } -\sec { t } \right] +c$

$=\cfrac { 1 }{ \sqrt { 3 } } \left[ \tan { \left( x+\cfrac { x }{ 4 } \right) } -\sec { \left( x+\cfrac { x }{ 4 } \right) } \right] +c$

Solve : $\int{x}{e^x}\sin x dx$

Now let

$u=x$

$du=dx$

Now

$\displaystyle dv=e^x\sin x\rightarrow v=\int e^x\sin x dx$

$\displaystyle \int e^x\sin x dx=\int \sin x d(e^x)$

Now

$\displaystyle v=\int e^x\sin x=e^x\sin x-\int e^x d(\sin x)=e^x\sin x=\int e^x \cos x$

$\displaystyle=e^x\sin x-e^x\cos x-\int e^x\sin xdx$

$v=e^x\sin x-e^x\cos x-v$

$\Rightarrow 2v=e^x\sin x-e^x\cos x-v$

$\Rightarrow v=\dfrac{e^x}{2}(\sin x-\cos x)$

Now

$\displaystyle \int xe^x\sin dx$ Let

$u=x\Rightarrow du=dx$

and

$v=\dfrac{e^x}{2}(\sin x-\cos x)$

$\displaystyle \int e^x\cos xdx=\dfrac{e^x}{2}(\sin x+\cos x)$

$\displaystyle\therefore \int xe^x\sin xdx=\dfrac{xe^x}{2}(\sin x-\cos x)-\int \dfrac{e^x}{2}(\sin x-\cos x)dx$

$\displaystyle =\dfrac{xe^x}{2}(\sin x-\cos x)-\dfrac{-1}{2}\int e^x\sin x+\dfrac{1}{2}\int e^x\cos dx$

$\displaystyle =\dfrac{xe^x}{2}(\sin x-\cos x)-\dfrac{e^x}{4}(\sin x-\cos x)+\dfrac{e^x}{4}(\sin x+\cos x)+c$

$\displaystyle I=\dfrac{xe^x}{2}(\sin x-\cos x)+\dfrac{e^x}{2}\cos x+c$

$\displaystyle\int { \sin { x\sin { \left( \cos { x } \right) } } } dx$

$I=\int { \sin { x } \sin { (\cos { x } ) } dx }$

Putting

$\cos { x } =t$

differentiating both sides,

$-\sin { x } dx=dt$

and substituting the values,

$I=-\int { \sin { t } dt } =-(-\cos { t } )=\cos { t }$

$=\cos { (\cos { x } ) } \quad [\therefore \cos { x } =t]$

Solve: $\displaystyle \int \dfrac{x \cdot \sin^{-1}x}{(1 - x^2)^{\frac{3}{2}}} dx$ .

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

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

Let

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

$\dfrac{dx}{(1-x^2)^{1/2}}=dt$

$=\displaystyle\int \dfrac{\sin t.t dt}{1-\sin^2 t}$

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

$=\displaystyle\int t \tan t \sec tdt$

$=t\displaystyle\int \tan t\sec t dt-\displaystyle\int \left[\left(\dfrac{dt}{dt}\right)\displaystyle\int \left(\sec t\tan t dt\right)\right]dt$ .

$=t\sec t-\displaystyle\int 1\sec t dt$

$=t\sec t-log |\sec t+\tan t|+c$

$=\sin^{-1}x\sec (\sin^{-1}x)-log|\sec(\sin^{-1}x)+\tan(\sin^{-1}x)|+c$

$\therefore \displaystyle\int \dfrac{x\sin^{-1}xdx}{(1-x^2)^{3/2}}=\sin^{-1}x\sec(\sin^{-1}x)-log|\sec(\sin^{-1}x)+\tan(\sin^{-1}x)|+c$ .

1116345_1157554_ans_8ac6f9b46f91494aba56ede84940ae5f.jpg

Solve: $\displaystyle \int\dfrac{1}{\sqrt{1+4x^2}}dx$

$\int { \dfrac { 1 }{ \sqrt { 1+4{ { x }^{ 2 } } } } dx } \\ =\int { \dfrac { 1 }{ \sqrt { 4\left( \dfrac { 1+4{ x }^{ 2 } }{ 4 } \right) } } dx } \\ =\int { \left[ \dfrac { 1 }{ 2 } .\dfrac { 1 }{ \sqrt { \dfrac { 1 }{ 4 } +{ x }^{ 2 } } } \right] dx } \\ =\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ \sqrt { { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }+{ x }^{ 2 } } } dx } \\ =\dfrac { 1 }{ 2 } \left[ \dfrac { 1 }{ \dfrac { 1 }{ 2 } } { \tan }^{ -1 }\left( \dfrac { x }{ \dfrac { 1 }{ 2 } } \right) \right] +C\\ ={ \tan }^{ -1 }2x+C$

solve it.
$I = \int {{x^2}\cos x} dx$

$I=\displaystyle\int{{x}^{2}\cos{x}dx}$

Let

$u={x}^{2}\Rightarrow\,du=2x\,dx$

$dv=\cos{x}dx\Rightarrow\,v=\sin{x}$

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

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

Let

$u=x\Rightarrow\,du=dx$

$dv=\sin{x}dx\Rightarrow\,v=-\cos{x}$

$I={x}^{2}\sin{x}-2\left[-x\cos{x}+\displaystyle\int{\cos{x}dx}\right]$

$I={x}^{2}\sin{x}+2x\cos{x}-2\displaystyle\int{\cos{x}dx}$

$I={x}^{2}\sin{x}+2x\cos{x}-2\sin{x}+c$

Interget that :-
$\int {x\sqrt {x + 2} }$ $dx$

$\displaystyle\int{x\sqrt{x+2}\,dx}$

Let

$t=x+2\Rightarrow\,dt=dx$

$=\displaystyle\int{\left(t-2\right)\sqrt{t}\,dt}$

$=\displaystyle\int{\left({t}^{1+\frac{1}{2}}-2{t}^{\frac{1}{2}}\right)dt}$

$=\displaystyle\int{\left({t}^{\frac{3}{2}}-2{t}^{\frac{1}{2}}\right)dt}$

$=\dfrac{{t}^{\frac{3}{2}+1}}{\dfrac{3}{2}+1}-\dfrac{2{t}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+c$

$=\dfrac{{t}^{\frac{5}{2}}}{\dfrac{5}{2}}-\dfrac{2{t}^{\frac{3}{2}}}{\dfrac{3}{2}}+c$

$=\dfrac{2}{5}{t}^{\frac{5}{2}}-\dfrac{4}{3}{t}^{\frac{3}{2}}+c$

$=\dfrac{2}{5}{\left(x+2\right)}^{\frac{5}{2}}-\dfrac{4}{3}{\left(x+2\right)}^{\frac{3}{2}}+c$

Evaluate
$\displaystyle \int {x.\,\,\,{{\sin }^2}x\,\,dx}$

solution :

$\rightarrow \int x sin^{2}x dx$

$\int sin^{2}x\,dx = \int \frac{1}{2}(1-cos2x)dx = \frac{1}{2}(x-\frac{sin 2x}{2})+c$

we take

$u = x_{1}$

$v = sin^{2}x$

${u}' = 1$

${v}' =\frac{1}{2}(x-\frac{sin2x}{2})$

w.r.t

$\int uv = uv - \int {u}'v$

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

$= \frac{x}{2}(x-\frac{sin2x}{2})-\frac{1}{2}(x^{2}+\frac{cos^{2}x}{2})$

on simpiflying we get

Ans

$= \frac{x^{2}}{4}-x\frac{sin(2x)}{4} - \frac{cos(2x)}{8}$

1211640_1180491_ans_948f7e846a6d463783c763a7e835a1a6.jpg

Evaluate $\int {x.\,\,\,\log x\,\,dx}$

$\displaystyle\int{x\log{x}dx}$

Let

$u=\log{x}\Rightarrow\,du=\dfrac{1}{x}dx$

$dv=x\,dx\Rightarrow\,v=\dfrac{{x}^{2}}{2}$

$=\dfrac{{x}^{2}\log{x}}{2}-\displaystyle\int{\dfrac{{x}^{2}}{2}\dfrac{1}{x}dx}$

$=\dfrac{{x}^{2}\log{x}}{2}-\dfrac{1}{2}\displaystyle\int{xdx}$

$=\dfrac{{x}^{2}\log{x}}{2}-\dfrac{{x}^{2}}{4}+c$

Solve:
$\int {{{\cos }^3}x{e^{\log \left( {\sin x} \right)}}dx}$

$I=\displaystyle \int \cos^3x\ e^{\log (\sin c)}dx$

put

$\sin x=t$

$\cos x \ dx=dt$

$\begin{matrix} \Rightarrow I=\int (1-t^{ 2 })e^{ \log (t) }dt= & \int e^{ \log (t) }dt- & \int t^{ 2 }e^{ \log (t) }dt \\ & \downarrow & \downarrow \\ & \left( A \right) & \left( B \right) \end{matrix}$

$(B)$

$k=\displaystyle \int t^2\ e^{\log (t)}dt$

$k=e^{\log (t)}\dfrac {t^3}{3} \displaystyle \int \dfrac {t^3}{3}\times e^{\log (t)}\times \dfrac {1}{t}dt$

$k=e^{\log (t)}\dfrac {t^3}{3} \displaystyle \int \dfrac {t^2}{3} e^{\log (t)}dt$

$k=e^{\log (t)}\dfrac {t^3}{3}-\dfrac {1}{3}k$

$\dfrac {4k}{3}=e^{\log (t)}\dfrac {t^3}{3}$

$k=\dfrac {t^3}{4}e^{\log (t)}$

$L=\displaystyle \int e^{\log (t)}dt$

$L=e^{\log (t)}xt-\displaystyle \int t\times e^{\log (t)}\times \dfrac {1}{t}dt$

$L=t\ e^{\log t}-L$

$\Rightarrow \ 2L=te^{\log t}$

$\Rightarrow \ L=\dfrac {1}{2}[t\ e^{\log t}]$

$\therefore \ I=A-B=\dfrac {1}{2}t e^{\log t}-\dfrac {1}{4}t^3 e^{\log t}$

$=\dfrac {t^{\log t}}{2} \left [1-\dfrac {t^2}{2}\right]$

$\dfrac {\sin x}{2}e^{\log (\sin x)} \left [1-\dfrac {\sin^2x}{2}\right]$

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

$I=\dfrac {\sin x}{4}e^{\log (\sin x)} [1+\cos^2x]$

$\therefore \ \displaystyle \int \cos^3 x e^{\log (\sin x)}dx=\dfrac {\sin x}{4}[1+\cos^2x]e^{\log (\sin x)}+C$

Integrate:
$\int \frac{1-V}{1+V^{2}}dv.$

$\int \frac{1-v}{1+v^{2}}dv$

let

$v = tan\theta$ ___(1)

$dv = sec^{2}\theta d\theta$

$\int \frac{1-tan\theta }{1+tan^{2}\theta }\times sec^{2}\theta d\theta$

$\int \frac{1-tan\theta }{sec^{2}\theta }\times sec^{2}\theta d\theta$

$\int d\theta -\int tan\theta d\theta$

$\theta -(-ln(cos\theta ))+c$

$[\because \int tan\theta d\theta = ln\, tan\theta ]$

$\theta +ln(cos\theta )+c$ ___(2)

from (1)

$\theta = tan^{-1}v$

putting in (1)

$tan^{-1}v+ln(tan^{-1}v)+c$ Ans

1112037_1200531_ans_932ea4254c584cfbadc7e3d3c507a1d9.jpg

$\displaystyle\int \:e^x\left(x^2-x+1\right)dx$

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

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

Consider

$\displaystyle\int{{x}^{2}{e}^{x}dx}$ is of the form

$\displaystyle\int{u.dv}=uv-\int{v.du}$

Take

$u={x}^{2}\Rightarrow du=2x dx$ and

$dv={e}^{x} dx\Rightarrow v={e}^{x}$

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

$={x}^{2}{e}^{x}-3\int{x{e}^{x}dx}+\int{{e}^{x}dx}$

Consider

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

Take

$u=x\Rightarrow du=dx$ and

$dv={e}^{x}dx\Rightarrow v={e}^{x}$

$\Rightarrow {x}^{2}{e}^{x}-3\displaystyle\int{x{e}^{x}dx}+\int{{e}^{x}dx}={x}^{2}{e}^{x}-3\left(x{e}^{x}-\int{{e}^{x}dx}\right)+\int{{e}^{x}dx}$

$={x}^{2}{e}^{x}-3x{e}^{x}+4\displaystyle\int{{e}^{x}dx}$

$={x}^{2}{e}^{x}-3x{e}^{x}+4{e}^{x}$

$=\left({x}^{2}-3x+4\right){e}^{x}+c$ where

$c$ is the constant of integration.

integrate :
$\int {{{\tan }^{ - 1}}xdx} .$

$I=\int { \tan ^{ -1 }{ x } } dx=\int { 1.\tan ^{ -1 }{ x } } dx$

Integration by parts

$I=x\tan ^{ -1 }{ x } -\int { \cfrac { x }{ 1+{ x }^{ 2 } } } dx=x\tan ^{ -1 }{ x } -\cfrac { 1 }{ 2 } \int { \cfrac { 2x }{ 1+{ x }^{ 2 } } } dx=x\tan ^{ -1 }{ x } -\cfrac { 1 }{ 2 } \left( \cfrac { n }{ 1+{ x }^{ 2 } } \right) +C$

$\int cos^{2}\theta d\theta$ = ?

@page { margin: 2cm } p { margin-bottom: 0.25cm; line-height: 120% }

$\begin{matrix} \int { { { \cos }^{ 2 } }\theta d\theta } \\ \Rightarrow \int { \dfrac { { \left( { 1-\cos 2\theta } \right) d\theta } }{ 2 } } \\ \Rightarrow \int { \dfrac { 1 }{ 2 } } d\theta -\dfrac { 1 }{ 2 } \cos 2\theta d\theta \\ \Rightarrow \dfrac { \theta }{ 2 } -\dfrac { 1 }{ 2 } \dfrac { { \sin 2\theta } }{ 1 } +C \\ \Rightarrow \dfrac { \theta }{ 2 } -\dfrac { { \sin 2\theta } }{ 4 } +C \\ \end{matrix}$

$\int \dfrac {\sec^{8}x}{cosec x}dx$ .

$I = \displaystyle\int \dfrac{\sec^{8}x}{cosec\,x}dx$

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

$u = \cos\,x \Rightarrow du = -\sin\,x\,dx$

$\Rightarrow I =\displaystyle \int \dfrac{-du}{u^{8}}$

$= \dfrac{u^{-7}}{7}+c$

$= \dfrac{(\cos\,x)^{-7}}{7}+c$

$= \dfrac{1}{7\cos^{7}x}+c$

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

$\\\int x sin^{-1}x dx\\=sin^{-1}x\times(\frac{x^2}{2})-\int(\frac{1}{\sqrt{1-x^2}})\times(\frac{x^2}{2}) dx\\=(\frac{x^2}{2})sin^{-1}x-(\frac{1}{2})\int(\frac{x^2}{\sqrt{1-x^2}})dx\\=(\frac{x^2}{2})sin^{-1}x+(\frac{1}{2})\int\left[(\frac{(1-x^2)-1}{\sqrt{1-x^2}})\right]dx\\=(\frac{x^2}{2})sin^{-1}x+(\frac{1}{2})\int\sqrt{1-x^2}dx-(\frac{1}{2})\int(\frac{1}{\sqrt{1-x^2}})dx\\=(\frac{x^2}{2})sin^{-1}x+(\frac{1}{2})\left[(\frac{1}{2})\left(x\sqrt{1-x^2}+sin^{-1}x\right)\right]-(\frac{1}{2})sin^{-1}x+C\\=(\frac{2x^2-1}{4})sin^{-1}x+(\frac{x}{4})\sqrt{1-x^2}+C$

Evaluate: $\int (3x^2 - 5)^2 dx$

$\displaystyle\int (3x^2-5)^2dx$

$=\displaystyle\int (9x^4+25-30x^2)dx$

$=\dfrac{9x^5}{5}+25x-\dfrac{30}{3}x^3+c$

$=\dfrac{9x^5}{5}-10x^3+25x+c$

$\therefore \displaystyle\int (3x^2-5)^2dx=\dfrac{9x^5}{5}-10x^3+25x+c$ .

Find the area bounded by the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ and the ordinates $x=0$ and $x=ae$ , where $b^{2}=a^{2}(1-e^{2})$ and $e<1$

Required area

$A$ is given by

$A=2$ (Area of shaded region in first quadrant)

$\Rightarrow A=2 \displaystyle\int_{0}^{ae}|y|\ dx=2\displaystyle\int_{0}^{ae}y\ dx$

$[\because y\ge 0\therefore |y|=y]$

$\Rightarrow A=2\dfrac{b}{a}\left[\displaystyle\int_{0}^{ae}\sqrt{a^{2}-x^{2}}dx\right]$

$\Rightarrow A=\dfrac{2b}{a}\left[\dfrac{1}{2}\times \sqrt{a^{2}-x^{2}}+\dfrac{1}{2}a^{2}\sin^{-1}\dfrac{x}{a}\right]_{ae}^{0}$

$\Rightarrow A=\dfrac{2b}{a}\left[\dfrac{ae}{2}\sqrt{a^{2}-a^{2}e^{2}}+\dfrac{1}{2}a^{2}\sin^{-1}e\right]$

$\Rightarrow A=\dfrac{b}{a}\left(a^{2}e\sqrt{1-e^{2}}+a^{2}\sin^{-1}e\right)=ab\left(e\sqrt{1-e^{2}}+\sin^{-1}e\right)$
1405206_1221533_ans_b2397e0878414150960e70f42f4e7b22.png

$\dfrac { d y } { d x } = ( 4 x + y + 1 ) ^ { 2 }$

$\cfrac{dy}{dx}=(4x+y+1)^2$

Let,

$(4x+y+1)^2=z$

So,

$4dx+dy=dz$

$\implies dy=dz-4dx$

So,

$\cfrac{dz-4dx}{dx}=x^2\implies \cfrac{dz}{dx}=z^2+4$

So,

$\cfrac{dz}{z^2+4}=dx$

On integrating on both sides

$\cfrac{1}{2}\tan^{-1}(\cfrac{z}{2})=x+c$

Now as

$(4x+y+1)^2=z$

So,

$\cfrac{1}{2}\tan^{-1}(\cfrac{(4x+y+1)}{2})=x+c$

Evaluate: $\displaystyle\int {{x \over {\sqrt {x + 4} }}dx}$

Solution:

Let

$x+4=t\quad \Rightarrow \ t-4=x$

$\Rightarrow \ dx=dt$

$I=\displaystyle \int {\dfrac {t-4}{t^{1/2}}dt}$

$=\displaystyle \int {(t^{1/2}-4t^{-1/2})dt}$

$=\dfrac {2}{3}t^{3/2}-8t^{1/2}+C$

$=\dfrac {2}{3}(x+4)^{3/2}-8(x+4)^{1/2}+C$

Evaluate:
$\int {\frac{1}{{\sqrt {1 - x - {x^2}} dx}}}$

$\begin{array}{l} Given, \\ \int { \frac { 1 }{ { \sqrt { 1-x-{ x^{ 2 } } } dx } } } \\ Now, \\ =\int { \frac { { dx } }{ { \sqrt { \left( { 1-\frac { 1 }{ 4 } -\left( { { x^{ 2 } }+x+\frac { 1 }{ 4 } } \right) } \right) } } } } \\ =\int { \frac { { dx } }{ { \sqrt { { { \left( { \frac { { \sqrt { 5 } } }{ 2 } } \right) }^{ 2 } }-{ { \left( { x+\frac { 1 }{ 2 } } \right) }^{ 2 } } } } } } \\ ={ { { Sin } }^{ -1 } }\left( { \frac { { 2x+1 } }{ { \sqrt { 5 } } } } \right) +C \end{array}$

$\int \dfrac { \sin 2 x } { 1 + \sin ^ { 2 } x }$

$\int { \dfrac { \sin2x }{ 1+{ \sin }^{ 2 }x } } dx$

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

Take

$u=1+{ \sin }^{ 2 }x$

$du=2sinxcosxdx$

$\therefore$

$\int { \dfrac { 1 }{ u } } du=log\left( 1+{ \sin }^{ 2 }x \right) +C$

$\displaystyle\int \dfrac{(x^2+\sin^2x)\sec^2x}{1+x^2}\cdot dx$ .

1184754_1224066_ans_89d4761ade1d4ad1b46141e190023ec6.jpg

Solve:
$\displaystyle \int_{0}^{2\pi}{e^{x}.\sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)dx}$

1129934_1229563_ans_14f1df6f05304f609229c6418272a589.jpg

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

Now,

$\displaystyle\int e^x(x^2+2x)\ dx$

$=x^2e^x-\displaystyle\int e^x(2x)\ dx$

$+\displaystyle\int e^x(2x)\ dx$ [ Using method of by parts]

$=x^2e^x+c$ . [ Where

$c$ is integrating constant]

Evaluate: $\displaystyle \int \frac { 1 } { \sqrt { x ^ { 2 } + 8 x + 25 } } d x$

$I =\displaystyle \int \frac{dx}{\sqrt{x^{2}+8x+25}}$

$=\displaystyle \int \frac{dx}{\sqrt{(x+4)^{2}+9}}$

$u = x+4 \Rightarrow du = dx$

$I = \int \dfrac{du}{\sqrt{u^{2}+9}} = ln\left | 3\sqrt{u^{2}+9}+3|u| \right |+c$

$= ln\left | 3\sqrt{x^{2}+8x+25}+3|x+4| \right |+c$

$I = \int \frac { 1 } { \sqrt { 2 x ^ { 2 } + 3 x + 8 } } d x$

$I=\int { \frac { 1 }{ \sqrt { 2{ x }^{ 2 }+3x+8 } } dx } \\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1 }{ \sqrt { { x }^{ 2 }+\frac { 3 }{ 2 } x+4 } } dx } \\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1 }{ \sqrt { { x }^{ 2 }+2\times \frac { 3 }{ 4 } x+\frac { 9 }{ 16 } +4-\frac { 9 }{ 16 } } } dx } \\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1 }{ \sqrt { { \left( x+\frac { 3 }{ 4 } \right) }^{ 2 }+\frac { 55 }{ 16 } } } dx } \\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1 }{ \sqrt { { \left( x+\frac { 3 }{ 4 } \right) }^{ 2 }+{ \left( \frac { \sqrt { 55 } }{ 4 } \right) }^{ 2 } } } dx }$

let

$x+\frac { 3 }{ 4 } =u$

$dx=du$

$I=\frac { 1 }{ \sqrt { 2 } } \int { \frac { 1 }{ \sqrt { { u }^{ 2 }+{ \left( \frac { \sqrt { 55 } }{ 4 } \right) }^{ 2 } } } du } \\ =\frac { 1 }{ \sqrt { 2 } } \ln { \left[ u+\sqrt { { u }^{ 2 }+{ \left( \frac { \sqrt { 55 } }{ 4 } \right) }^{ 2 } } \right] } +c$

putting

$u=x+\frac { 3 }{ 4 }$

$I=\frac { 1 }{ \sqrt { 2 } } \ln { \left[ x+\frac { 3 }{ 4 } +\sqrt { { \left( x+\frac { 3 }{ 4 } \right) }^{ 2 }+{ \left( \frac { \sqrt { 55 } }{ 4 } \right) }^{ 2 } } \right] } +c\\ =\frac { 1 }{ \sqrt { 2 } } \ln { \left[ x+\frac { 3 }{ 4 } +\sqrt { { x }^{ 2 }+\frac { 3 }{ 2 } x+\frac { 9 }{ 16 } +\frac { 55 }{ 16 } } \right] } +c\\ =\frac { 1 }{ \sqrt { 2 } } \ln { \left[ x+\frac { 3 }{ 4 } +\sqrt { { x }^{ 2 }+\frac { 3 }{ 2 } x+4 } \right] } +c$

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

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

let

$I=\int { { e }^{ x }\cfrac { x+1-1 }{ { (x+1) }^{ 2 } } dx }$

$\Rightarrow I=\int { { e }^{ x }\{ \cfrac { 1 }{ x+1 } +\cfrac { -1 }{ { (x+1) }^{ 2 } } \} dx } \\ \Rightarrow I=\int { { e }^{ x }\cfrac { 1 }{ x+1 } dx } +\int { { e }^{ x }\cfrac { (-1) }{ { (x+1) }^{ 2 } } dx } \\ I=\cfrac { 1 }{ x+1 } { e }^{ x }-\int { \cfrac { (-1) }{ { (x+1) }^{ 2 } } { e }^{ x }dx } +\int { \cfrac { { e }^{ x }(-1) }{ { (x+1) }^{ 2 } } dx } +C\\ I=\cfrac { 1 }{ x+1 } { e }^{ x }+C$

$\displaystyle \int x \sin \dfrac {x}{2}=?$

Sol:

$\displaystyle \int x\sin \dfrac {x}{2}dx$

Let

$I=\displaystyle \int x\sin \dfrac {x}{2}dx$

We know that

$I=\displaystyle \int \Pi dx =I\displaystyle \int \Pi dx-\int \left (\dfrac {dI}{dx}\displaystyle \int \Pi dx\right)dx+C$

Now, using

$LAT \varepsilon$

$I=\displaystyle \int { x\sin { \dfrac { x }{ 2 } dx } =x } \displaystyle \int { \sin { \dfrac { x }{ 2 } dx\quad - } } \left[ \displaystyle \int { \left( \dfrac { dx }{ dx } \displaystyle \int { \sin { \dfrac { x }{ 2 } } dx } \right) } dx \right] +C$

$=x\dfrac { \left[ -\cos { \dfrac { x }{ 2 } } \right] }{ \dfrac { 1 }{ 2 } } -\displaystyle \int { \dfrac { \left( -\cos { \dfrac { x }{ 2 } } \right) }{ \dfrac { 1 }{ 2 } } } dx+C$

$=-2x\cos \dfrac {x}{2}+\displaystyle \int 2\cos \dfrac {x}{2}dx+C$

$=2x\cos \dfrac {x}{2}+2\displaystyle \int \cos \dfrac {x}{2}dx+C$

$=-2cos \dfrac {x}{2}+2\left [\dfrac {\dfrac {\sin x}{2}}{\dfrac {1}{2}}\right ]+C$

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

Hence, this is the answer.

Evaluate: $\displaystyle \int \tan ^ { 3 } 2 x \sec 2 x d x$

$I=\displaystyle \int { \tan ^{ 3 }{ 2x } \sec { 2x } } \ dx$

$I\displaystyle =\int { \tan ^{ 2 }{ 2x } \tan { 2x } \sec { 2x } } .\cfrac { 1 }{ 2 } d(2x)$

$I\displaystyle =\cfrac { 1 }{ 2 } \int { \left( \sec ^{ 2 }{ 2x } -1 \right) } \tan { 2x } \sec { 2x } d(2x)\quad$

$u=\sec { 2x }$

$du=\sec { 2x } \tan { 2x } d(2x)$

$\displaystyle =\cfrac { 1 }{ 2 } \int { \left( { u }^{ 2 }-1 \right) } du\quad \quad$

$=\cfrac { 1 }{ 2 } \left( \cfrac { { u }^{ 3 } }{ 3 } -u \right) =\cfrac { 1 }{ 2 } \left( \cfrac { \sec ^{ 3 }{ 2x } }{ 3 } -\sec { 2x } \right)$

Solve:
$\int \frac{dx}{\sqrt{4-x^{2}}}$

$\int \dfrac{dx}{\sqrt4-x^2} \ \ Let \ x=2\sin \theta$

$\dfrac{dx}{d \theta}=2 \cos \theta$

$=\int \dfrac{dx}{\sqrt{4-4 \sin^2} \theta}$

$=\int \dfrac{2\cos \theta d\theta}{\sqrt{4-4 \sin^2} \theta}$

$=\int { \dfrac { 2\cos { \theta d\theta } }{ \sqrt { 4\left( 1-{ \sin }^{ 2 }\theta \right) } } }$

$=\int { \dfrac { 2\cos { \theta d\theta } }{ \sqrt { 4\cos ^{ 2 }\theta } } }$

$=\int { \dfrac { 2\cos { \theta d\theta } }{ 2\cos { \theta } } }$

$=\int d\theta=\theta$

$\therefore \theta=\dfrac{1}{2}\sin^{-1}x+c.$ Ans

Solve
$I= \displaystyle\int \dfrac{x+2}{x^{2}+5x+6}dx$

We have,

$I=\displaystyle \int\dfrac{x+2}{x^2+5x+6}dx$

$I=\displaystyle \int\dfrac{x+2}{x^2+3x+2x+6}dx$

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

$I=\displaystyle \int\dfrac{1}{(x+3)}dx$

on Integrating and we get,

$I=\log(x+3)+c \ \ \because \displaystyle \int \dfrac{1}{x}dx=\log2$

Hence, this is the answer.

$\int {\frac{{1 - \sin x}}{{\cos x}}} dx$ =?

$\begin{array}{l} We\, have \\ \int { \frac { { 1-\sin x } }{ { \cos x } } } dx \\ Now, \\ \frac { { 1-\sin x } }{ { \cos x } } =\frac { { \left( { 1-\sin x } \right) } }{ { \cos x } } .\frac { { \left( { 1+\sin x } \right) } }{ { \left( { 1+\sin x } \right) } } \\ =\frac { { 1-{ { \sin }^{ 2 } }x } }{ { \cos \left( { 1+\sin x } \right) } } \\ =\frac { { \cos x } }{ { 1+\sin x } } \\ Then \\ \int { \frac { { 1-\sin x } }{ { \cos x } } } dx \\ =\int { \frac { { 1-\sin x } }{ { \cos x } } } dx \\ =\ln { \left( { 1+\sin x } \right) } +C \end{array}$

Evaluate: $\int x \sec ^ { 2 } x d x$

$\textbf{Step-1: Comparing with uv form. }$

$\text{We know that}$

$\int{uv}\: dx=u \int{v} \: dx-\int{\dfrac{du}{dx}}.\int{v}\: dx$

$\text{Hence, in the given question,}$

$u=x,v=\sec ^2{x}$

$\textbf{Step-2: Solving the integral. }$

$\text{By using}$

$u.v$

$\text{rule,}$

$\int{x.\sec^2{x}} \:dx=x. \int{\sec^2{x}}\: dx-\int{1}.\int{\sec^2{x}}\: dx$

$=x.\tan{x}-\int{\tan{x}}\: dx$

$=x.\tan{x}-\log(\sec{x})+C$

$\textbf{Thus, the solution for the integral is:}$

$\mathbf{x.tan{x}-log{sec{x}}+C.}$

Integrate: $\sin ^ { 3 } ( 2 x + 1 )$

$I=\displaystyle \int \sin^3 (2x + 1) dx$

Let,

$2x + 1 = t$

$2 \, dx = dt$

$I=\dfrac{1}{2} \displaystyle \int \sin^3 (t) dt$

We know that,

$\sin (3t) = 3 \sin t - 4 \sin^3 t$

$4 \sin^3 t = 3 \sin t - \sin (3t)$

$\sin^3 t = \dfrac{1}{4} [3 \sin t - \sin (3t)]$

$I=\dfrac{1}{2} \left[\displaystyle \int \dfrac{1}{4} (3 \sin t \, dt) - \int \dfrac{1}{4} \sin (3t) dt \right]$

$I=\dfrac{1}{2} \times \dfrac{1}{4} \left[3 \cos t - \dfrac{\cos 3t}{3} \right] + c$ . substitute

$t = 2x + 1$

$I=\dfrac{1}{24} [ 9 \cos (2x + 1) - \cos (6x + 3) ] + c$ .

Evaluate: $\int x \sin 2 x d x$

$I=\int x\sin 2x$

We know the formula (by Parts)

$\displaystyle\int uv=u\int v-\int \left(\int v\right)\dfrac{du}{dn}dx$

$u=x;v\sin 2x$

$I=x\displaystyle \int \sin 2x-\int \left(\int \sin 2x\right)\dfrac{d}{dx}(x)dx$

$I=x\dfrac{(-\cos (2x))}{2}-\displaystyle\int \left(\dfrac{-\cos 2x}{2}\right)1dn$

$I=\dfrac{-x\cos 2x}{2}+\dfrac{\sin 2x}{2.2}+C$

$I=\dfrac{\sin 2x}{4}-\dfrac{x\cos 2x}{2}+C$ .

Solve
$\int (4x^{3}-\dfrac{3}{x^{4}}) dx$

$\int \left( 4x^3-\dfrac{3}{x^4}\right)dx$

$=4\int x^3dx-3\int \dfrac{1}{x^4}dx$

$=4\dfrac{x^4}{4}-\dfrac{3d\left( x^{-4}\right)}{dx}$

$=4\times \dfrac{x^4}{4}-(3)(-4)x^{-4}$

$=x^4+\dfrac{12}{x^4}+C$

Hence, solved.

Evaluate : $\int { \sqrt { \dfrac { x }{ { x }^{ 3 }{ a }^{ 3 } } } dx }$

Consider the given integral.

$I=\int \sqrt{\dfrac{x}{x^3a^3}}dx$

$I=\int \sqrt{\dfrac{1}{x^2a^3}}dx$

$I={\dfrac{1}{a\sqrt a}}\int \dfrac{1}{x} dx$

$I={\dfrac{1}{a\sqrt a}} log_e x+C$

Hence, this is the answer.

Evaluate: $\displaystyle\int \dfrac { \cos x + x \sin x } { x ( x + \cos x ) } d x$

$I=\displaystyle\int \dfrac{\cos x+x\sin x}{x(x+\cos x)}dx$

$I=\displaystyle\int \dfrac{(\cos x+x)+(x\sin x-x)}{x(x+\cos x)}dx$

$I=\displaystyle\int \dfrac{1}{x}\dfrac{(x+\cos x)}{(x+\cos x)}+\displaystyle\int \dfrac{x(\sin x-1)}{x(x+\cos x)}dx$

$I=\displaystyle\int \dfrac{1}{x}dx+\displaystyle\int \dfrac{(\sin x-1)}{x+\cos x}dx$

Let

$x+\cos x=u$

$du=(1-\sin x)dx$

$I=ln|x|+\displaystyle\int \dfrac{-(1-\sin x)}{u}\dfrac{du}{(1-\sin x)}$

$I=\ln|x|-\ln|u|+c$

$I=\ln|x|-\ln|x+\cos x|+c$ .

Evaluate
$\int \dfrac {x+2}{2X^{2}+6x+5}dx$

1213729_1264624_ans_1bb043eb360c4b4a9a565f6ce83cbfd4.pdf

Solve: $\displaystyle \int{\dfrac{dx}{x(a+bx^{n})^{2}}}$

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

$put\, \, a+b{ x^{ n } }=t \\ bn.{ x^{ n-1 } }dx=dt \\$

Therefore,

$I=\dfrac { 1 }{ { n.b } } \int _{ }^{ }{ \dfrac { { dt } }{ { \dfrac { { \left( { t-a } \right) } }{ b } .{ t^{ 2 } } } } } \\$

$I=\dfrac { 1 }{ n } \int _{ }^{ }{ \dfrac { { dt } }{ { { t^{ 2 } }\left( { t-a } \right) } } } \\ =\dfrac { 1 }{ n } \int _{ }^{ }{ \left\{ { -\dfrac { 1 }{ { { a^{ 2 } }.t } } -\dfrac { 1 }{ { a{ t^{ 2 } } } } +\dfrac { 1 }{ { { a^{ 2 } }\left( { t-a } \right) } } } \right\} } dt \\ =\dfrac { 1 }{ n } \left[ { -\dfrac { 1 }{ { { a^{ 2 } } } } \log t+\dfrac { 1 }{ { at } } +\dfrac { 1 }{ { { a^{ 2 } } } } \log \left( { t-a } \right) } \right] +C \\ =\dfrac { 1 }{ n } \left[ { -\dfrac { 1 }{ { { a^{ 2 } } } } \log \left( { a+b{ x^{ n } } } \right) +\dfrac { 1 }{ { a\left( { a+b{ x^{ n } } } \right) } } +\dfrac { 1 }{ { { a^{ 2 } } } } \log \left( { b{ x^{ n } } } \right) } \right] +C$

Hence, this is the answer.

Solve
$\int \csc x dx$

$\displaystyle\int csc xdx$

$=\displaystyle\int csc x\dfrac{cscx+\cot x}{csc x+\cot x}\cdot dx$

$u=cscx+\cot x$

$du=-cscx\cot x-\csc^2x$

$=\displaystyle\int \dfrac{csc^2x+cscx\cot x}{csc x+\cot x}\cdot dx$

$=-\displaystyle\int\dfrac{csc^2x+cscx\cot x}{csc x+\cot x}\cdot dx$

$=-\displaystyle\int \dfrac{du}{u}=-\displaystyle\int \dfrac{1}{u}du$

$=-log|u|+c$

$=-log|csc +\cot x|+c$ .

1202604_1282373_ans_5776c0f5bc66425aa42a859ccbcd605b.jpg

Prove that:
$\displaystyle \int {\sin 2x}\ dx$

$\int sin\,2x \,dx = \int sinU.\frac{dU}{2} = \frac{1}{2} \int sinU.dU = \frac{1}{2}(-cos\: u) = -\frac{1}{2} cos\,2x + c$

$[\because U = 2x$

$dU = 2 \,dx$

$\frac{dU}{2} = dx$ ]

1210265_1284464_ans_e919d2f2055743d795a7f057c560fc52.jpg

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

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

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

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

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

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

$-\displaystyle\int \left( \dfrac{1}{(\log x)^2} \right) . dx$ [ Using method of by parts]

$= \left(\dfrac{x}{\log x} \right)+c$ [ Where

$c$ is integrating constant]

Find: $\int { \frac { cos\theta }{ (4+sin^{ 2 }\theta )(5-4{ cos }^{ 2 }\theta ) } d\theta }$

$\displaystyle\int \dfrac{\cos \theta}{(4+\sin ^2 \theta)(5-4\cos^2 \theta)}d \theta=\int \dfrac{\cos \theta}{(4+\sin ^2 \theta)(5-4+4\sin ^2 \theta)}d\theta=\int \dfrac{\cos \theta}{(4+\sin^2 \theta)(1+4\sin ^2 \theta)}d\theta$

Put

$t=\sin \theta\implies d t=\cos \theta d\theta$

$\implies \displaystyle\int \dfrac{d t}{(t^2+4)(4 t^2+1)}=\int \dfrac{A d t}{t^2+4}+\int \dfrac{B dt}{4 t^2+1}$

$\implies A(4 t^2+1)+B(t^2+4)=1\implies 4 A+B=0,A+4 B=1$

$\implies A=-\dfrac{1}{15},B=\dfrac{4}{15}$

$\implies -\dfrac{1}{15}\displaystyle\int \dfrac{d t}{t^2+4}+\dfrac{1}{15}\int \dfrac{d t}{t^2+1/4}=-\dfrac{1}{30}\text{tan}^{-1}\bigg(\dfrac{t}{2}\bigg)+\dfrac{2}{15}\text{tan}^{-1}(2 t)+C=-\dfrac{1}{30}\text{tan}^{-1}\bigg(\dfrac{\sin \theta}{2}\bigg)+\dfrac{2}{15}\text{tan}^{-1}(2\sin \theta)+C$

$\displaystyle\int { \dfrac { dx }{ { e }^{ x }+{ e }^{ -x } } }$ equals

$\begin{array}{l} \int { \dfrac { { dx } }{ { { e^{ x } }+{ e^{ -x } } } } } & \\ \Rightarrow \int { \dfrac { { { e^{ x } } } }{ { { e^{ 2x } }+1 } } } dx & put;{ e^{ x } }=t \\ \Rightarrow \int { \dfrac { 1 }{ { { t^{ 2 } }+1 } } dt } & { e^{ x } }dx=dt \\ \Rightarrow { \tan ^{ -1 }t }+c & \\ \Rightarrow { \tan ^{ -1 } }{ e^{ x } }+c & \end{array}$

Hence, this is the answer.

Evaluate $\int \dfrac{dx}{sin^2 X cos^2 X}$

$\displaystyle\int \dfrac{dx}{\sin^2x\cos^2x}=\displaystyle\int cosec^2x\sec^2xdx=\displaystyle\int (1+\cot^2x)\sec^2xdx$

$=\displaystyle\int \sec^2xdx+\displaystyle\int cosec^2xdx=\tan x-\cot x+c$ .

1181992_1288908_ans_dec76ca1ceef44099e673c5ba9f1fbce.jpg

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

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

$I=\dfrac{1}{2}\displaystyle \int \left(x+\dfrac{3}{2}+\dfrac{7}{4(x-\dfrac{1}{2})}\right)dx$

$I=\dfrac{1}{2}\left(\dfrac{x^2}{2}+\dfrac{3x}{2}+\dfrac{7\ln(x-\dfrac{1}{2})}{4}\right)+C$

Hence, this is the answer.

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

We have,

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

Let

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

$2xdx=dt$

$xdx=\dfrac{1}{2}dt$

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

$\Rightarrow \dfrac{1}{2}\displaystyle\int{\sqrt{t}dt}+2\int{\sqrt{{{x}^{2}}+{{1}^{2}}}}dx$

$\Rightarrow \dfrac{1}{2} \int{{{t}^{\dfrac{1}{2}}}}dt+2\left[ \dfrac{1}{2}\sqrt{{{x}^{2}}+{{1}^{2}}}+\dfrac{1}{2}\log \left( x+\sqrt{{{x}^{2}}+1} \right) \right]+C$

$\Rightarrow \dfrac{1}{2} \dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}+\left[ \sqrt{{{x}^{2}}+1}+\log \left( x+\sqrt{{{x}^{2}}+1} \right) \right] +C$

$\Rightarrow \dfrac{1}{3}{{t}^{\dfrac{3}{2}}}+\left[ \sqrt{{{x}^{2}}+1}+\log \left( x+\sqrt{{{x}^{2}}+1} \right) \right]+C$

$\Rightarrow \dfrac{1}{3}{{\left( \sqrt{{{x}^{2}}+1} \right)}^{\dfrac{3}{2}}}+\left[ \sqrt{{{x}^{2}}+1}+\log \left( x+\sqrt{{{x}^{2}}+1} \right) \right] +C$ .

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

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

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

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

Now

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

Put

$x^{2}-1$

differentiating wrt.x

2x .dx = dt

$\therefore xdx = \frac{dt}{2}$

$= \int \frac{x^{2}.xdx}{x^{2}-1} = \int \frac{t.dt}{2(t-1)}= \frac{1}{2}\int \frac{t}{t-1}$

$= \frac{1}{2}[\int \frac{t-1}{t-1}dt + \int \frac{1}{t-1}dt]$

$= \frac{1}{2}[\int dt + log(t-1)]$

$= \frac{t}{2} + \frac{1}{2}log(t-1)$

$= \frac{x^{2}}{2} + \frac{1}{2}log (x^{2}-1)+c$

$\therefore \int \frac{x^{3}+x+1}{x^{2}-1} = \int \frac{x^{3}dx}{x^{2}-1}+\int \frac{dx}{x-1}$

$= \frac{x^{2}}{2} + \frac{1}{2}log (x^{2}-1) + log(x-1) + c$

$= \frac{x^{2}}{2} + log\sqrt{x^{2}-1} + log (x-1) + c$

$= \frac{x^{2}}{2} + log (\sqrt{x^{2}-1}(x-1)) +c$

1226807_1297482_ans_36233909020b4df1a61420f36c87ce0a.jpg

Solve : $\displaystyle \int \dfrac{dx}{\sqrt{x^{2}-3x+2}}$

$\int \frac{dx}{\sqrt{x^{2}-3x+2}}=\int \frac{dx}{\sqrt{x^{2}-3x+\frac{9}{4}-\frac{9}{4}+2}}$

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

$=log (x-\frac{3}{2}+\sqrt{(x-\frac{3}{2})^{2}-\frac{1}{4}})+c$

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

$=log (x-\frac{3}{2}+\sqrt{x^{2}-3x+2})+c$

1233212_1303457_ans_dff5ff0dc4b84a53933744a7c20914aa.jpg

Solve
$f(x)=\displaystyle\int _{ 0 }^{ 2 }{ t\left( \sin { x } -\sin { t } \right) dt }$

Given,

$f(x)=\int _0^2t\left(\sin \left(x\right)-\sin \left(t\right)\right)dt$

apply integration by parts,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

Here,

$u=\left(\sin \left(x\right)-\sin \left(t\right)\right),v=t$

$f(x)=\left[\frac{1}{2}t^2\left(\sin \left(x\right)-\sin \left(t\right)\right)-\int \:-\frac{1}{2}t^2\cos \left(t\right)dt\right]^2_0$

$=\left[\frac{1}{2}t^2\left(\sin \left(x\right)-\sin \left(t\right)\right)-\left(-\frac{1}{2}t^2\sin \left(t\right)-t\cos \left(t\right)+\sin \left(t\right)\right)\right]^2_0$

$=\left[\frac{1}{2}t^2\sin \left(x\right)+t\cos \left(t\right)-\sin \left(t\right)\right]^2_0$

$=2\sin \left(x\right)+2\cos \left(2\right)-\sin \left(2\right)$

Integration
$\displaystyle\int \dfrac{x^{2}+3}{x^{6}(x^{2}+1)}dx$

$= \int \frac{(x^{2}+3)}{x^{b}(x^{2}+1)}dx$

Adding and subtrating

$2x^{2}$ in numerator

$I = \int [\frac{3(x^{2}+1)}{x^{6}(x^{2}+1)}-\frac{2x^{2}}{x^{6}(x^{2}+1)}]dx$

$=\int \frac{3}{x^{6}}dx - \int \frac{2}{x^{4}(x^{2}+1)}dx$

Again adding a subtracting

$2x^{2}$ in numerator

$I = \int \frac{3}{x^{6}}dx -[\int \frac{2(x^{2}+1)}{x^{4}(x^{2}+1)}dx - \frac{2x^{2}dx}{x^{4}(x^{2}+1)}]$

$=\int \frac{3}{x^{6}}dx -\int \frac{2}{x^{4}} + \int \frac{2dx}{x^{2}(x^{2}+1)}$

$= \int \frac{3}{x^{6}}dx-\int \frac{2}{x^{4}} +\int \frac{2dx}{x^{2}}-\int \frac{2dx}{x^{2}+1}$

$= \frac{-3}{5x^{5}} + \frac{2}{3x^{2}} - \frac{2}{x} - 2tan^{-1} (x)+c$

1234923_1303735_ans_580b3af6dd2643e390275031be43cc4b.jpg

Solve:
$\displaystyle \int{\dfrac{1}{x}\sqrt{\dfrac{x-1}{x+1}}dx}$

$\begin{array}{l} I=\int { \dfrac { 1 }{ x } \sqrt { \dfrac { { x-1 } }{ { x+1 } } } dx } \\ I=\int { \dfrac { 1 }{ x } \sqrt { \dfrac { { x-1 } }{ { x+1 } } \times \dfrac { { x+1 } }{ { x+1 } } } dx } \\ \int { \dfrac { 1 }{ x } \dfrac { { \sqrt { { x^{ 2 } }-1 } } }{ { x+1 } } dx } \, \, \, \, \, \, \, \left( \begin{array}{l} x=\sec \theta \\ dx=\sec \theta .\tan \theta \end{array} \right) \\ \int { \dfrac { 1 }{ { \sec \theta } } .\dfrac { { \sqrt { { { \sec }^{ 2 } }\theta -1 } } }{ { \sec \theta +1 } } .\sec \theta .\tan \theta } \, \, \, \, \, \left( \begin{array}{l} { \sec ^{ 2 } }\theta -{ \tan ^{ 2 } }\theta =1 \\ { \sec ^{ 2 } }\theta -1={ \tan ^{ 2 } }\theta \end{array} \right) \\ \int { \dfrac { { \sqrt { { { \tan }^{ 2 } }\theta } } }{ { \sec \theta +1 } } .\tan \theta } \\ \int { \dfrac { { { { \tan }^{ 2 } }\theta } }{ { \sec \theta +1 } } d\theta } =\int { \dfrac { { { { \sec }^{ 2 } }\theta -1 } }{ { \sec \theta +1 } } d\theta } \\ I=\int { \dfrac { { \left( { \sec \theta +1 } \right) \left( { \sec \theta -1 } \right) } }{ { \left( { \sec \theta +1 } \right) } } d\theta } \\ \int { \sec \theta \, d\theta - } \int { 1d\theta } \\ \ln { \left( { \sec \theta +\tan \theta } \right) } -\theta +c \\ I=\ln { \left( { x+\sqrt { { x^{ 2 } }-1 } } \right) } -{ \sec ^{ -1 } }x+c. \end{array}$

Hence, this is the answer.

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

Given

$\int { \dfrac { 1 }{ { x\sqrt { { x^{ 2 } }-1 } } } dx } \to \left( i \right)$

Let

$x=\sec \theta \Rightarrow \theta ={ \sec ^{ -1 } }x$

Differentiate

$dx=\sec \theta \tan \theta d\theta$ (by standard formulae)

Put in equation (i)

$\begin{array}{l} \Rightarrow \int { \dfrac { 1 }{ { \sec \theta \sqrt { { { \sec }^{ 2 } }\theta -1 } } } \times \sec \theta \tan \theta d\theta } & \\ \Rightarrow \int { \dfrac { { \tan \theta d\theta } }{ { \sqrt { { { \sec }^{ 2 } }\theta -1 } } } } & \left[ \begin{array}{l} { \sec ^{ 2 } }\theta -1={ \tan ^{ 2 } }\theta \\ \sqrt { { { \sec }^{ 2 } }\theta -1 } =\tan \theta \end{array} \right] \\ \Rightarrow \int { \dfrac { { \tan \theta d\theta } }{ { \tan \theta } } } =\int { d\theta } =\theta ={ \sec ^{ -1 } }x+c & \end{array}%$

(Where

$c$ is integral constant)

Evaluate
$\int{e^{x} \left(\dfrac{1+\sin x}{1+\cos x}\right)}$

$\begin{array}{l} \int { { e^{ x } }\left( { \cfrac { { 1+\sin x } }{ { 1+\cos x } } } \right) dx } & \\ \Rightarrow \int { \left[ { \cfrac { 1 }{ { 1+\cos x } } +\cfrac { { \sin x } }{ { 1+\cos x } } } \right] { e^{ x } }dx } & \\ \Rightarrow \int { \left[ { \cfrac { 1 }{ { 2{ { \cos }^{ 2 } }\cfrac { x }{ 2 } } } +\cfrac { { 2\sin \cfrac { x }{ 2 } .\cos \cfrac { x }{ 2 } } }{ { 2{ { \cos }^{ 2 } }\cfrac { x }{ 2 } } } } \right] { e^{ x } }dx } & \\ \Rightarrow \int { \left[ { \cfrac { { { { \sec }^{ 2 } }\cfrac { x }{ 2 } } }{ 2 } +\tan \cfrac { x }{ 2 } } \right] { e^{ x } }dx } & \\ \Rightarrow \int { \left[ { f'\left( x \right) +f\left( x \right) } \right] { e^{ x } }dx } & \left[ { \int { { e^{ x } }\left[ { f\left( x \right) +f'\left( x \right) } \right] dx={ e^{ x } }f\left( x \right) +c } } \right] \\ \Rightarrow { e^{ x } }.f\left( x \right) +c & \\ { e^{ x } }.\tan \cfrac { x }{ 2 } +c. & \end{array}$

Hence, this is the answer.

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

1346750_1315975_ans_19991b6b9a1c4dafa9c08b2b68613244.jpg

Evaluate
$\int \dfrac{\log (1+x)}{(1+x)}dx$

$\begin{array}{l} \int { \dfrac { { \log \left( { 1+x } \right) } }{ { 1+x } } dx } \\ u=\log \left( { 1+x } \right) \, \, \, \, du=\dfrac { 1 }{ { 1+x } } dx \\ \int { u\, \, du=\dfrac { { { u^{ 2 } } } }{ 2 } +c } \\ =\dfrac { 1 }{ 2 } { \left( { \log \left( { 1+x } \right) } \right) ^{ 2 } }+c \end{array}$

Hence, this is the answer.

Integrate:
$\displaystyle \int{\dfrac{\sqrt{1-x^{2}}+\sqrt{1+x^{2}}}{\sqrt{1-x^{4}}}}dx=$

$I= \int { \dfrac { { \sqrt { 1-{ x^{ 2 } } } +\sqrt { 1+{ x^{ 2 } } } } }{ { \sqrt { 1-{ x^{ 4 } } } } } } dx$

$I= \int { \dfrac { { \sqrt { 1-{ x^{ 2 } } } } }{ { \sqrt { 1-{ x^{ 4 } } } } } } dx+\int { \dfrac { { \sqrt { 1+{ x^{ 2 } } } } }{ { \sqrt { 1-{ x^{ 4 } } } } } } dx \\$

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

$I=\log \left| { x+\sqrt { { x^{ 2 } }+1 } } \right| +{ \sin ^{ -1 } }+C$

Hence, this is the answer.

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

$y = \int \frac{xe^{x}}{(1+x)^{2}}dx$

$y=\int [\frac{1}{1+x}-\frac{1}{(1+x)^{2}}]dx$

We know

$\int e^{x}[f(x)-f(x)]=e^{x}f(x)$

then

$y=\int e^{x}[\frac{1}{1+x}-\frac{1}{(1+x)^{2}}]dx=\frac{e^{x}}{1+x}+c$

1212968_1365828_ans_3df70d4bb4c2431dbe5b8392c7b6c920.JPG

Evaluate
$\displaystyle\int { \dfrac { { x }^{ 3 }+{ 4x }^{ 2 }-7x+5 }{ x+2 } } dx$

$\displaystyle\int \dfrac{x^3+4x^2-7x+5}{x+2}dx$

$x^2+2x-11$

________________

$x+2)x^3+4x^2-7x+5$

$x^3+2x^2$

________________

$2x^2-7x$

$2x^2-4x$

______________

$-11x+5$

$-11x-22$

____________

$27$

____________

$\displaystyle\int x^2+2x-11+\dfrac{27}{x+2}dx$

$\Rightarrow\dfrac{x^3}{3}+\dfrac{2x^2}{2}-11x+27 log(x+2)+c$

$\Rightarrow\dfrac{x^3}{3}+x^2-11x+27 log(x+2)+c$ .

$\int \dfrac{x^2dx}{(x sin x+ cosx)^2}$

Let

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

$=\displaystyle\int{\dfrac{x\cos{x}}{{\left(x\sin{x}+\cos{x}\right)}^{2}}\dfrac{x}{\cos{x}}dx}$

Integrating by parts, taking

$\dfrac{x}{\cos{x}}$ as the first function and

$\dfrac{x\cos{x}}{{\left(x\sin{x}+\cos{x}\right)}^{2}}$ as the second function.

Now, let us first evaluate

$\displaystyle\int{\dfrac{x\cos{x}dx}{{\left(x\sin{x}+\cos{x}\right)}^{2}}}$

Put

$t=x\sin{x}+\cos{x}$

$\Rightarrow dt=\left(\sin{x}+x\cos{x}-\sin{x}\right)dx=x\cos{x}dx$ we get

$\displaystyle\int{\dfrac{x\cos{x}dx}{{\left(x\sin{x}+\cos{x}\right)}^{2}}}$

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

$=\dfrac{-1}{t}=\dfrac{-1}{x\sin{x}+\cos{x}}$

Hence,

$I=\dfrac{x}{\cos{x}}.\dfrac{-1}{x\sin{x}+\cos{x}}-\displaystyle\int{\dfrac{\cos{x}+x\sin{x}}{{\cos}^{2}{x}}\times\dfrac{-1}{x\sin{x}+\cos{x}}dx}$

$=\dfrac{x}{\cos{x}}.\dfrac{-1}{x\sin{x}+\cos{x}}-\displaystyle\int{{\sec}^{2}{x}dx}$

$=\dfrac{-x}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+\tan{x}+c$

$=\dfrac{-x}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+\dfrac{\sin{x}}{\cos{x}}+c$

$=\dfrac{-x+x{\sin}^{2}{x}+\sin{x}\cos{x}}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+c$

$=\dfrac{x\left(1-{\sin}^{2}{x}\right)+\sin{x}\cos{x}}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+c$

$=\dfrac{x{\cos}^{2}{x}+\sin{x}\cos{x}}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+c$

$=\dfrac{{\cos}{x}\left(\sin{x}-x\cos{x}\right)}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+c$

$=\dfrac{\left(\sin{x}-x\cos{x}\right)}{\left(x\sin{x}+\cos{x}\right)}+c$

$\therefore \displaystyle\int{\dfrac{{x}^{2}dx}{{\left(x\sin{x}+\cos{x}\right)}^{2}}}=\dfrac{\left(\sin{x}-x\cos{x}\right)}{\left(x\sin{x}+\cos{x}\right)}+c$

Write the value of $\int \dfrac{dx}{x^2+16}$ .

$I=\int \dfrac{dx}{x^2+16}$

$I=\int \dfrac{dx}{x^2+4^2}$

$I=\dfrac{1}{4} tan^{-1} \dfrac{x}{4}+C$

Evaluate:
$\displaystyle\int e^x\left[\tan^{-1} x+\dfrac{1}{1+x^2}\right] dx$ .

Now,

$\displaystyle\int e^x\left[\tan^{-1} x+\dfrac{1}{1+x^2}\right] dx$

$=\displaystyle\int e^x.\tan^{-1} x dx$

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

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

$+\displaystyle\int \dfrac{e^x}{1+x^2}dx$ [ Using method of by parts]

$=e^x\tan^{-1} x+c$ . [ Where

$c$ is integrating constant]

$\displaystyle\int { \sqrt { 4-{ x }^{ 2 } } } dx$ is equals to

$I=\displaystyle\int{\sqrt{4-{x}^{2}}dx}$

Take

$x=2\sin{\theta}\Rightarrow dx=2\cos{\theta}d{\theta}$

$\sqrt{4-{x}^{2}}=\sqrt{4-4{\sin}^{2}{\theta}}=\sqrt{4\left(1-{\sin}^{2}{\theta}\right)}=2\sqrt{{\cos}^{2}{\theta}}=2\cos{\theta}$

$\displaystyle\int{\sqrt{4-{x}^{2}}dx}$

$=\displaystyle\int{2\cos{\theta}.2\cos{\theta}d{\theta}}$

$=2 \displaystyle \int{2{\cos}^{2}{\theta}d{\theta}}$

$=2 \displaystyle \int{\left(1+\cos{2\theta}\right)d{\theta}}$ since

$1+\cos{2\theta}=2{\cos}^{2}{\theta}$

$=2\left(\theta+\dfrac{\sin{2\theta}}{2}\right)+c$

$=2\left({\sin}^{-1}{x/2}+\dfrac{\sin{2{\sin}^{-1}{x/2}}}{2}\right)+c$ since

$x=2\sin{\theta}\Rightarrow \theta={\sin}^{-1}{x/2}$

Solve
$\int { \cfrac { { x }^{ 2 }-1 }{ ({ x }+1) } } dx$ .

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

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

$I=\int (x-1)dx$

$I=\dfrac{x^2}{2}-x+C$

Evaluate:
$\displaystyle\int\dfrac{a^x}{a^x+1} dx,a>0$

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

$=\dfrac{1}{\log_e a}\displaystyle\int\dfrac{a^x.\log_e a}{a^x+1} dx$

$=\dfrac{1}{\log_e a}\displaystyle\int\dfrac{d(a^x)}{a^x+1}$

$=\dfrac{1}{\log_e a}\log|(a^x+1)|+c.$ [ Where

$c$ is integrating constant]

Find the value of $\displaystyle\int { { x }^{ 2 }\left( 1-\dfrac { 1 }{ { x }^{ 2 } } \right) dx }$ .

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

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

$=\dfrac{x^3}{3}-x+C$

Find the value of $\displaystyle\int\dfrac{1-\cos 2x}{1+\cos 2x}dx$

Now,

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

$=\displaystyle\int\dfrac{2\sin^2 x}{2\cos^2 x}dx$ [ Since

$1-\cos 2x=2\sin^2 x$ and

$1+\cos 2x=2\cos^2 x$ ]

$=\displaystyle\int\tan^2 xdx$

$=\sec x+c$ [ Where

$c$ is integrating constant]

Evaluate $\int \dfrac{(\log x)^2}{x}dx$

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

Let

$\log x =t$

$\dfrac{1}{x}dx=dt$

Therefore,

$I=\int t^2 dt$

$I=\dfrac{t^3}{3}+C$

$I=\dfrac{(\log x)^3}{3}+C$

Solve
$I=\displaystyle \int (\sin 3x+ \cos 4\ x\ )dx$

Now,

$I=\displaystyle \int (\sin 3x+ \cos 4\ x\ )dx$

or,

$I=-\dfrac{\cos 3x}{3}+\dfrac{\sin 4x}{4}+c$ [ Where

$c$ is integrating constant].

$\displaystyle \int 2x^3+9x^2-8x+5 dx$

$\displaystyle \int 2x^3+9x^2-8x+5 dx\\\displaystyle \int 2x^3 dx+\int 9x^2 dx-\int 8x dx+\int 5 dx\\\dfrac{2x^4}{4}+\dfrac{9x^3}{3}-\dfrac{8x^2}{2}+5x+c\\\dfrac{x^4}{2}+3x^3-4x^2+5x+c$

$\displaystyle \int \dfrac {x^2}{x^3+64} dx$

$\displaystyle \int \dfrac{x^2}{x^3+64}dx\\t=x^3+64\implies dt=3x^2dt\\\displaystyle \dfrac 13\int \dfrac 1t dt=\dfrac 13\log t+c$

$\displaystyle \int \dfrac{8x+5}{4x^2+5x+6} dx$

Let

$t=4x^2+5x+6\\dt=8x+5dx\\\implies \displaystyle \int \dfrac 1tdt=\log t\\\log (4x^2+5x+6)+c$

$\displaystyle \int \dfrac 2x+\dfrac 3x dx$

$\displaystyle \int \dfrac 2x+\dfrac 3xdx\\\displaystyle \int \dfrac 5xdx\\5\log x+c$

If $\displaystyle \int {\left (\dfrac { x-1 }{ { x }^{ 2 } } \right ){ e }^{ x }dx=f\left( x \right) { e }^{ x }+c }$ , then write the value of f(x).

$\displaystyle\int \dfrac{x-1}{x^2}\times e^{x} d x=\int e^{x}\bigg(\dfrac{1}{x}-\dfrac{1}{x^2}\bigg) dx$

This is in the form of

$\displaystyle\int e^{x}(g(x)+g'(x)) d x=e^{x}g(x)+C$

Here

$g(x)=\dfrac{1}{x}$ and

$g'(x)=-\dfrac{1}{x^2}$

$\implies \displaystyle\int (\dfrac{x-1}{x^2})e^{x} d x=\dfrac{e^{x}}{x}+C$

$\implies f(x)=\dfrac{1}{x}$

Integrate:
$\int { x } \sin { 2x } dx$

$\displaystyle\int x\sin 2 x d x$

Differentiating by parts

$\displaystyle\int x\sin 2 x=x\int \sin 2 x d x-\int \dfrac{d (x)}{d x}\bigg(\int \sin 2 x d x\bigg)d x$

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

$\displaystyle \int 4x^3+2x^3-3x+6 dx$

$\displaystyle \int 4x^3+2x^3-3x+6 dx\\\displaystyle \int 6x^3-3x+6 dx\\3\displaystyle \int 2x^3-x+2 dx\\3\left(\dfrac {x^4}{2}-\dfrac{x^2}{2}+2x\right)\\\dfrac{3x^4-3x^2+6x}{2}$

$\displaystyle \int \dfrac{ dx}{\sqrt{1 + x}}$

$\displaystyle \int \dfrac{dx}{\sqrt{1 + x}}\\t=1+x\\dx=dt\\\displaystyle \int \dfrac{dt}{\sqrt t}=2\sqrt t\\2\sqrt{1+x}+c$

Evaluate $\displaystyle \int { \dfrac { 1-4x }{ \sqrt { 6+x-{ 2x }^{ 2 } } } } dx$

Let

$t=6+x-2x^2\\dt=1-4x dx$

$\displaystyle \int \dfrac 1{\sqrt t} dt\\2\sqrt t+c\\2\sqrt{6+x-2x^2}+c$

$\displaystyle \int \dfrac {\sin x\cos x}{\tan x \cot x}dx$

$\displaystyle \int \dfrac{\sin x\cos x}{\tan x\cot x }dx\\\displaystyle \int {\sin x\cos x}dx\\\dfrac {\sin ^2x}{2}+c$

$\int { \frac { dx }{ \sqrt { x } -\sqrt { x-1 } } }$

$\displaystyle\int{\dfrac{dx}{\sqrt{x}-\sqrt{x-1}}}$

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

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

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

$=\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x-1}}$

Solve
$\int { { x }^{ 3 } } \log xdx$

Integrating by parts,

Take

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

$u=\log{x}\Rightarrow du=\dfrac{dx}{x}$

$\displaystyle\int{{x}^{3}\log{x}dx}=\log{x}\times\dfrac{{x}^{4}}{4}-\dfrac{1}{4}\displaystyle\int{{x}^{4}\times\dfrac{1}{x}dx}$

$=\dfrac{{x}^{4}\log{x}}{4}-\dfrac{1}{4}\displaystyle\int{{x}^{3}dx}$

$=\dfrac{{x}^{4}\log{x}}{4}-\dfrac{1}{4}\dfrac{{x}^{4}}{4}+c$

$=\dfrac{{x}^{4}\log{x}}{4}-\dfrac{{x}^{4}}{16}+c$

Simplify : $\displaystyle\int{\dfrac{2dx}{\sqrt{x}}}$

$I=\displaystyle\int{\dfrac{2dx}{\sqrt{x}}}$

$=2\displaystyle\int{{x}^{\frac{-1}{2}}dx}$

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

$=2\times 2\sqrt{x}+c$

$=4\sqrt{x}+c$

Find: $\int { { x }^{ 2 }.\log { x } dx }$ .

$\displaystyle\int x^2.\log x d x$

Integrating by parts

$\implies \displaystyle\int (x^2)(\log x) d x=\log x\int x^2 d x-\int \dfrac{d(\log x)}{d x}\times \bigg(\int x^2 d x\bigg)d x=\dfrac{x^3}{3}\log x-\int \dfrac{1}{x}\times \dfrac{x^3}{3} d x=\dfrac{x^2}{3} \bigg(\log x-\dfrac{1}{3}\bigg)+C$

Match the correct pair.

$(A)$

$I=\int \text{cosec} xdx$

$I=\int \dfrac{\csc x(\csc x+\cot x)}{\csc x+\cot s}dx$

$I=\int \dfrac{\csc^2 x+\csc x.\cot x}{\csc x+\cot s}dx$

Let

$t=\csc x +\cot x$

$dx={\csc ^2 x+\csc x\cot x}dt$

Therefore,

$I=-\int \dfrac{1}{t}dt$

$I=-\log(t)+C$

$I=-\log(\csc x+\cot x)+C$

Hence, this is the answer.

$(B)$

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

Let

$t=\sqrt {x^2-1}$

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

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

Therefore,

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

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

$I=\tan^{-1} {\sqrt {x^2-1}}+C$

Hence, this is the answer.

$(C)$

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

Let

$x=a\sin t$

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

$dx=a\cos t dt$

Therefore,

$I=a^2\int \cos^2 t dt$

$I=\dfrac{a^2}{2}\int (1+\cos 2t) dt$

$I=\dfrac{a^2}{2}\left[t+\dfrac{\sin 2t}{2}\right]+C$

$I=\dfrac{a^2}{2}\left[t+\cos t\sin t\right]+C$

$I=\dfrac{a^2}{2}\left[\sin^{-1}\dfrac{x}{a}+\cos \sin^{-1} \dfrac{x}{a}\sin \sin ^{-1}\dfrac{x}{a}\right]+C$

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

Hence, this is the answer.

$(D)$

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

Let

$t=\dfrac{x}{a}$

$dt=\dfrac{dx}{da}$

Therefore,

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

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

$I=\sin^{-1}\dfrac{x}{a}+C$

Hence, this is the answer.

$(E)$

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

Let

$t=\dfrac{x}{a}$

$dt=\dfrac{dx}{da}$

Therefore,

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

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

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

Hence, this is the answer.

$\int x log x dx =?$

$\displaystyle \int x\log x d x$

Integrating by parts

$\implies \log x\displaystyle\int x d x-\int \bigg(\frac{d(\log x)}{d x}\bigg)\bigg(\int x d x\bigg)d x=\log x\bigg(\dfrac{x^2}{2}\bigg)-\int \dfrac{1}{x}\times \dfrac{x^2}{2} d x=\dfrac{x^2}{2}\log x-\dfrac{1}{2}\int x d x=\dfrac{x^2}{2}\log x-\dfrac{x^2}{4}+C$

Evaluate : $\displaystyle \;\int {{x^4}\left( {1 + \log x} \right)dx}$

1310310_1534430_ans_5aea7c5933d44e83a0644b8484a19c84.jpeg

Find :-
$\int {\left[ {\log (\log x) + \frac{1}{{(\log {x^2})}}} \right]dx}$

1230316_1462847_ans_bb7ceb8357af48ccaae8a5b163004587.jpg

Find : $\displaystyle\int (log \, x)^2 dx$

Let

$I=\displaystyle\int 1.(\log x)^2 d x$

Integrating by parts

$I=(\log x)^2\displaystyle\int 1 d x-\int \dfrac{d(\log x)^2}{d x}\bigg(\int 1 dx\bigg) d x=x(\log x)^2-2\int \log x d x=x(\log x)^2-2\log x\int dx+2\int \dfrac{d(\log x)}{d x}\bigg(\int d x\bigg)d x$

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

Evaluate the given integral
$\int { x.cosec ^{ 2 }{ x } } dx$

$I=\displaystyle\int{x.{cosec}^{2}{x}dx}$

Integrating by parts,we get

Let

$u=x\Rightarrow\,du=dx$

$dv={cosec}^{2}{x}dx\Rightarrow\,v=-\cot{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$I=-x\cot{x}-\displaystyle\int{-\cot{x}dx}$

$I=-x\cot{x}+\displaystyle\int{\cot{x}dx}$

We know that

$\displaystyle\int{\cot{x}dx}=\log{\left|\sin{x}\right|}+c$

$\therefore\,I=-x\cot{x}+\log{\left|\sin{x}\right|}+c$

Evaluate the given integral.
$\int { x\sin { x } \cos { x } } dx$

$I=\displaystyle\int{x\sin{x}\cos{x}\,dx}$

$\Rightarrow\,I=\dfrac{1}{2}\displaystyle\int{x\times \,2\sin{x}\cos{x}\,dx}$

$\Rightarrow\,I=\dfrac{1}{2}\displaystyle\int{x\sin{2x}\,dx}$

Integrating by parts, we get

$u=x\Rightarrow\,du=dx$

$dv=\sin{2x}\,dx\Rightarrow\,v=\dfrac{-1}{2}\cos{2x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=\dfrac{1}{2}\left[\dfrac{-x}{2}\cos{2x}-\displaystyle\int{\dfrac{-1}{2}\cos{2x}dx}\right]$

$\Rightarrow\,I=\dfrac{1}{2}\left[\dfrac{-x}{2}\cos{2x}+\dfrac{1}{2}\displaystyle\int{\cos{2x}dx}\right]$

$\Rightarrow\,I=\dfrac{1}{2}\left[\dfrac{-x}{2}\cos{2x}+\dfrac{1}{2}\dfrac{1}{2}\sin{2x}\right]+c$

$\therefore\,I=\dfrac{-x}{4}\cos{2x}+\dfrac{1}{8}\sin{2x}+c$

Evaluate the given integral.
$\int { { x }^{ n }\log { x } } dx$

$I=\displaystyle\int{{x}^{n}\log{x}dx}$

Integrating by parts, we get

Let

$u=\log{x}\Rightarrow\,du=\dfrac{dx}{x}$

$dv={x}^{n}\,dx\Rightarrow\,v=\dfrac{{x}^{n+1}}{n+1}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=\dfrac{{x}^{n+1}\log{x}}{n+1}-\displaystyle\int{\dfrac{{x}^{n+1}}{n+1}\times\dfrac{dx}{x}}$

$\Rightarrow\,I=\dfrac{{x}^{n+1}\log{x}}{n+1}-\dfrac{1}{n+1}\displaystyle\int{{x}^{n}\,dx}$

$\Rightarrow\,I=\dfrac{{x}^{n+1}\log{x}}{n+1}-\dfrac{1}{n+1}\times\dfrac{{x}^{n+1}}{n+1}+c$

$\therefore\,I=\dfrac{{x}^{n+1}\log{x}}{n+1}-\dfrac{1}{{\left(n+1\right)}^{2}}{x}^{n+1}+c$

Evaluate the given integral.
$\int { \sin { x } \log { \left( \cos { x } \right) } } dx\quad$

$I=\displaystyle\int{\sin{x}\log{\left(\cos{x}\right)}dx}$

Let

$t=\cos{x}\Rightarrow\,dt=-\sin{x}dx$

$I=-\displaystyle\int{\log{t}dt}$

Integrating by parts, we get

Let

$u=\log{t}\Rightarrow\,du=\dfrac{1}{t}dt$

$dv=dt\Rightarrow\,v=t$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=t\log{t}-\displaystyle\int{t\times\dfrac{1}{t}dt}$

$\Rightarrow\,I=t\log{t}-\displaystyle\int{dt}$

$\Rightarrow\,I=t\log{t}-t+c$

$\Rightarrow\,I=\cos{x}\log{\cos{x}}-\cos{x}+c$ where

$t=\cos{x}$

$\therefore\,I=\cos{x}\left(\log{\cos{x}}-1\right)+c$

Evaluate the given integral.
$\int { { x }^{ 2 }\cos { x } } dx$

$I=\displaystyle\int{{x}^{2}\cos{x}dx}$

Integrating by parts

Let

$u={x}^{2}\Rightarrow\,du=2x\,dx$

$dv=\cos{x}dx\Rightarrow\,v=\sin{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I={x}^{2}\sin{x}-\displaystyle\int{2x\sin{x}\,dx}$

$\Rightarrow\,I={x}^{2}\sin{x}-2\displaystyle\int{x\sin{x}\,dx}$

Let

$u=x\Rightarrow\,du=dx$ and

$dv=\sin{x}\,dx\Rightarrow\,v=-\cos{x}$

$\Rightarrow\,I={x}^{2}\sin{x}-2\left[-x\cos{x}+\displaystyle\int{\cos{x}dx}\right]$

$\Rightarrow\,I={x}^{2}\sin{x}-2\left[-x\cos{x}+\sin{x}\right]+c$

$\therefore\,I={x}^{2}\sin{x}+2x\cos{x}-2\sin{x}+c$

Evaluate the given integral.
$\int { x\cos { x } } dx\quad$

$I=\displaystyle\int{x.\cos{x}dx}$

Integrating by parts, we get

$u=x\Rightarrow\,du=dx$ and

$dv=\cos{x}dx\Rightarrow\,v=\sin{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$I=x\sin{x}-\displaystyle\int{\sin{x}\,dx}$

$=x\sin{x}+\cos{x}+c$

Evaluate the given integral.
$\int { x\sin { 2x } } dx\quad$

$I=\displaystyle\int{x\sin{2x}dx}$

Integrating by parts, we get

$u=x\Rightarrow\,du=dx$ and

$dv=\sin{2x}dx\Rightarrow\,v=\dfrac{-1}{2}\cos{2x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=\dfrac{-x}{2}\cos{2x}+\dfrac{1}{2}\displaystyle\int{\cos{2x}dx}$

$\Rightarrow\,I=\dfrac{-x}{2}\cos{2x}+\dfrac{1}{4}\sin{2x}+c$

Evaluate the given integral.
$\int { { e }^{ x } } \left( \tan { x } -\log { \cos { x } } \right) dx$

$I=\displaystyle\int{{e}^{x}\left(\tan{x}-\log{\cos{x}}\right)dx}$

$=\displaystyle\int{{e}^{x}\tan{x}\,dx}-\displaystyle\int{{e}^{x}\log{\cos{x}}dx}$

Consider

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

Let

$u=\log{\cos{x}}\Rightarrow\,du=\dfrac{1}{\cos{x}}\left(-\sin{x}\right)dx=-\tan{x}\,dx$

$dv={e}^{x}\,dx\Rightarrow\,v={e}^{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,\displaystyle\int{{e}^{x}\log{\cos{x}}dx}={e}^{x}\log{\cos{x}}-\displaystyle\int{-{e}^{x}\tan{x}\,dx}$

$\Rightarrow\,\displaystyle\int{{e}^{x}\log{\cos{x}}dx}={e}^{x}\log{\cos{x}}+\displaystyle\int{{e}^{x}\tan{x}\,dx}$

$\therefore\,I=\displaystyle\int{{e}^{x}\tan{x}\,dx}-\displaystyle\int{{e}^{x}\log{\cos{x}}dx}$

$=\displaystyle\int{{e}^{x}\tan{x}\,dx}-{e}^{x}\log{\cos{x}}-\displaystyle\int{{e}^{x}\tan{x}\,dx}+c$

$=-{e}^{x}\log{\cos{x}}+c$

$={e}^{x}\log{{\left(\cos{x}\right)}^{-1}}+c$

$={e}^{x}\log{\left(\dfrac{1}{\cos{x}}\right)}+c$

$={e}^{x}\log{\sec{x}}+c$

Evaluate the given integral.
$\displaystyle \int { \cos ^{ -1 }{ \left( \cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) } } dx$

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

Let

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

$I=\displaystyle\int{{\cos}^{-1}{\left(\dfrac{1-{\tan}^{2}{\theta}}{1+{\tan}^{2}{\theta}}\right)}{\sec}^{2}{\theta}d\theta}$

We know that

$\dfrac{1-{\tan}^{2}{\theta}}{1+{\tan}^{2}{\theta}}=\cos{2\theta}$

$I=\displaystyle\int{{\cos}^{-1}{\cos{2\theta}}{\sec}^{2}{\theta}d\theta}$

$I=2\displaystyle\int{\theta{\sec}^{2}{\theta}d\theta}$

Let

$u=\theta\Rightarrow\,du=d\theta$

$dv={\sec}^{2}{\theta}d\theta\Rightarrow\,v=\tan{\theta}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$I=2\theta\tan{\theta}-2\displaystyle\int{\tan{\theta}d\theta}$

$I=2\theta\tan{\theta}-2\log{\left|\sec{\theta}\right|}+c$

We have

$x=\tan{\theta}\Rightarrow\,{x}^{2}={\tan}^{2}{\theta}$

$\Rightarrow\,1+{x}^{2}=1+{\tan}^{2}{\theta}={\sec}^{2}{\theta}$

$\Rightarrow\,\sec{\theta}=\sqrt{1+{x}^{2}}$

$\therefore\,I=2x{\tan}^{-1}{x}-2\log{\left|\sqrt{1+{x}^{2}}\right|}+c$

$\therefore\,I=2x{\tan}^{-1}{x}-\log{\left|1+{x}^{2}\right|}+c$

Evaluate the given integral.
$\int { \log _{ 10 }{ x } } dx$

$I=\displaystyle\int{\log_{10}{x}\,dx}$

Integrating by parts, we get

Let

$u=\log_{10}{x}\Rightarrow\,du=\log{10}\dfrac{1}{x}dx$

$dv=dx\Rightarrow\,v=x$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$I=x\log_{10}{x}-\dfrac{1}{\log{10}}\int{x\times\dfrac{1}{x}dx}$

$I=x\log_{10}{x}-\int{dx}$

$I=x\log_{10}{x}-x+c$

$\therefore\,I=\dfrac{1}{\log{10}}\left\{x\left(\log{x}-1\right)\right\}+c$

Evaluate the given integral.
$\int { { e }^{ x } } \left( \sec { x } (1+\tan { x } ) \right) dx$

$I=\displaystyle\int{{e}^{x}\sec{x}\left(1+\tan{x}\right)dx}$

$I=\displaystyle\int{{e}^{x}\sec{x}dx}+\displaystyle\int{{e}^{x}\sec{x}\tan{x}\,dx}$

Consider

$\displaystyle\int{{e}^{x}\sec{x}\tan{x}\,dx}$

Let

$u={e}^{x}\Rightarrow\,du={e}^{x}\,dx$

$dv=\sec{x}\tan{x}\,dx\Rightarrow\,v=\sec{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\therefore\,\displaystyle\int{{e}^{x}\sec{x}\tan{x}\,dx}={e}^{x}\sec{x}-\displaystyle\int{{e}^{x}\sec{x}\,dx}$

Now,

$I=\displaystyle\int{{e}^{x}\sec{x}dx}+{e}^{x}\sec{x}-\displaystyle\int{{e}^{x}\sec{x}\,dx}+c$

$\therefore\,I={e}^{x}\sec{x}+c$

Evaluate $\int x\cdot log (x+1) dx$

Let

$I=\int x\log(x+1)dx$

Let

$u=\log(x+1)$ and

$v=x$

$\displaystyle \int uv \,dx=u\int v dx-\int \left[\dfrac{du}{dx}\int v dx\right]dx$

$\displaystyle I=\log(x+1)\int x\,dx-\int\left[\dfrac{d\log(x+1)}{dx}\int x\, dx\right]dx$

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

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

Now, we can write

$x^2=(x-1)(x+1)+1$

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

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

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

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

$I=\log(x+1)\dfrac{x^2}{2}-\dfrac{x^2}{4}+\dfrac{x}{2}-\dfrac{1}{2}\log|x+1|+C$

$\text { Evaluate: } \displaystyle \int \dfrac{e^{x}}{\sqrt{5-4 e^{x}-e^{2 x}}} \mathrm{d} \mathrm{x}$

$\begin{array}{l}\text { Let } I=\displaystyle \int \dfrac{e^{x}}{\sqrt{5-4 e^{x}-e^{2 z}}} \mathrm{d} \mathrm{x} \\\text { Put } \quad e^{x}=t \quad \Rightarrow \quad e^{x} d x=d t, \text { we get } \\\therefore \quad I=\displaystyle \int \dfrac{\mathrm{dt}}{\sqrt{5-4 t-t^{2}}}=\displaystyle \int \dfrac{\mathrm{dt}}{\sqrt{-\left(t^{2}+4 t-5\right)}}=\int \dfrac{\mathrm{dt}}{\sqrt{-\left(t^{2}+2 \cdot t \cdot 2+2^{2}-9\right)}} \\=\displaystyle \int \dfrac{\mathrm{dt}}{\sqrt{3^{2}-(t+2)^{2}}}\end{array}$

Using

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

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

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

Find $\displaystyle \int x^2 \,tan^{-1} \,xdx.$

$I = \displaystyle \int x^2 tan^{-1}\,x dx.$

Using integration by parts

$= tan^{-1} x \displaystyle \int x^2 dx - \displaystyle \int \dfrac{1}{1 + x^2}. \dfrac{x^3}{3} dx$

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

$= \dfrac{x^3}{3} tan^{-1}x - \dfrac{x^2}{6} + \dfrac{1}{6} log|1 + x^2| + C.$

Solve:
$\int x\sin x \ dx$

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

Using

$\int uv \ dx=u \int v \ dx-\int [ \dfrac {du}{dx} \cdot \int v \ dx] \cdot dx$

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

$=-x\cos x+\sin x+c$

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

$I = \displaystyle \int_{0}^{1} x(tan^{-1} x)^2 dx$
Integrating by parts, we have

$I = \left ( \dfrac{x^2}{2} (tan^{-1}x)^2 \right )_{0}^{1} - \displaystyle \int_{0}^{1} 2tan^{-1}x \dfrac{1}{1 + x^2} \dfrac{x^2}{2} dx$

$I = \dfrac{\pi^2}{32} - \displaystyle \int_{0}^{1} tan^{-1}x \dfrac{x^2}{1 + x^2} dx$

$I = \dfrac{\pi^2}{32} - I_1$ where

$I_1 = \displaystyle \int_{0}^{1} \dfrac{x^2}{1 + x^2} tan^{-1}xdx$
Now,

$I_1 = \displaystyle \int_{0}^{1} \dfrac{x^2 + 1 - 1}{1 + x^2} tan^{-1} xdx$

$I_1 = \displaystyle \int_{0}^{1} tan^{-1} xdx - \displaystyle \int_{0}^{1} \dfrac{1}{1 + x^2} tan^{-1}xdx$

$I_1 = I_2 - \dfrac{1}{2} \left [ (tan^{-1} x)^2 \right ]_{0}^{1} = I_2 - \dfrac{\pi^2}{32}$ where

$I_2 = \displaystyle \int_{0}^{1} tan^{-1} xdx$

Using by parts in

$I_2$

$I_2 = (xtan^{-1}x)_{0}^{1} - \displaystyle \int_{0}^{1} \dfrac{1}{1 + x^2} xdx$

$I_2 = \dfrac{\pi}{4} - \dfrac{1}{2} \left [ log |1 + x^2| \right [_{0}^{1} = \dfrac{\pi}{4} - \dfrac{1}{2} log 2$
Thus,

$I_1 = \dfrac{\pi}{4} - \dfrac{1}{2} log 2 - \dfrac{\pi^2}{32}$
Therefore,

$I = \dfrac{\pi^2}{32} - \dfrac{\pi}{4} + \dfrac{1}{2} log 2 + \dfrac{\pi^2}{32} = \dfrac{\pi^2}{16} - \dfrac{\pi}{4} + \dfrac{1}{2} log 2$

$I = \dfrac{\pi^2 - 4\pi}{16} + log \sqrt{2}$

$\displaystyle \int xe^{x}\cos x dx=\frac{e^{x}}{2}\left [ \left ( x-1 \right )\sin x+x\cos x \right ]$ . If this is true enter 1, else enter 0.

$\displaystyle \cos x$ is R.P. of

$\displaystyle e^{ix}$ i.e. of

$\displaystyle \cos x+i\sin x\therefore I=R.P.$ of

$\displaystyle \int xe^{x}e^{ix}dx=\int xe^{\left ( 1+i \right) x}dx$
Intergrating by parts

$\displaystyle I=x.\frac{e^{1+i}x}{\left ( 1+i \right )}-\frac{1}{\left ( 1+i \right )}\int e^{\left ( 1+i \right) x}.1dx$

$\displaystyle I=x.\frac{e^{1+i}x}{\left ( 1+i \right )}-\frac{1}{\left ( 1+i \right )} e^{\left ( 1+i \right) x}$

$\displaystyle = \frac{e^{\left ( 1+i \right )x}}{\left ( 1+i \right )^{2}}\left [ x\left ( 1+i \right )-1 \right ]=\frac{e^{x}.e^{ix}}{2i}\left [ x+xi-1 \right ]$

$\displaystyle = \frac{e^{x}\left [ \left ( x-1 \right ) +ix\right ]\left ( \cos x+i\sin x \right )}{2i}$
Real part of above is imaginary part of N'

$\displaystyle I=\frac{e^{x}}{2}\left [ \left ( x-1 \right )\sin x+x\cos x \right ]$

$f$ the graph of the antiderivative $F(x)$ of $\displaystyle f\left ( x \right )=\log \left ( \log x \right )+\left ( \log x \right )^{-2}$
passes through $(e, 1998-e)$ then the term independent of $x$ in $F(x)$ is

Antiderivative of

$f(x)=F(x)=\int \left( \log \left( \log x \right) +\left( \log x \right) ^{ -2 } \right) dx+C$

$=x\log \left( \log x \right) -\int \dfrac { x }{ x\log x } dx+\int \left( \log x \right) ^{ -2 }dx+C\\ =x\log \left( \log x \right) -\left[ x\left( \log x \right) ^{ -1 }+\int \left( \log x \right) ^{ -2 }dx \right] +\int \left( \log x \right) ^{ -2 }dx+C\\ =x\log \left( \log x \right) -x\left( \log x \right) ^{ -1 }+C$
For point

$\left( e,1998-e \right)$ , we get

$1998-e=e.0-e+C\Rightarrow C=1998$

$\displaystyle \int \sqrt{x^{6}+1}.\frac{\log \left ( x^{6}+1 \right )-6\log x}{x^{10}}dx=\frac{1}{6}\left [ \frac{2}{3}t^{3/2}\log t -\frac{2}{3}\int t^{3/2}\frac{1}{t} \right ]$ where $t=1+\frac{1}{x^{6}}$ . If this is true enter 1, else enter 0.

Let

$\displaystyle I=\int \sqrt { x^{ 6 }+1 } .\frac { \log \left( x^{ 6 }+1 \right) -6\log x }{ x^{ 10 } } dx$

Write

$\displaystyle \frac { 1 }{ x^{ 10 } } =\frac { 1 }{ x^{ 7 }.x^{ 3 } } =\frac { 1 }{ x^{ 7 }.\sqrt { x^{ 6 } } }$

We get

$\displaystyle I=\int \sqrt { \frac { x^{ 6 }+1 }{ x^{ 6 } } } \log \frac { x^{ 6 }+1 }{ x^{ 6 } } \cdot \frac { 1 }{ x^{ 7 } } dx$

Put

$\displaystyle \frac { x^{ 6 }+1 }{ x^{ 6 } } =1+\frac { 1 }{ x^{ 6 } } =t\Rightarrow \frac { -6 }{ x^{ 7 } } dx=dt$

Therefore

$\displaystyle I=\int -\frac { 1 }{ 6 } \sqrt { t } \log tdt=\frac { -1 }{ 6 } \left[ \frac { 2 }{ 3 } t^{ 3/2 }\log t-\frac { 2 }{ 3 } \int t^{ 3/2 }\frac { 1 }{ t } \right] dt$

$\displaystyle =-\frac { 1 }{ 6 } \left[ \frac { 2 }{ 3 } t^{ 3/2 }\log t-\frac { 2 }{ 3 } \int t^{ 3/2 }\frac { 1 }{ t } \right]$

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

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

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

$\displaystyle =\frac { 1 }{ 2 } \log { \left( 1+{ x }^{ 2 } \right) } +c$

Evaluate: $\int \cos^{-1} (\sin x)dx.$

Let

$\cos^{-1}(\sin x) = \theta$

$\Rightarrow \sin x = \cos \theta$

$\Rightarrow \sin x = \sin \left(\dfrac{\pi}{2} - \theta\right)$

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

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

$\therefore \int \cos^{-1}(\sin x)dx = \displaystyle \int \left(\cfrac{\pi}{2} - x\right)dx$

$= \displaystyle \int \cfrac{\pi}2 dx - \int xdx$

$=\cfrac{\pi x}{2} - \cfrac{x^2}2+c$ , where

$c$ is a constant of integration.

Evaluate : $\displaystyle\int \dfrac{\sin(x-a)}{\sin(x+a)}dx$ .

Let

$\displaystyle I=\int \dfrac{\sin (x-a)}{\sin (x+a)}dx$

Put

$x + a = t$ , we have

$dx = dt$

$I=\displaystyle \int\dfrac{\sin (t-2a)}{\sin t}dt$

$I=\displaystyle \int\left(\dfrac{\sin t \cos 2a}{\sin t} -\dfrac{\cos t\sin 2a}{\sin t}\right )dt$

$\displaystyle I=\cos 2a\int dt-\sin 2a\int \cot tdt$

$I=t\cos 2a-\sin 2a.\log \left | \sin t \right |+C$

$I=(x+a)\cos 2a-\sin 2a.\log \left | \sin (x+a) \right |+C$

Find: $\displaystyle \int x \tan^{-1} x dx$

$\displaystyle \int { f\left( x \right) \dot { g\left( x \right) } = f\left( x \right) g\left( x \right) -\int { \dot { f\left( x \right) } g\left( x \right) } }$

$f\left( x \right) =\tan ^{ -1 }{ x }$ and

$\dot { g\left( x \right) } =x\quad \\ g(x)=\cfrac { { x }^{ 2 } }{ 2 }$

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

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

Prove that: $\displaystyle \int \sqrt {a^{2} - x^{2}}dx = \dfrac {x}{2}\sqrt {a^{2} - x^{2}} + \dfrac {a^{2}}{2}\sin^{-1} \left (\dfrac {x}{a}\right ) + c$

Let

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

$= \int \sqrt {a^{2} - x^{2}} \cdot 1 dx$
On integrating by parts, we get

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

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

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

$= x\cdot \sqrt {a^{2} - x^{2}} -\displaystyle \int \left [\sqrt {a^{2} - x^{2}} - \dfrac {a^{2}}{\sqrt {a^{2} - x^{2}}} \right ]dx$

$= x\cdot \sqrt {a^{2} - x^{2}} -\displaystyle \int \sqrt {a^{2} - x^{2}} dx + a^{2} \int \dfrac {1}{\sqrt {a^{2} - x^{2}}} dx$

$= x\cdot \sqrt {a^{2} - x^{2}} - I + a^{2} \cdot \sin^{-1} \left (\dfrac {x}{a}\right ) + c_{1}$

$2I = x \sqrt {a^{2} - x^{2}} + a^{2}\cdot \sin^{-1} \left (\dfrac {x}{a}\right ) + c_{1}$

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

$\therefore \displaystyle \int \sqrt {a^{2} - x^{2}} dx = \dfrac {x}{2}\sqrt {a^{2} - x^{2}} + \dfrac {a^{2}}{2}\sin^{-1} \left (\dfrac {x}{a}\right ) + c$ where,

$c = \dfrac {c_{1}}{2}$ .

Prove that $\displaystyle \int \sqrt{x^2 - a^2} dx = \dfrac{x}{2} \sqrt{x^2 - a^2} - \dfrac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + c$ .

Given,

$\displaystyle \int \sqrt{x^2 - a^2} dx$

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

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

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

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

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

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

$\displaystyle \int \sqrt{x^2 - a^2} dx = \dfrac{1}{2} \sqrt{x^2 - a^2}(x) + \dfrac{1}{2} a^2 \log |x + \sqrt{x^2 - a^2}| + c$ ....where

$c$ is constant term.

Solve $\int { \sqrt { \cfrac { \sin { (x-a) } }{ \sin { (x+a) } } } } dx$

1306037_781868_ans_807e120130fa4c219ef3e2995009e9c3.jpg

Evaluate $\displaystyle\int\displaystyle\frac{\cos 2x}{\sqrt{1+\sin 2x}}dx$ .

We need to evaluate

$\int \dfrac{{cos\:2x}}{{\sqrt{1+sin\:2x}}} dx$

Consider

$\int \dfrac{{cos\:2x}}{{\sqrt{1+sin\:2x}}} dx$

Let

$u=1+sin\:2x$

$\Rightarrow du=2cos\:2x\:dx$

$\Rightarrow cos\:2x\:dx=\dfrac{du}{2}$

$\therefore \int \dfrac{cos\:2x}{\sqrt{1+sin\:2x}}dx=\int \dfrac{du}{2\sqrt{u}}$

$=\dfrac{1}{2}\int \dfrac{du}{\sqrt{u}}$

$=\dfrac{1}{2}\left[\dfrac{u^{1/2}}{(1/2)}\right]$

$=u^{1/2}$

$=\sqrt{u}$

Substitute back

$u=1+sin\:2x$ we get

$=\sqrt{1+sin\:2x}+c$ where

$c$ is integration constant.

Thus

$\int \dfrac{{cos\:2x}}{{\sqrt{1+sin\:2x}}} dx=\sqrt{1+sin\:2x}+c$

Evaluate: $\displaystyle\int \tan^{-1}\sqrt{x}dx$ .

Use integration by parts, assume

$tan^{-1}\sqrt{x}$ as first function and

$1$ as second function.

$I=\int tan^{-1}\sqrt{x}dx=tan^{-1}\sqrt{x}\int 1dx - \int \dfrac{dtan^{-1}\sqrt{x}}{dx} xdx$

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

$I_2=\int \dfrac{\sqrt{x}}{2(x+1)} dx$

Put

$u=\sqrt{x}\rightarrow du=\dfrac{1}{2\sqrt{x}}dx$

$I_2=\int \dfrac{u^2}{u^2+1}du=\int (1-\dfrac{1}{u^2+1})du=u-tan^{-1}u+C$

Undo

$u=\sqrt{x}$

$I_2=\sqrt{x}-tan^{-1}\sqrt{x}+C$

$\therefore I=xtan^{-1}\sqrt{x}+ \sqrt{x}-tan^{-1}\sqrt{x}+C$

$I=(x+1)tan^{-1}\sqrt{x}+\sqrt{x}+C$

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

Given :

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

Let :

$t={ x }^{ { x }^{ 2 } }\\ \log _{ e }{ t } ={ x }^{ 2 }\log { x }$

Differentiating with respect to

$x$

$\cfrac { 1 }{ t } dt=2\times \ln { x } +{ x }^{ 2 }\times \cfrac { 1 }{ x } \\ dt={ x }^{ { x }^{ 2 }+1 }(2\ln { x } +1)dx\\ \int { { x }^{ { x }^{ 2 }+1 }(2\ln { x } +1)dx } =\int { dt } =t={ x }^{ { x }^{ 2 } }+C$

Hence the correct answer is

${ x }^{ { x }^{ 2 } }+C$

Prove that $\int \sin^{17}x dx = -\dfrac {\sin^{16}x \cos x}{17} + \dfrac {16}{17}\int \sin^{15}x dx + c$ .

$\int { \sin ^{ 17 }{ x } } dx$

Integration by parts when

$\sin ^{ 16 }{ x }$ is the differentiating term and

$\sin { x }$ is integrating term

$=\sin ^{ 16 }{ x } \int { \sin { x } dx } -\int { \frac { d }{ dx } } \left( \sin ^{ 16 }{ x } \right) \left( \int { \sin { x } dx } \right) dx\\ =\sin ^{ 16 }{ x } \left( -\cos { x } \right) +\int { 16 } \sin ^{ 15 }{ x } \cos ^{ 2 }{ x } dx\\ =-\sin ^{ 16 }{ x } \cos { x } +16\int { \sin ^{ 15 }{ x } \left( 1-\sin ^{ 2 }{ x } \right) dx } \\ =-\sin ^{ 16 }{ x } \cos { x } +16\int { \sin ^{ 15 }{ x } } -16\int { \sin ^{ 17 }{ x } dx } \\ \\ 17\int { \sin ^{ 17 }{ x } dx } =-\sin ^{ 16 }{ x } \cos { x } +16\int { \sin ^{ 15 }{ x } } \\ \int { \sin ^{ 17 }{ x } dx } =-\cfrac { \sin ^{ 16 }{ x } \cos { x } }{ 17 } +\cfrac { 16 }{ 7 } \int { \sin ^{ 15 }{ x } } +C$

Hence proved

$\int { \cfrac { { e }^{ x }\left( { x }^{ 3 }-x+2 \right) }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } } } dx=$

For this problem we'll use a very interesting result that derivative of $(x+1){ e }^{ x }$ equals to $(x+2){ e }^{ x }$ .

The given integral can be broken to use above result.

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

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

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

Using integration by parts and taking $({ x }{ e }^{ x }+{ e }^{ x })$ as first and $\dfrac { 2x }{ { (x }^{ 2 }+1)^{ 2 } }$ as second function, we get:

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

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

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

$={ e }^{ x }\dfrac { (x+1) }{ { (x }^{ 2 }+1) } +C.$

Hence the given integral equals to ${ e }^{ x }\dfrac { (x+1) }{ { (x }^{ 2 }+1) } +C.$

Integrate $\int_{0}^{2[x]} \left \{\dfrac {x}{2}\right \} dx$

$\displaystyle\int_{0}^{2[x]}\left\{\dfrac{x}{2}\right\}dx=2[x]\int_{0}^{1}\left\{\dfrac{x}{2}\right\}dx$ As

$\left\{\dfrac{x}{2}\right\}=\dfrac{x}{2}\forall x\epsilon [0,1]$

$=2[x]\displaystyle\int_{0}^{1}\dfrac{x}{2}dx=\dfrac{2}{4}[x] [x^{2}]_{0}^{1}$

$=\dfrac{1}{2}[x]$

Integrate $\displaystyle \int{ \frac{1}{x^{1/2} + x^{1/3}} dx}$

$\Rightarrow \ LCM$ of

$2$ and

$3=6$

Thus

$x=t^6$ equation

$(1)$

$dx=6t^5\ dt$ equation

$(2)$

$\therefore \ \displaystyle \int \dfrac {dx}{x^{1/2}+x^{1/3}}=I$

Now

$I=\displaystyle \int \dfrac {6t^2 \ dt}{t^3 +t^2}$

$=\displaystyle \int \dfrac {6t^5 dt}{t^2 (t+1)}$

$=6\displaystyle \int \dfrac {t^3 dt}{t+1}$

$=6\displaystyle \int \dfrac {(t^3 +1)-1\ dt}{t+1}$

$=6\displaystyle \int \left (t^2 -t+1-\dfrac {1}{t+1} \right)dt$

$=6\left [\dfrac {t^2}{3}-\dfrac {t^2}{2}+t-\log |t+1|\right]+C$

$=6\left [\dfrac {\sqrt x}{3}-\dfrac {x^{1/3}}{2}+x^{1/6}-\log |x^{1/6 +1}|\right]+C$

Ans:

$6\left [\dfrac {\sqrt x}{3}-\dfrac {x^{1/3}}{2}+x^{1/6}-\log |x^{1/6 +1}\right]+C$

Evaluate $\int_{0}^{4}(|x| + |x - 2| + |x - 4|)$ dx.

$\displaystyle \int_{0}^{4} (\left | x \right |+ \left | x-2 \right |+ \left | x-4 \right |)dx$

$\displaystyle \int_{0}^{4} \left | x \right |+ \int_{0}^{2} 2-x + \int_{0}^{4}x-2+\int_{0}^{4} 4-x$

$\displaystyle _{0}^{4}| \dfrac{x^2}{2}+ \left [ 2x- \dfrac{x^2}{2} \right ]_{0}^{2} + \left [ \dfrac{x^{2}}{2}-2x \right ]_{2}^{4} + \left [ 4x-\dfrac{x^2}{2} \right ]_{0}^{4}$

$= 8+4-2+8-8-2+4+16-8$

$= 32-12$

$= 20$

1175872_877236_ans_e5a992c3a649455992093e6466c12536.jpg

Evaluate : $\int^1_0$ $\frac{log (1 +x)}{1 + x^2}dx$

$\displaystyle\int_{0}^{1}\dfrac{\log (1+x)}{1+x^{2}}dx$

Put

$x=\tan$

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

limits are

$I=\displaystyle\int_{0}^{\pi/4}\log (1+\tan\theta)d\theta ...... (1)$

$\therefore I=\displaystyle\int_{0}^{\pi/4}\dfrac{\log (1+\tan\theta)}{(1+\tan^{2}\theta)}\sec^{2}\theta d\theta$

$(\because 1+\tan^{2}\theta=\sec^{2}\theta)$

Now we know that

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

$\therefore I=\displaystyle\int_{0}^{\pi/4}\log (1+\tan (\pi/4-\theta))d\theta$

and

$\tan \left(\dfrac{\pi}{4}-\theta\right)=\dfrac{\tan \dfrac{\pi}{4}-\tan\theta}{1+\tan\dfrac{\pi}{4}\tan\theta}=\dfrac{1-\tan\theta}{1+\tan\theta}$

$\therefore I=\displaystyle\int_{0}^{\pi/4}\log \left(1+\dfrac{1-\tan\theta}{1+\tan\theta}\right)d\theta$

$I=\displaystyle\int_{0}^{\pi/4}\log \left(\dfrac{2}{1+\tan\theta}\right)d\theta$

$I=\displaystyle\int_{0}^{\pi/4}[\log 2-\log (1+\tan\theta)]d\theta$

$I=\displaystyle\int_{0}^{\pi/4}\log 2 d\theta-I$

$2I=\log 2\displaystyle\int_{0}^{\pi/4}d\theta$

$I=\dfrac{1}{2}\log 2[0]_{0}^{\pi/4}$

$I=\dfrac{1}{2}\log 2\left[\dfrac{\pi}{4}-0\right]$

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

Evaluate $\displaystyle\int\dfrac{e^{-x}}{1+e^x}dx$

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

${ e }^{ x }=m$

${ e }^{ x }dx=dm$

$\Rightarrow \int { \dfrac { 1 }{ { m }^{ 2 }\left( m+1 \right) } } dm$

$=\int { \left[ \dfrac { 1 }{ m+1 } -\dfrac { 1 }{ m } +\dfrac { 1 }{ { m }^{ 2 } } \right] }$

$\Rightarrow In\left| m+1 \right| -In\left| m \right| -\dfrac { 1 }{ m } +c$

$\Rightarrow In\left| { e }^{ x }+1 \right| -In{ e }^{ x }-{ e }^{ -x }+c$

$=In\left| e^{ x }+1 \right| -x-{ e }^{ -x }+c$

Solve $\left[-\displaystyle \int^{\pi/2}_0\cos \left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)e^x\right]dx$

Given: $-\int _{0}^{\dfrac{\pi }{2}}\cos \left(\dfrac{\pi }{4}+\dfrac{x}{2}\right) e^x dx$

Apply the u-substitution method:

$u=\dfrac{x}{2}$

$= \int_{0}^{\dfrac{\pi }{2}} \cos \left(\dfrac{\pi }{4}+u\right)e^{2u}\cdot \:2du$

$= 2\cdot \int \:\cos \left(\dfrac{\pi }{4}+u\right)e^{2u}du$

Apply the integration by parts,

Let $u=e^{2u},\:v'=\cos \left(\dfrac{\pi }{4}+u\right)$

$= \left[ 2\left(e^{2u}\sin \left(\dfrac{\pi }{4}+u\right) \right] - \int \:e^{2u}\cdot \:2\sin \left(\dfrac{\pi }{4}+u\right)du\right)$

$= 2\left[ 2\left(e^{2u}\sin \left(\dfrac{\pi }{4}+u\right) \right] - 2\cdot \int e^{2u}\sin \left(\dfrac{\pi }{4}+u\right)du\right)$

Apply integration by parts $u=e^{2u},\:v'=\sin \left(\dfrac{\pi }{4}+u\right)$

$=2\left(e^{2u}\sin \left(\dfrac{\pi }{4}+u\right)-2\left(-e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)-\int \:-2e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)du\right)\right)$

$=2\left(e^{2u}\sin \left(\dfrac{\pi }{4}+u\right)-2\left(-e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)-\left(-2\cdot \int \:e^{2u}\cos \left(\dfrac{\pi }{4}+u\right)du\right)\right)\right)$

$= 2\cdot \dfrac{1}{5}e^{2u}\left(\sin \left(u+\dfrac{\pi }{4}\right)+2\cos \left(u+\dfrac{\pi }{4}\right)\right)$

$=\dfrac{2}{5}e^x\left(\sin \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)+2\cos \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)\right)$

$= \left[ \dfrac{2}{5}e^x\left(\sin \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)+2\cos \left(\dfrac{x}{2}+\dfrac{\pi }{4}\right)\right) \right] _{0}^{\dfrac{\pi }{2}}$

$=\dfrac{2e^{\dfrac{\pi }{2}}}{5}-\dfrac{3\sqrt{2}}{5}$

$= -\dfrac{2e^{\dfrac{\pi }{2}}}{5}+\dfrac{3\sqrt{2}}{5}$

Thus, the factor of the equation is $-\dfrac{2e^{\dfrac{\pi }{2}}}{5}+\dfrac{3\sqrt{2}}{5}$ .

$\int { { x }^{ n }\log { x } dx }$

The first preference is to be given for logarithmic function than algebraic function from 'ILATE'

$\int u.v dx =u\int vdx-\int\left(\left(\dfrac{du}{dx} \right)\int vdx \right)dx$

$u=logx$

$v=x^n$

$\int x^nlogxdx=logx\int x^n dx-\int\left(\left(\dfrac{d(logx)}{dx}\right)\int x^ndx \right)dx$

$=\dfrac{x^{n+1}.logx}{n+1}-\int \left ( \dfrac{1}{x}.\dfrac{x^{n+1}}{n+1} \right)dx$

$=\dfrac{x^{n+1}.logx}{n+1}- \dfrac{1}{n+1} \int x^n dx$

$=\dfrac{x^{n+1}.logx}{n+1}-\dfrac{x^{n+1}}{(n+1)^2}+c$

$\int { \sin { 4x } .{ e }^{ \tan ^{ 2 }{ x } } } dx$

Let $I = \int {\sin 4x{e^{{{\tan }^2}x}}dx}$

$I = \int {\sin 2\left( {2x} \right){e^{{{\tan }^2}x}}dx}$

$= \int {2\sin 2x\cos 2x{e^{{{\tan }^2}x}}dx}$

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

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

$= \int {\dfrac{{4\tan x\left( {1 - {{\tan }^2}x} \right)}}{{{{\left( {{{\sec }^2}x} \right)}^2}}}{e^{{{\tan }^2}x}}dx}$

Put ${\tan ^2}x = t$ , then,

$2\tan x{\sec ^2}xdx = dt$

$2\tan xdx = \frac{{dt}}{{{{\sec }^2}x}}$

$I = \int {\dfrac{{2\left( {1 - t} \right)}}{{{{\left( {{{\sec }^2}x} \right)}^3}}}{e^t}dt}$

$= \int {\dfrac{{2\left( {1 + 1 - 1 - t} \right)}}{{{{\left( {1 + {{\tan }^2}x} \right)}^3}}}{e^t}dt}$

$= 2\int {\dfrac{{2 - \left( {1 + t} \right)}}{{{{\left( {1 + t} \right)}^3}}}{e^t}dt}$

$= 2\int {\left[ {\dfrac{2}{{{{\left( {1 + t} \right)}^3}}} - \frac{{1 + t}}{{{{\left( {1 + t} \right)}^3}}}} \right]{e^t}dt}$

$= 2\int {\left[ {\dfrac{2}{{{{\left( {1 + t} \right)}^3}}} - \frac{1}{{{{\left( {1 + t} \right)}^2}}}} \right]{e^t}dt}$

Applying the formula $\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx} = {e^x}f\left( x \right) + c$

$I = 2{e^t}\left( { - \dfrac{1}{{{{\left( {1 + t} \right)}^2}}}} \right) + c$

$= - \dfrac{{2{e^{{{\tan }^2}x}}}}{{{{\left( {1 + {{\tan }^2}x} \right)}^2}}} + c$

Integrate the following functions:
$x^{2}\cdot e^{3x}$

$I=\int x^{2}e^{3x}dx$

By applying integration by parts

According to it let $u$ and $v$ be two functions defined in $x$ then,

$\int u\left ( x \right )v\left (x \right )dx=u\left ( x \right )\int v\left ( x \right )dx-\int\left ( \dfrac{\mathrm{d} \left ( u\left ( x \right ) \right )}{\mathrm{d} x}\int v\left ( x \right )dx \right )dx$ (1)

Let $v=x^{2}$ and $u=e^{3x}$

Then according to the formulae in equation (1)

$\int x^{2}e^{3x}dx=x^{2}\int e^{3x}dx-\int\left ( \dfrac{\mathrm{d} \left (x ^{2} \right )}{\mathrm{d} x}\int e^{3x}dx \right )dx$ (2)

Since $\int e^{3x}dx=\dfrac{e^{3x}}{3}$ and $\dfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{2} \right )=2x$

Equation (2) is reduced to,

$\int x^{2}e^{3x}dx=\dfrac{x^{2}e^{3x}}{3}-\int \left ( \dfrac{2xe^{3x}}{3} \right )dx$

$=\dfrac{x^{2}e^{3x}}{3}-\dfrac{2}{3}I_{1}$ (3)

Solve $I_{1}$

$I_{1}=\int xe^{3x}dx$

Let $v=x$ and $u=e^{3x}$

Using formula in equation (1)

$\int xe^{3x}=x\int e^{x}dx-\int \left ( \dfrac{\mathrm{d} \left ( x \right )}{\mathrm{d} x}\int e^{3x}dx \right )dx$ (4)

Since $\int e^{3x}dx=\frac{e^{3x}}{3}$ and $\dfrac{\mathrm{d} \left ( x \right )}{\mathrm{d} x}=1$

Equation (4 )is reduced to,

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

$=\dfrac{x^{2}e^{3x}}{3}-\dfrac{e^{3x}}{9}$

Substitute the value of $I_{2}$ in equation (3)

$I=\dfrac{x^{2}e^{3x}}{3}-\dfrac{2}{3}\left ( \dfrac{x^{2}e^{3x}}{3}-\dfrac{e^{3x}}{9} \right )$

$=\dfrac{x^{2}e^{3x}}{3}-\dfrac{2x^{2}e^{3x}}{9}+\dfrac{2e^{3x}}{27}$

$=\dfrac{x^{2}e^{3x}}{9}+\dfrac{2e^{3x}}{27}$

Thus, $I=\int x^{2}e^{3x}dx$ is $\dfrac{x^{2}e^{3x}}{9}+\dfrac{2e^{3x}}{27}+C$ where $C$ is integration constant.

$\int {{{dx} \over {2{x^2} + 2x + 5}}}$

1314722_1034380_ans_8272ea6dd8b14c7dafcf7db7206f84bd.jpeg

Integrate $\int { { e }^{ x }{ a }^{ x }dx }$

Using integration by parts

$\Rightarrow I=e^{x}. \int a^{x} dx+\int \dfrac {d} {dx} (e^{x}). \int a^{x} dx. dx$

$\Rightarrow I=e^{x} a^{x} log(a) +\int e^{x} a^{x} log(a)$

$\Rightarrow I=e^{x} a^{x} log(a) +log(a) \int e^{x} a^{x} dx$

$\Rightarrow I=e^{x} a^{x} log(a) +log(a) (I)$

$\Rightarrow I=\dfrac {e^{x} a^{x} log(a)} {1-log(a)}$

Solve $\displaystyle \int_0^{\pi/2} \dfrac{x\,\,\sin\,x \,cos\,x}{cos^4x+\sin^4x}dx$

$I=\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{x\sin x \cos x}{\sin^4 x+\cos^4 x}}dx$

Using ,

$\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$

$I=\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{\left(\dfrac{\pi}{2}-x\right)\sin{\left(\dfrac{\pi}{2}-x\right)} \cos{\left(\dfrac{\pi}{2}-x\right)}}{\sin^4{\left(\dfrac{\pi}{2}-x\right)}+\cos^4{\left(\dfrac{\pi}{2}-x\right)}}}dx$

We know,

$\cos{\left(\dfrac{\pi}{2}-x\right)}=\sin{x}$ and

$\sin{\left(\dfrac{\pi}{2}-x\right)}=\cos{x}$

Putting in

$I$

$\Rightarrow \displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{\left(\dfrac{\pi}{2}-x\right)\cos x \sin x}{\cos^4 x+\sin^4 x}}dx$

$I=\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{\dfrac{\pi}{2}\sin x \cos x}{\sin^4 x+\cos^4 x}}dx-\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{x\sin x \cos x}{\sin^4 x+\cos^4 x}}dx$

$I=\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{\dfrac{\pi}{2}\sin x \cos x}{\sin^4 x+\cos^4 x}}dx-I$

$2I=\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{\dfrac{\pi}{2}\sin x \cos x}{\sin^4 x+\cos^4 x}}dx$

$I=\dfrac{\pi}{4}\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{\sin x \cos x}{\sin^4 x+\cos^4 x}}dx$

Dividing Numerator and Denominator with

$\cos^4{x}$

$I=\dfrac{\pi}{4}\displaystyle\int_{0}^{\dfrac{\pi}{2}} {\dfrac{\dfrac{\sin x}{\cos^3 x}}{\dfrac{\sin^4 x}{\cos^4 x}+1}}dx$

$I=\dfrac{\pi}{4}\displaystyle\int_{0}^{\dfrac{\pi}{2}}\dfrac{\tan x\sec^2 x}{tan^4 x+1}dx$

Substitue

$\tan^2 x=t$

Then

$2\tan x \sec^2 x dx=dt$

When

$x=0,t=0$ and

$x=\dfrac{\pi}{2},t=\infty$

$I=\dfrac{\pi}{4}\displaystyle\int_{0}^{\infty}\dfrac{\dfrac{dt}{2}}{t^2+1}$

$I=\dfrac{\pi}{8}\displaystyle\int_{0}^{\infty}\dfrac{dt}{t^2+1}$

$I=\dfrac{\pi}{8}[\tan^{-1} t]_0^{\infty}$

$I=\dfrac{\pi}{8}[tan^{-1}{\infty}- tan^{-1}{0}]$

$I=\dfrac{\pi}{8}[\dfrac{\pi}{2}-0]$

$I=\dfrac{\pi^2}{16}$

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

$\begin{array}{l} \int _{ }^{ }{ \sqrt { 2x } \cos 2x } =\int _{ }^{ }{ \sqrt { 2 } \sqrt { x } \cos \left( { 2x } \right) dx } \\ =\frac { { \sqrt { x } \sin \left( { 2x } \right) } }{ 2 } -\int _{ }^{ }{ \frac { { \sin \left( { 2x } \right) } }{ { 4\sqrt { x } } } dx } \\ =\frac { { \sqrt { x } \sin \left( { 2x } \right) } }{ 2 } -\frac { { \sqrt { \pi } { { S } }\left( { \frac { { 2\sqrt { x } } }{ { \sqrt { \pi } } } } \right) } }{ 4 } \left( { Where\, S\, is\, Fresnal\, Integral } \right) \end{array}$

$\int _{ 0 }^{ \pi /2 }{ \underbrace { { x }^{ 2 } }_{ I } \underbrace { \csc ^{ 2 }{ xd } }_{ II } } =I$

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

$I = \int_{0}^{\pi / 2} \underset {I}{x^{2}} \underset {II}{cosec^{2}x} dx$

$=-x^{2} \cot x \int_{0}^{\pi/2} - 2x \int_{0}^{\pi/2} - \cot x\ dx$

$= -\left (\dfrac {\pi}{2}\right )^{2} (0) + 2\int \underset {I}{x} \underset {II}{\cot x} dx$

$I = 2\times ln\ \sin x]_{0}^{\pi/2} - 2\int_{0}^{\pi/2} ln \sin x\ dx$

$I = 2\times ln\ 1 - 2 \int_{0}^{\pi/2} ln\ \sin x\ dx$

Now take

$\int_{0}^{\pi/2} ln\ \sin x = I_{1} - (a)$

$I_{1} = \int_{0}^{\pi/2} ln\ \cos x - (b)$

Now

$(a) + (b)\Rightarrow$

$2I_{1} = \int_{0}^{\pi/2} (ln\ \cos x + ln\ \sin x) dx$

$\left \{ln\ a + ln\ b = ln\ ab\right \}$

$2I_{1} = \int_{0}^{\pi/2} ln (\sin x \cos x) dx$

$2I_{1} = \int_{0}^{\pi/2} ln \left (\dfrac {\sin 2x}{2}\right )dx$

$2I_{1} = \int_{0}^{\pi/2} ln\ \sin 2x\ dx - \int_{0}^{\pi/2} ln\ 2\ dx$

$2I_{1} = \int_{0}^{\pi/2} ln\ \sin 2x\ dx - \dfrac {\pi}{2} ln 2$

Now

$2x = t$

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

$2I_{1} = \dfrac {1}{2} \int_{0}^{\pi} ln\ \sin t\ dt - \dfrac {\pi}{2} ln 2$

$\because \int_{a}^{b} f(t) dt = \int_{a}^{b} f(x) dx$

$2I_{1} = \dfrac {1}{2} \int_{0}^{\pi} ln\ \sin x\ dx - \dfrac {\pi}{2} ln 2$

Let

$f(x) = ln \sin x$

$f(2a - x) = ln \sin (\pi - x) = ln \sin x = f(x)$

$\therefore \int_{0}^{2a} f(x) dx = 2\int_{0}^{a} f(x) dx$

Now

$2I_{1} = \dfrac {2}{3} \int_{0}^{\pi/2} ln \sin x\ dx - \dfrac {\pi}{2} ln 2$

$I_{1} = -\dfrac {\pi}{2} ln 2$

$I = -2I_{1}$

$I = +\pi ln 2$

$I = \pi ln 2$ .

Solve $\displaystyle\int \dfrac{\sin x\cos x}{5\sin^2 x+3\cos^2 x}dx$ .

We are given,

$I=\displaystyle \int \dfrac {\sin x.\cos x}{5\sin ^2 x+3\cos^2 x}dx$

$I=\displaystyle \int \dfrac {\sin x.\cos x}{5\sin^2 x+3(1-\sin^2 c)}dx$

(as we know

$\sin^2 \theta +\cos^2 \theta =1$ )

$=\displaystyle \int \dfrac {\sin x.\cos x.dx}{5\sin^2x +3\sin^2x}$

$=\displaystyle \int \dfrac {\sin x. \cos x dx}{3+2\sin^2 x}$

Let

$\sin^2 x=t$ , differentiating it with respect to

$x$ we get

$23\sin x \cos x \ dx=dt$

$\therefore \ \sin x\cos x\ dx=\dfrac {dt}{2}$

$\displaystyle \int \dfrac {dt}{2(3+2t)}$

$=\displaystyle \int \dfrac {dt}{6+4t}$

$I=\dfrac {1}{4} \log |6+4t| +C$

$I=\displaystyle \int \dfrac {\sin x.\cos x}{5\sin^2 x +3\cos^2 x}=\dfrac {1}{4} \log |6+4\sin^2 x|+C$

Evaluate $\int {\dfrac{{dx}}{{\left( {2x\, - \,7} \right)\,\sqrt {\left( {x\, - \,3} \right)\left( {x\, - \,4} \right)} }}}$

$\int{\dfrac{dx}{(2x-7)\sqrt{(x-3)(x-4)}}}$

$\int{\dfrac{dx}{(2x-7)\sqrt{x^2-7x+12}}}$

Multiply and divide with

$2$

$\int{\dfrac{2dx}{2(2x-7)\sqrt{x^2-7x+12}}}$

Take the factor

$2$ inside the square root and multiply with the quadratic expression

$\int{\dfrac{2dx}{2(2x-7)\sqrt{4x^2-28x+48}}}$

Add and subtract

$1$ inside the square root

$\int{\dfrac{2dx}{(2x-7)\sqrt{4x^2-28x+48+1-1}}}$

$\int{\dfrac{2dx}{(2x-7)\sqrt{4x^2-28x+49-1}}}$

This can be simplified as

$\int{\dfrac{2dx}{(2x-7)\sqrt{(2x-7)^2-1}}}$

This is in the form of

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

Hence the required integral can be simplified as

$\Rightarrow\int{\dfrac{2dx}{(2x-7)\sqrt{(2x-7)^2-1}}}=\dfrac{1}{2}sec^{-1}(2x-7)+C$

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

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

This can be written as

$\Rightarrow \int{x\sqrt{\dfrac{(1+x)(1+x)}{(1-x)(1+x)}}}dx$

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

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

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

$\Rightarrow \int{\dfrac{x}{\sqrt{1-x^2}}}dx+\int{\dfrac{x^2}{\sqrt{1-x^2}}}dx$

Take

$x=sin\theta \space \Rightarrow dx=cos\theta d\theta$

$\Rightarrow \int{\dfrac{sin\theta cos\theta}{\sqrt{1-sin^2\theta}}}d\theta+\int{\dfrac{sin^2\theta cos\theta }{\sqrt{1-sin^2\theta}}}d\theta$

$\Rightarrow \int{\dfrac{sin\theta cos\theta}{cos\theta}}d\theta+\int{\dfrac{sin^2\theta cos\theta }{cos\theta}}d\theta$

$\Rightarrow \int{sin\theta}d\theta+\int{sin^2\theta }d\theta$

$\Rightarrow \int{sin\theta}d\theta+\dfrac{1}{2}\int{2sin^2\theta }d\theta$

$\Rightarrow -cos\theta+\dfrac{1}{2}\int{(1-cos2\theta) }d\theta$

$\Rightarrow -cos\theta+\dfrac{1}{2}\int{1}d\theta -\dfrac{1}{2}\int{cos2\theta}d\theta$

$\Rightarrow -cos\theta+\dfrac{1}{2}\theta -\dfrac{1}{4}sin2\theta$

$\Rightarrow -\sqrt{1-x^2}+\dfrac{1}{2}sin^{-1}x-\dfrac{1}{2}x\sqrt{1-x^2}$

$\Rightarrow -\sqrt{1-x^2}(1+\dfrac{x}{2})+\dfrac{1}{2}sin^{-1}x$

Solve $\displaystyle\int{\dfrac{6{u}^{3}du}{1+u}}$

1410470_1064100_ans_c62b933cf5ee41d388a1246da58dd725.jpg

$\int \frac{dx}{5+4Cosx}$

1321153_1078495_ans_6ed905cfa6984c6cae78f387a57e58a8.jpeg

$\displaystyle\int\sqrt{a^2-x^2}dx$ .

1321150_1078023_ans_296881e1f313444e93c34cbe0692ab74.jpg

Solve $\int { \dfrac { \sin ^{ -1 }{ \sqrt { x } } -\cos ^{ -1 }{ \sqrt { x } } }{ \sin ^{ -1 }{ \sqrt { x } } +\cos ^{ -1 }{ \sqrt { x } } } } dx$

$\int{\dfrac{\sin^{-1}\sqrt x-\cos ^{-1}\sqrt x}{\sin^{-1}\sqrt x+\cos^{-1}\sqrt x}}dx$

We know that

$\sin^{-1}\sqrt x+\cos ^{-1}\sqrt x=\dfrac{\pi}{2}$ .......

$(1)$

Also,

$\cos ^{-1}\sqrt x=\dfrac{\pi}{2}-\sin ^{-1}\sqrt x$ .....

$(2)$

Using

$(1)$ and

$(2)$ we get

$\Rightarrow \int{\dfrac{\sin^{-1}\sqrt x-(\dfrac{\pi}{2}-\sin ^{-1}\sqrt x)}{\pi/2}}dx$

$\Rightarrow \dfrac{2}{\pi}\int{(2\sin^{-1}\sqrt x-\dfrac{\pi}{2}})dx$

$\Rightarrow \dfrac{4}{\pi}\int{\sin^{-1}\sqrt x}\space dx-\int 1\space dx$

Let

$x=\sin^2\theta\space \Rightarrow dx=2\sin \theta \cos \theta d\theta$

$\Rightarrow \dfrac{4}{\pi}\int{\sin^{-1}( \sin \theta)2\sin \theta \cos \theta }d\theta-x+C$

$\Rightarrow \dfrac{4}{\pi}\int{ \theta \sin 2 \theta }d\theta-x+C$

$\Rightarrow \dfrac{4}{\pi}[\dfrac{-\theta \cos 2 \theta }{2}+\int {1\times \dfrac{\cos 2\theta}{2}}d\theta ]-x+C$

Integrate by parts

$\Rightarrow \dfrac{4}{\pi}[\dfrac{-\theta \cos 2 \theta }{2}+\dfrac{\sin 2\theta}{4} ]-x+C$

$\Rightarrow \dfrac{4}{4\times \pi}[-2\sin^{-1}\sqrt x(1-2x)+2\sqrt x \sqrt {1-x}]-x+C$

$\Rightarrow \dfrac{2}{\pi}[\sqrt {x-x^2}-(1-2x)\sin^{-1}\sqrt x]-x+C$

Evaluate:
$\displaystyle\int { \left( { e }^{ a\ln { x } }+{ e }^{ x\ln { a } }+\sin { \alpha } \right) dx }$

$\displaystyle\int{\left({e}^{a\ln{x}}+{e}^{x\ln{a}}+\sin{\alpha}\right)dx}$

$=\displaystyle\int{{e}^{a\ln{x}}dx}+\displaystyle\int{{e}^{x\ln{a}}dx}+\displaystyle\int{\sin{\alpha}dx}$

$=\displaystyle\int{{e}^{\ln{{x}^{a}}}dx}+\displaystyle\int{{e}^{\ln{{a}^{x}}}dx}+\displaystyle\int{\sin{\alpha}dx}$

$=\displaystyle\int{{x}^{a}dx}+\displaystyle\int{{a}^{x}dx}+\sin{\alpha}\displaystyle\int{dx}$

$=\dfrac{{x}^{a+1}}{a+1}+\dfrac{{a}^{x}}{\ln{a}}+x\sin{\alpha}+c$

Solve the problem:-
$\int {x{{\sin }^{ - 1}}x\,dx}$

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

Let

$u={\sin}^{-1}{x}\Rightarrow\,du=\dfrac{dx}{\sqrt{1-{x}^{2}}}$

$I=\dfrac{{x}^{2}}{2}{\sin}^{-1}{x}-\displaystyle\int{\dfrac{{x}^{2}}{2\sqrt{1-{x}^{2}}}dx}$ [using ILATE rule]

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

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

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

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

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

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

$\displaystyle\int{\sqrt{1-{x}^{2}}dx}=\dfrac{x}{2}\sqrt{1-{x}^{2}}+\dfrac{1}{2}{\sin}^{-1}{x}$

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

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

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

Integrate the following expression with respect to $x$ . $\dfrac{1}{x^{4}+x^{2}+1}$

We have,

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

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

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

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

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

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

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

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

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

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

Hence, this is the answer.

Integrate with respect to $x$ :
$e^{x}(\sec^{2}x+\tan x)$

Consider the given integral.

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

$I=\int{{{e}^{x}}{{\sec }^{2}}}xdx+\int{{{e}^{x}}\tan xdx}$

$I={{e}^{x}}\tan x-\int{{{e}^{x}}\tan xdx}+\int{{{e}^{x}}\tan xdx}$

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

Hence, this is the answer.

Find the integral of $\frac{1}{{\sqrt {{a^2} - {x^2}} }}$ with respect to x and hence evaluate $\int {\frac{{dx}}{{\sqrt {7 - 6x - {x^2}} }}}$

2042670_1083209_ans_d2c6c2985a9449fa910e7d2a596cf95f.png

$\displaystyle \int(\cot x+x)\cot^{2}x.dx$

$\int { (x+cotx){ cot }^{ 2 } } xdx$

Writing

${ cot }^{ 2 }x={ csc }^{ 2 }x-1$

$I=\int { (x+cotx) } { csc }^{ 2 }xdx-\int { (x+cotx)dx }$

$I=\int { { xcsc }^{ 2 }xdx } +\int { cotx{ csc }^{ 2 }xdx } -\int { (x+cotx)dx }$

Integrating the first by parts

$I=x(−cotx)+\int { cotxdx-\frac { { cot }^{ 2 }x }{ 2 } -\frac { { x }^{ 2 } }{ 2 } -\int { cotxdx } }$

$I=-xcotx-\frac { { cot }^{ 2 }x }{ 2 } -\frac { { x }^{ 2 } }{ 2 } +C$

Evaluate:
$\displaystyle\int \dfrac{dx}{\sqrt{1+4x^2}}$ .

Taking

$\frac{1}{2}$ common factor

$\begin{array}{l} \frac { 1 }{ 2 } \int { \frac { { dx } }{ { \sqrt { \left( { \frac { 1 }{ 4 } +{ x^{ 2 } } } \right) } } } } \, \, \, \, \, \, \, \, \, \left[ { \, \int { \frac { { dx } }{ { \sqrt { { a^{ 2 } }+{ x^{ 2 } } } } } =\log \, \left| { x+\sqrt { { a^{ 2 } }+{ x^{ 2 } } } } \right| } } \right] \\ \frac { 1 }{ 2 } \int { \frac { { dx } }{ { \sqrt { { { \left( { \frac { 1 }{ 2 } } \right) }^{ 2 } }+{ x^{ 2 } } } } } } \\ \frac { 1 }{ 2 } \log \left| { x+\sqrt { \frac { 1 }{ 4 } +{ x^{ 2 } } } } \right| +C \\ \frac { 1 }{ 2 } \log \left| { x+\sqrt { \frac { { 1+4{ x^{ 2 } } } }{ 2 } } } \right| +C \end{array}$

Solve : $I = \int \dfrac{dx}{\sqrt{3x + 4} - \sqrt{3x + 1}}$

$\displaystyle I=\int \frac{dx}{\sqrt{3x+4}-\sqrt{3x+1}}$

$t=\dfrac{1}{\sqrt{3x+4}-\sqrt{3x+1}}\,\, \dfrac{1}{t}=\sqrt{3x+4}-\sqrt{3x+1}$

$t=\dfrac{\sqrt{3x+4}+\sqrt{3x+1}}{(3x+4-3x-1)}$

$t=\dfrac{1}{3}(\sqrt{3x+4}+\sqrt{3x+1})$

$\Rightarrow 3t+\dfrac{1}{t}=2\sqrt{3x+4}$

$\Rightarrow \left [\dfrac{1}{2}\left[3t+\dfrac{1}{4}\right]\right]^2=3x+4$

$\dfrac{1}{4}\times 2(3t+\dfrac{1}{t})\times (3-\dfrac{1}{t^2})dt=3dx$

$dx=\dfrac{1}{6}\left(\dfrac{3t^2+1}{t}\right)\left(\dfrac{3t^2-1}{t^2}\right)dt$

$\boxed{dx=\dfrac{1}{6}(\dfrac{9t^2-1}{t^3})dt}$

$\displaystyle I=\int t\times \dfrac{1}{6}\dfrac{(9t^4-1)}{t^3}dt$

$\displaystyle =\int \frac{1}{6}\dfrac{(9t^4-1)}{t^2}dt$

$\displaystyle =\dfrac{1}{6}\int 9t^2dt=\dfrac{1}{6}\int \dfrac{1}{t^2}dt$

$=\dfrac{3}{2}\dfrac{t^3}{3}-\dfrac{1}{6}\dfrac{t^{2+1}}{(-2+1)}+C$

$=I=\dfrac{t^3}{2}+\dfrac{1}{6t}+C$

$=I=\boxed{\dfrac{2}{2\sqrt{3x+4}-\sqrt{3x+1}3}}+\dfrac{1}{6}(\sqrt{3x+4}-\sqrt{3x+1})+C$

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

Let $I=\int { { sin }^{ 2 }(x) } dx$

We use the Rule of Integration by Parts,which is,

$\int { uvdx } =u\int { vdx } −\int { \left[ \dfrac { du }{ dx } \int { vdx } \right] } dx$

$u=x,so\dfrac { du }{ dx } =1;v={ cos }^{ 2 }x,so,\int { vdx } =\dfrac { { sin }^{ 2 } }{ { x }^{ 2 } } .$

$I=\dfrac { x }{ 2 } sin2x−\dfrac { 1 }{ 2 } \int { sin2xdx } =\dfrac { x }{ 2 } sin2x-$ $\dfrac { 1 }{ 2 } \int { \dfrac { -cos2x }{ 2 } }$ $=\dfrac{1}{4} 2x.sin2x+cos2x+c$

Evaluate the following:
$\int x \sin ^{2} x.dx$ ?

We start by integrating the function ${ sin }^{ 2 }(x)$

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

so then you do the original integral by party.you take:

$u=x$

$u=1$

$v={ sin }^{ 2 }(x)=\dfrac { 1 }{ 2 } (x-sin{ (2x) }^{ 2 })$

we know that $\int { u.v } =uv−\int { uv }$

$\int { { sin }^{ 2 }(x) } dx={ x }^{ 2 }(x-sin{ (2x) }^{ 2 }-\int { \dfrac { 1 }{ 2 } (x-sin{ (2x) }^{ 2 }) } =\dfrac { x }{ 2 } \left( \dfrac { x−sin(2x) }{ 2 } \right) -\dfrac { 1 }{ 2 } \left( { x }^{ 2 }+\dfrac { cos(2x) }{ 4 } \right)$

Which then can be simplified to

$\dfrac { { x }^{ 2 } }{ 4 } -x\dfrac { sin(2x) }{ 4 } -\dfrac { cos(2x) }{ 8 }$

$\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{x\,dx}{1+\sin{x}}dx}$

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

$=\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{x\,dx}{1+\sin{x}}}$ ......

$(1)$

Replace

$x\rightarrow \dfrac{\pi}{4}+\dfrac{3\pi}{4}-x=\pi-x$

$= \displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{\left(\pi-x\right)\,dx}{1+\sin{\left(\pi-x\right)}}}$

$= \displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{\left(\pi-x\right)\,dx}{1+\sin{x}}}$ ......

$(1)$

Adding

$(1)$ and

$(2)$ we get

$2I=\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{x\,dx}{1+\sin{x}}}+\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{\left(\pi-x\right)\,dx}{1+\sin{x}}}$

$2I=\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{\left(x+\pi-x\right)dx}{1+\sin{x}}}$

$2I=\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{\left(\pi\right)dx}{1+\sin{x}}}$

$2I=\pi\displaystyle\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\dfrac{dx}{1+\sin{x}}}$

Let

$t=x-\dfrac{\pi}{2}$

$2I=\pi\displaystyle\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}{\dfrac{dt}{1+\cos{t}}}$

$\Rightarrow 2I=2\pi\displaystyle\int_{0}^{\frac{\pi}{4}}{\dfrac{dt}{1+\cos{t}}}$

$\Rightarrow 2I=2\pi\displaystyle\int_{0}^{\frac{\pi}{4}}{{\sec}^{2}{\dfrac{t}{2}}dt}$

$\Rightarrow 2I=2\pi\left[\tan{\dfrac{t}{2}}\right]_{0}^{\frac{\pi}{4}}$

$\Rightarrow 2I=2\pi\tan{\dfrac{\pi}{8}}$

$\Rightarrow 2I=2\pi\left(\sqrt{2}-1\right)$

$\therefore I=\pi\left(\sqrt{2}-1\right)$

Integrate:
$\displaystyle \int \cos x \log \cos x\ dx$

The equation of the value is,

$u=ln(cos(x)),$

$du=\frac { -sin(x) }{ cos(x) }$

$dv=cos(x),$ ,

$v=sin(x)I$

Therefore

$u*v-int(v*du)$

$=ln(cos(x))*sin(x)-sin(x)+ln(sec(x)+tan(x))$

now evaluating at

$\pi/2$ weget,

$=lncos(\frac { \pi }{ 2 } )-1+lnsec(\frac { \pi }{ 2 } )+lntan(\frac { \pi }{ 2 } )$

$=\frac { lncos(\frac { \pi }{ 2 } )-1+ln(1+sin)(\frac { \pi }{ 2 } ) }{ (cos\frac { \pi }{ 2 } ) }$

$=ln(1+sin(\frac { \pi }{ 2 } ))-1=ln(2)-1$

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

We are write denominator as

$1+x+{x}^{2}=1+\dfrac{1}{4}+2. \dfrac{1}{2}.x+{x}^{2}-\dfrac{1}{4}=\dfrac{3}{4}+{\left(\dfrac{1}{2}+x\right)}^{2}$

$\therefore \displaystyle \int{\dfrac{1}{1+x+{x}^{2}}}dx=\displaystyle \int{\dfrac{1}{\dfrac{3}{4}+{\left(x+\dfrac{1}{2}\right)}^{2}}}dx$

$=\displaystyle \int { \dfrac { 1 }{ \dfrac { 3 }{ 4 } \left[ 1+\dfrac { 4 }{ 3 } { \left( x+\dfrac { 1 }{ 2 } \right) }^{ 2 } \right] } } dx=\dfrac { 4 }{ 3 } \displaystyle \int { \dfrac { 1 }{ 1+{ \left[ \dfrac { 2 }{ \sqrt { 3 } } +\left( x+\dfrac { 1 }{ 2 } \right) \right] }^{ 2 } } } dx$

$=\dfrac { 4 }{ 3 } \displaystyle \int { \dfrac { 1 }{ 1+{ \left( \dfrac { 2 }{ \sqrt { 3 } } x+\dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 } } } dx=\dfrac { 4 }{ 3 } .\dfrac { \sqrt { 3 } }{ 2 } \displaystyle \int { \dfrac { \dfrac { 2 }{ \sqrt { 3 } } }{ 1+{ \left( \dfrac { 2x }{ \sqrt { 3 } } x+\dfrac { 1 }{ \sqrt { 3 } } \right) }^{ 2 } } } dx$

$=\dfrac { 2 }{ \sqrt { 3 } } arc \tan \left( \dfrac { 2x }{ \sqrt { 3 } } x+\dfrac { 1 }{ \sqrt { 3 } } \right) =\dfrac { 2 }{ \sqrt { 3 } } arc \tan \left( \dfrac { 2x+1 }{ \sqrt { 3 } } \right)$ Ans.

Solve $\int {\dfrac { 1-{ x }^{ 7 } }{ x\left( 1+{ x }^{ 7 } \right) } } dx$

$\displaystyle \int \frac{1-x^{7}}{x(1+x^{7})}dx = \int \frac{1+x^{7}-2x^{7}}{x(1+x^{7})}dx$

$=\displaystyle \int \frac{(4+x^{7})}{x(1+x^{7})}dx-2\int \frac{x^{6}}{(1+x^{7})}dx$

$\displaystyle =\int \frac{dx}{x}-2 \int \frac{x^{6}}{1+x^{7}}dx$

$\displaystyle lnx-2\int \frac{x^{6}}{1+x^{7}}dx$

$1+x^{7}=t$

$7x^{6}dx=dt$

$\displaystyle = lnx = \frac{2}{7}\int \frac{dt}{t}$

$\displaystyle = lnx = \frac{2}{7}lnt +C$

$\displaystyle = lnx - \frac{2}{7} ln (1+x^{7})+C = lnx - ln (1+x^{7})^{2/7}+C$

$= ln \left(\dfrac{x}{(1+x^{7})^{2/7}}\right)+C$

Evaluate: $\int \frac { t ^ { 4 } d t } { \sqrt { 1 - t ^ { 2 } } }$

$\int \cfrac{t^{4}dt}{\sqrt{1-t^{2}}}$

Say

$t=\sin\theta \Rightarrow dt=\cos\theta d\theta$

$=\int \cfrac{\sin^{4}\theta}{\sqrt{1-\sin^{2}\theta}}\cos\theta d\theta$

$=\int\cfrac{\sin^{4}\theta}{\sqrt \cos^{2}\theta}\cdot \cos\theta d\theta=\int \cfrac{\sin^{4}\theta}{\cos\theta}\cos\theta d\theta$

$=\int \sin^{4}\theta d\theta$

$=\cfrac{1}{4}\int (2\sin^{2}\theta)^2d\theta=\cfrac{1}{4}\int(1-\cos 2\theta)^{2}d\theta$

$=\cfrac{1}{4}[\int (1+\cos^{2}2\theta-2\cos2\theta)d\theta]$

$=\cfrac{1}{4}[\int d\theta+\int \cos^{2}2\theta d\theta-2\int\cos 2\theta d\theta]$

$=\cfrac{1}{4}[\int d\theta+\cfrac{1}{2}\int 2\cos^{2}2\theta d\theta-2\int\cos 2\theta d\theta]$

$=\cfrac{1}{4}[\int d\theta+\cfrac{1}{2}(1+\cos 4\theta)d\theta-2\int \cos 2\theta d\theta]$

$=\cfrac{!}{4}[\int d\theta+\cfrac{1}{2}\int d\theta+\cfrac{1}{2}\cos 4\theta d\theta-2\int \cos 2\theta d\theta]$

$=\cfrac{1}{4}[\theta+\cfrac{\theta}{2}+\cfrac{\sin 4\theta}{8}-\sin 2\theta]$

$=\cfrac{3\theta}{8}+\cfrac{\sin 4\theta}{32}-\cfrac{\sin 2\theta}{4}$

$=\cfrac{3\sin^{-1}t}{8}+\cfrac{4\sin\theta \cdot \cos\theta\cdot(1-2\sin^{2}\theta)}{32}-\cfrac{2\sin\theta\cdot \cos\theta}{4}$

$=\cfrac{3}{8}\sin^{-1}t+\cfrac{1}{8}t\sqrt{1-t^{2}}(1-2+2t^{2})-\cfrac{1}{2}t\sqrt{1-t^{2}}$

$=\cfrac{3}{8}\sin^{-1}t+\cfrac{1}{8}t\sqrt{1-t^{2}}(2t^{2}-1)-\cfrac{1}{2}t\sqrt{1-t^{2}}$

Evaluate $\int e ^ { 2 x } \sin x d x$ .

$I=\int e^{2x}sin x dx =?$

$I= -e^{2x} cos x+ \int e^{2x}(2) cos x dx$

$I=-e^{2x}cos x+ 2 e^{2x} sin x- \int 4e^{2x} sin x dx$

$I= -e^{2x} cos x+ 2 e^{2x} sin x- 4 I$

$5 I = -e^{2x} cos x+2e^{2x} sin x$

$I=\frac{1}{5}[2e^{2x} sin x-e^{2x} cos x]=\frac{1}{5} e^{2x}[2 sin x- cos x]$

Evaluate :
$\displaystyle \int \dfrac{x^{1/2}}{1+x^{3/4}}dx$

1960975_1211342_ans_df74719557ba492d8fde7f6104d08caf.jpg

Evaluate: $\displaystyle \int x \sin 3 x d x$

Given,

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

Let

$\displaystyle I=\int x \, \sin \, 3x \, dx$

$\Rightarrow$ integration by part

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$\displaystyle I= x \int \sin \, 3x - \int 1 \cdot \int \sin \, 3x dx \cdot dx$

$\displaystyle = \dfrac {x(- \cos 3x)}{3}+\int \dfrac {(\cos \, 3x)}{3}dx$

$I\displaystyle = -\dfrac {x}{3} \cos 3x+\dfrac {\sin \, 3x}{9}+C$

$\int { x } e ^ { 2 x } d x$

${\textbf{Step - 1: Apply integration by parts rule to find the integration of the function.}}$

$\therefore \int {x{e^{2x}}dx} .$

$= x\int {{e^{2x}}dx - \int {\left[ {\dfrac{d}{{dx}}x\int {{e^{2x}}dx} } \right]} } dx.$

$= x\dfrac{{{e^{2x}}}}{2} - \int {\left[ {\dfrac{{{e^{2x}}}}{2}} \right]dx} .$

${\textbf{Step - 2: Simplify the equation and deduce integration of }}\mathbf{e^2x .}$

$\therefore x\dfrac{{{e^{2x}}}}{2} - \dfrac{1}{4}\int {{e^{2x}}} d\left( {2x} \right).$

$= x\dfrac{{{e^{2x}}}}{2} - \dfrac{1}{4}{e^{2x}} + c{\text{ }}\left[ {\ c = {\text{integrating constant}}} \right].$

$= {e^{2x}}\left( {\dfrac{x}{2} - \dfrac{1}{4}} \right) + c.$

${\textbf{Hence , the value will be }}\mathbf{{e^{2x}}\left( {\dfrac{x}{2} - \dfrac{1}{4}} \right) + c.}$

Evaluate $\int \dfrac {\sin \theta}{\sin 3\theta} d\theta$ .

$\int { \cfrac { \sin { \theta } }{ \sin { 3\theta } } d\theta }$

$=\int { \cfrac { 1 }{ 3-4\sin ^{ 2 }{ \theta } } d\theta } =\int { \cfrac { 1 }{ 3\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } } d\theta } =\int { \cfrac { 1 }{ 2\cos { 2\theta } } d\theta }$

$=\cfrac { 1 }{ 2 } \int { \cfrac { 1 }{ 2\cos { x } +1 } dx } \left[ 2\theta =x;2d\theta =dx \right]$

We know

$\int { \cfrac { 1 }{ 1+a\cos { x } } dx } =\cfrac { -1 }{ \sqrt { { a }^{ 2 }-1 } } \log { \left| \cfrac { a+\cos { x } +\sqrt { { a }^{ 2 }-1 } \sin { x } }{ 1+a\cos { x } } \right| } +c$

$\Rightarrow \cfrac { 1 }{ 2 } \int { \cfrac { 1 }{ 1+2\cos { \theta } } d\theta } =\cfrac { -1 }{ 2\sqrt { 4-1 } } \log { \left| \cfrac { 2+\cos { 2\theta } +\sqrt { 4-1 } \sin { 2\theta } }{ 1+2\cos { 2\theta } } \right| } +c$

$=\cfrac { -1 }{ 2\sqrt { 3 } } \log { \left| \cfrac { 2+\cos { 2\theta } +\sqrt { 3 } \sin { 2\theta } }{ 1+2\cos { 2\theta } } \right| } +c$

Solve: $\displaystyle \int {\dfrac{{{x^2} - 1}}{{x\sqrt {1 + {x^4}} }}\;{\text{dx}}}$

$\begin{array}{l} \int { \frac { { { x^{ 2 } }-1 } }{ { x\sqrt { 1+{ x^{ 4 } } } } } dx } \\ \Rightarrow \int { \frac { x }{ { \sqrt { 1+{ x^{ 4 } } } } } } \, \, -\int { \frac { 1 }{ { x\sqrt { 1+{ x^{ 4 } } } } } dx } \\ \Rightarrow \frac { 1 }{ 2 } \int { \frac { { dt } }{ { \sqrt { 1+{ t^{ 2 } } } } } } -\frac { 1 }{ 2 } \int { \frac { 1 }{ { t\sqrt { 1+{ t^{ 2 } } } } } } dt \\ \Rightarrow \frac { 1 }{ 2 } \log \left| { t+\sqrt { { t^{ 2 } }+1 } } \right| \frac { { -1 } }{ 2 } \int { t\frac { 1 }{ { \sqrt { \left( { 1+{ t^{ 2 } } } \right) } } } dt } \\ let,-\sqrt { 1+{ t^{ 2 } } } =P \\ \Rightarrow \frac { 1 }{ { 2\sqrt { 1+{ t^{ 2 } } } } } .2t\, dt=dp \\ \Rightarrow \frac { 1 }{ 2 } \log \left| { { x^{ 2 } }+\sqrt { { x^{ 4 } }+1 } } \right| \frac { { -1 } }{ 2 } \int { \frac { 1 }{ { \left( { { P^{ 2 } }-1 } \right) } } dp } \\ \Rightarrow \frac { 1 }{ 2 } \log \left| { { x^{ 2 } }+\sqrt { 1+{ x^{ 4 } } } -\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \log } \right| \left. { \frac { { P-1 } }{ { P+1 } } } \right| +c \\ \Rightarrow \frac { 1 }{ 2 } \log \left| { { x^{ 2 } }+\sqrt { 1+{ x^{ 4 } } } } \right| -\frac { 1 }{ 4 } \log \left| { \frac { { P-1 } }{ { P+1 } } } \right| +c \\ =\frac { 1 }{ 2 } \log \left| { { x^{ 2 } }+\sqrt { { x^{ 4 } }+1 } -\frac { 1 }{ 4 } \log } \right| \left. { \frac { { \sqrt { 1+{ x^{ 4 } } } -1 } }{ { \sqrt { 1+{ x^{ 4 } } } +1 } } } \right| . \end{array}$

Evaluate: $I=\displaystyle \int x\sin{2x}\ dx$

$I=\displaystyle\int x\sin 2xdx$

Let

$U=x$ ; Let

$V'=\sin 2x$

$U'=1$ ;

$V=\dfrac{-\cos 2x}{2}$

$\therefore \displaystyle\int x\sin 2xdx$

$UV-\displaystyle\int U'V$

$=\dfrac{-1}{2}x\cos 2x-\displaystyle 1\cdot \dfrac{-\cos 2x}{2}$

$=\dfrac{-1}{2}x\cos 2x+\dfrac{1}{2}\displaystyle\int \cos 2x$

$=\dfrac{-1}{2}x\cos 2x+\dfrac{1}{4}\sin 2x+c$ .

1201605_1280372_ans_1076e68d8f324257a1fdc77457665892.jpg

Evaluate
$\int x^{2} \tan^{-1} xdx$

$\begin{array}{l} \int { { x^{ 2 } }{ { \tan }^{ -1 } }dx } =\frac { { { x^{ 3 } } } }{ 3 } { \tan ^{ -1 } }x-\int { \frac { 1 }{ { 1+{ x^{ 2 } } } } \times \frac { { { x^{ 3 } } } }{ 3 } dx } \\ { \tan ^{ -1 } }x\left( { \frac { { { x^{ 3 } } } }{ 3 } } \right) -\frac { 1 }{ 3 } \int { \frac { { { x^{ 3 } } } }{ { 1+{ x^{ 2 } } } } dx } \\ 1+{ x^{ 2 } }=t\, \, \, \, \, \, 2dx=dt \\ dx=\frac { { dt } }{ { 2x } } \\ { \tan ^{ -1 } }\left( { \frac { { { x^{ 3 } } } }{ 3 } } \right) -\frac { 1 }{ 3 } \int { \frac { { { x^{ 3 } } } }{ t } dt } \\ { \tan ^{ -1 } }\left( { \frac { { { x^{ 3 } } } }{ 3 } } \right) -\frac { 1 }{ 6 } \left( { 1+{ x^{ 2 } }-\log \left| { 1+{ x^{ 2 } } } \right| +c } \right) \end{array}$

Evaluate:
$\displaystyle \int{\sqrt{\dfrac{a+x}{a-x}}}dx$ .

$\begin{array}{l} \int { \sqrt { \frac { { a+x } }{ { a-x } } } dx } \\ x=a\cos 2\theta \\ dx=-2\sin 2\theta d\theta \\ \sqrt { \frac { { a+a\cos 2\theta } }{ { a-a\cos 2\theta } } } \left( { -2a\sin 2\theta d\theta } \right) \\ \sqrt { \frac { { a\left( { 1+\cos 2\theta } \right) } }{ { a\left( { 1-\cos 2\theta } \right) } } } \left( { -2asin2\theta d\theta } \right) \\ \sqrt { \frac { { \cos 2\theta } }{ { \sin 2\theta } } } \left( { -2\sin 2\theta d\theta } \right) \\ \int { \cot \theta } \left( { -2a\sin 2\theta d\theta } \right) \\ =-2a\int { 2\cos 2\theta d\theta } \\ =-2a\int { \left( { 1+\cos 2\theta } \right) d\theta } \\ =-2a\left( { \theta +\frac { { \sin 2\theta } }{ 2 } } \right) \\ =-2a\theta -a\sin 2\theta \\ =-a{ \cos ^{ -1 } }\frac { x }{ a } -\sqrt { { a^{ 2 } }-{ a^{ 2 } }{ { \cos }^{ 2 } }2\theta } \\ =-a{ \cos ^{ -1 } }\frac { x }{ a } -\sqrt { { a^{ 2 } }-{ x^{ 2 } } } +c \end{array}$

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

$\int { \frac { 1 }{ { \left( { x+1 } \right) } } .\frac { 1 }{ { \left( { { x^{ 2 } }-1 } \right) } } \, \, dx } \\ \Rightarrow \int { \frac { 1 }{ { { { \left( { x+1 } \right) }^{ 2 } } } } .\frac { 1 }{ { \left( { x-1 } \right) } } .dx } \\ Let \\ \Rightarrow \frac { 1 }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x-1 } \right) } } =\frac { A }{ { \left( { x-1 } \right) } } +\frac { B }{ { \left( { x+1 } \right) } } +\frac { C }{ { { { \left( { x+1 } \right) }^{ 2 } } } } \\ \Rightarrow \frac { 1 }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x-1 } \right) } } =\frac { { { { \left( { x+1 } \right) }^{ 3 } }A+B{ { \left( { x+1 } \right) }^{ 2 } }\left( { x-1 } \right) +C\left( { { x^{ 2 } }-1 } \right) } }{ { \left( { x-1 } \right) { { \left( { x+1 } \right) }^{ 2 } } } } \\ \Rightarrow \frac { 1 }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x-1 } \right) } } =\frac { { { x^{ 3 } }A+A+3{ x^{ 2 } }A+3Ax+{ x^{ 3 } }+B{ x^{ 2 } }-Bx+C{ x^{ 2 } }-C } }{ { { { \left( { x+1 } \right) }^{ 2 } }\left( { x-1 } \right) } } \\ \Rightarrow \left( { A+B } \right) { x^{ 3 } }=0\, \, \, \to \left( i \right) \\ \Rightarrow A-B-C=1\, \, \to \left( { ii } \right) \\ \Rightarrow 3A-B=0\, \, \to \left( { iii } \right) \\ \Rightarrow 3A+B+C=0\, \, \to \left( { iv } \right) \\ \Rightarrow adding\, \, equation\, \, \left( { ii } \right) \, \, and\, \, \left( { iv } \right) ,\, \, \\ \Rightarrow 4A=1,\, \, A=\frac { 1 }{ 4 } ,\, \, B=\frac { 3 }{ 4 } \, \, and\, \, C=\frac { { -3 } }{ 4 } \\ \Rightarrow \frac { 1 }{ 4 } \int { \frac { 1 }{ { \left( { x-1 } \right) } } \, \, dx+\frac { 3 }{ 4 } \int { \frac { 1 }{ { \left( { x+1 } \right) } } \, \, dx\frac { { -3 } }{ 4 } \int { \frac { 1 }{ { { { \left( { x+1 } \right) }^{ 2 } } } } \, \, dx } } } \\ \Rightarrow \frac { 1 }{ 4 } \, \, \, \log \left( { x-1 } \right) +\frac { 3 }{ 4 } \, \, \log \left( { x+1 } \right) +\frac { 3 }{ 4 } \frac { 1 }{ { { { \left( { x+1 } \right) }^{ 2 } } } } +C\, \, \, \, Ans. \\$

Evaluate the given integral.
$\int { \sec ^{ -1 }{ \sqrt { x } } } dx$

$I=\displaystyle\int{{\sec}^{-1}{\sqrt{x}}dx}$

Let

$t=\sqrt{x}\Rightarrow\,dt=\dfrac{1}{2\sqrt{x}}dx\Rightarrow\,dx=2t\,dt$

$=\displaystyle\int{{\sec}^{-1}{t}2t\,dt}$

$=2\displaystyle\int{t{\sec}^{-1}{t}\,dt}$

Let

$u={\sec}^{-1}{t}\Rightarrow\,du=\dfrac{1}{t\sqrt{{t}^{2}-1}}$

$dv=t\,dt\Rightarrow\,v=\dfrac{{t}^{2}}{2}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$=2\left[\dfrac{{t}^{2}{\sec}^{-1}{t}}{2}-\displaystyle\int{\dfrac{{t}^{2}}{2}\times\dfrac{1}{t\sqrt{{t}^{2}-1}}dt}\right]$

$={t}^{2}{\sec}^{-1}{t}-\displaystyle\int{\dfrac{t\,dt}{\sqrt{{t}^{2}-1}}}$

$={t}^{2}{\sec}^{-1}{t}-\dfrac{1}{2}\displaystyle\int{\dfrac{2t\,dt}{\sqrt{{t}^{2}-1}}}$

Let

$w={t}^{2}-1\Rightarrow\,dw=2t\,dt$

$={t}^{2}{\sec}^{-1}{t}-\dfrac{1}{2}\displaystyle\int{{w}^{-\frac{1}{2}}dw}$

$=x{\sec}^{-1}{\sqrt{x}}-\dfrac{1}{2}\displaystyle\int{{w}^{-\frac{1}{2}}dw}$ ......where

$t=\sqrt{x}$

$=x{\sec}^{-1}{\sqrt{x}}-\dfrac{1}{2}\dfrac{{w}^{-\frac{1}{2}+1}}{\dfrac{-1}{2}+1}+c$

$=x{\sec}^{-1}{\sqrt{x}}-\dfrac{1}{2}\dfrac{{w}^{\frac{1}{2}}}{\dfrac{1}{2}}+c$

$=x{\sec}^{-1}{\sqrt{x}}-\sqrt{w}+c$

$=x{\sec}^{-1}{\sqrt{x}}-\sqrt{w}+c$ .........where

$w={t}^{2}-1$

$=x{\sec}^{-1}{\sqrt{x}}-\sqrt{{t}^{2}-1}+c$ ......where

$t=\sqrt{x}$

$=x{\sec}^{-1}{\sqrt{x}}-\sqrt{x-1}+c$

Integrate:
$\displaystyle \int{\dfrac{dx}{3x^{2}+13x-10}}$

$\displaystyle\int{\dfrac{dx}{3{x}^{2}+13x-10}}$

$=\displaystyle\int{\dfrac{dx}{3{x}^{2}+15x-2x-10}}$

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

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

Consider

$\dfrac{1}{\left(x+5\right)\left(3x-2\right)}=\dfrac{A}{x+5}+\dfrac{B}{3x-2}$

$\Rightarrow 1=A\left(3x-2\right)+B\left(x+5\right)$

Put

$x=-5$

$\Rightarrow 1=A\left(3\times -5-2\right)+0$

$\Rightarrow 1=A\left(-15-2\right)$

$\therefore A=\dfrac{-1}{17}$

Put

$x=0$

$\Rightarrow 1=A\left(0-2\right)+B\left(0+5\right)$

$\Rightarrow -2A+5B=1$

$\Rightarrow -2\times \dfrac{-1}{17}+5B=1$

$\Rightarrow \dfrac{2}{17}+5B=1$

$\Rightarrow 5B=1-\dfrac{2}{17}=\dfrac{17-2}{17}=\dfrac{15}{17}$

$\therefore B=\dfrac{3}{17}$

$\therefore \displaystyle\int{\dfrac{dx}{3{x}^{2}+13x-10}}$

$=\dfrac{-1}{17}\displaystyle\int{\dfrac{dx}{x+5}}+\dfrac{3}{17}\displaystyle\int{\dfrac{dx}{3x-2}}$

$=\dfrac{-1}{17}\log{\left(x+5\right)}+\dfrac{3}{17}\times \dfrac{1}{3}\log{\left(3x-2\right)}$

$=\dfrac{-1}{17}\log{\left(x+5\right)}+\dfrac{1}{17}\log{\left(3x-2\right)}+c$

$=\dfrac{1}{17}\log{\left(\dfrac{3x-2}{x+5}\right)}+c$

Evaluate the given integral.
$\int { { x }^{ 2 }\cos ^{ 2 }{ x } } dx$

$I=\displaystyle\int{{x}^{2}{\cos}^{2}{x}dx}$

$\Rightarrow\,I=\dfrac{1}{2}\displaystyle\int{{x}^{2}\times 2{\cos}^{2}{x}dx}$

We know that

$\cos{2x}=2{\cos}^{2}{x}-1$

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

$\Rightarrow\,I=\dfrac{1}{2}\displaystyle\int{{x}^{2}dx}-\dfrac{1}{2}\displaystyle\int{{x}^{2}\cos{2x}\,dx}$

Let

${I}_{1}=\dfrac{1}{2}\displaystyle\int{{x}^{2}dx}=\dfrac{1}{2}\dfrac{{x}^{3}}{3}+{c}_{1}=\dfrac{1}{6}{x}^{3}+{c}_{1}$

Let

${I}_{2}=\dfrac{1}{2}\displaystyle\int{{x}^{2}\cos{2x}\,dx}$

Integrating by parts, we get

Let

$u={x}^{2}\Rightarrow\,du=2x\,dx$ and

$dv=\cos{2x}\,dx\Rightarrow\,v=\dfrac{1}{2}\sin{2x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

${I}_{2}=\dfrac{1}{2}{x}^{2}\sin{2x}-\displaystyle\int{\dfrac{1}{2}\sin{2x}\times\,2x\,dx}$

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

Let

$u=x\Rightarrow\,du=dx$ and

$dv=\sin{2x}\,dx\Rightarrow\,v=\dfrac{-1}{2}\cos{2x}$

${I}_{2}=\dfrac{1}{2}{x}^{2}\sin{2x}+\dfrac{1}{2}x\cos{2x}-\dfrac{1}{2}\displaystyle\int{\cos{2x}\,dx}$

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

$\therefore\,I={I}_{1}-{I}_{2}$

$\Rightarrow\,I=\dfrac{1}{6}{x}^{3}+{c}_{1}-\dfrac{1}{2}{x}^{2}\sin{2x}-\dfrac{1}{2}x\cos{2x}+\dfrac{1}{4}\sin{2x}-{C}_{2}$

$\Rightarrow\,I=\dfrac{1}{6}{x}^{3}-\dfrac{1}{2}{x}^{2}\sin{2x}-\dfrac{1}{2}x\cos{2x}+\dfrac{1}{4}\sin{2x}+\left({c}_{1}-{C}_{2}\right)$

$\Rightarrow\,I=\dfrac{1}{6}{x}^{3}-{x}^{2}\sin{x}\cos{x}-\dfrac{1}{2}x\cos{2x}+\dfrac{1}{4}\sin{2x}+C$ .......where

$C=\left({c}_{1}-{C}_{2}\right)$

Evaluate the given integral.
$\int { \left( x+1 \right) { e }^{ x }\log { \left( x{ e }^{ x } \right) } } dx$

Let

$t=x{e}^{x}\Rightarrow\,dt=\left(x{e}^{x}+{e}^{x}\right)dx$

$\Rightarrow\,dt=\left(x+1\right){e}^{x}\,dx$

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

$=\displaystyle\int{\log{t}\,dt}$

Integrating by parts, we get

Let

$u=\log{t}\Rightarrow\,du=\dfrac{1}{t}dt$

$dv=dt\Rightarrow\,v=t$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$I=t\log{t}-\displaystyle\int{t\times{\dfrac{1}{t}}dt}$

$=t\log{t}-\displaystyle\int{dt}$

$=t\log{t}-t+c$

$=x{e}^{x}\log{x{e}^{x}}-x{e}^{x}+c$ .........where

$t=x{e}^{x}$

Evaluate: $\displaystyle \int \frac { \log x } { ( 1 + \log x ) ^ { 2 } } d x$

$\int \frac{logx}{(1+logx)^{2}}dx$

adding and subtracting 1 from numerator

$\int \frac{1-1+log x}{(1+log x)^{2}}dx$

$\int \frac{1+log x}{(1+log x)^{2}}dx-\int \frac{1}{(1+ log x)^{2}}dx$

$\int \frac{1}{1+log x}dx-\int \frac{1}{(1+ log x)}^{2}dx$

For the integral

$\int \frac{1}{1+log x}dx$

integrate by parts within the sum :

$\int fg'=fg-\int f'g$

$f=\frac{1}{1+log x}dx, g'=1$

$f'=-\frac{1}{(1+ log x)^{2}}g=x$

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

$=\frac{x}{log (x)+1}$

1211362_1505696_ans_2c550abed16b46ff90c579cfa53e20f2.png

Prove that:
$\displaystyle \int \dfrac {x^{2}dx}{(x\sin x+\cos x)^{2}}$

Let

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

$=\displaystyle\int{\left(x\sec{x}\right)\left(\dfrac{x\cos{x}}{{\left(x\sin{x}+\cos{x}\right)}^{2}}\right)dx}$

Using the rule of integration by parts, we get

$\dfrac{d}{dx}\left(\dfrac{1}{x\sin{x}+\cos{x}}\right)$

$=\dfrac{-1}{{\left(x\sin{x}+\cos{x}\right)}^{2}}\dfrac{d}{dx}\left(x\sin{x}+\cos{x}\right)$

$=\dfrac{-1}{{\left(x\sin{x}+\cos{x}\right)}^{2}}\left(x\cos{x}+\sin{x}-\sin{x}\right)$

$=\dfrac{-1}{{\left(x\sin{x}+\cos{x}\right)}^{2}}\left(x\cos{x}\right)$

$=\dfrac{-x\cos{x}}{{\left(x\sin{x}+\cos{x}\right)}^{2}}$

$\displaystyle\int{\dfrac{x\cos{x}}{{\left(x\sin{x}+\cos{x}\right)}^{2}}dx}=\dfrac{-1}{x\sin{x}+\cos{x}}$ ..........

$(1)$

Also,

$\dfrac{d}{dx}\left(x\sec{x}\right)=x\sec{x}\tan{x}+\sec{x}$

$=x\times\dfrac{1}{\cos{x}}\times\dfrac{\sin{x}}{\cos{x}}+\dfrac{1}{\cos{x}}$

$\Rightarrow \dfrac{d}{dx}\left(x\sec{x}\right)=\dfrac{x\sin{x}+\cos{x}}{{\cos}^{2}{x}}$ .........

$(2)$

Now, in integration by parts, we take,

$u=x\sec{x}$ and

$dv=\dfrac{x\cos{x}}{{\left(x\sin{x}+\cos{x}\right)}^{2}}$

$\therefore du=\dfrac{x\sin{x}+\cos{x}}{{\cos}^{2}{x}}$ and

$v=\dfrac{-1}{x\sin{x}+\cos{x}}$

$\therefore I=\left(x\sec{x}\right)\left(\dfrac{-1}{x\sin{x}+\cos{x}}\right)-\displaystyle\int{\dfrac{x\sin{x}+\cos{x}}{{\cos}^{2}{x}}\times\dfrac{-1}{x\sin{x}+\cos{x}}dx }$

$=\dfrac{-x}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+\displaystyle\int{{\sec}^{2}{x}dx}$

$=\dfrac{-x}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+\tan{x}$

$=\dfrac{-x}{\cos{x}\left(x\sin{x}+\cos{x}\right)}+\dfrac{\sin{x}}{\cos{x}}$

$=\dfrac{-x+\sin{x}\left(x\sin{x}+\cos{x}\right)}{\cos{x}\left(x\sin{x}+\cos{x}\right)}$

$=\dfrac{-x+x{\sin}^{2}{x}+\sin{x}\cos{x}}{\cos{x}\left(x\sin{x}+\cos{x}\right)}$

$=\dfrac{-x\left(1-{\sin}^{2}{x}+\sin{x}\cos{x}\right)}{\cos{x}\left(x\sin{x}+\cos{x}\right)}$

$=\dfrac{-x{\cos}^{2}{x}+x\sin{x}\cos{x}}{\cos{x}\left(x\sin{x}+\cos{x}\right)}$

$=\dfrac{\cos{x}\left(-x\cos{x}+x\sin{x}\right)}{\cos{x}\left(x\sin{x}+\cos{x}\right)}$

$=\dfrac{\left(-x\cos{x}+x\sin{x}\right)}{\left(x\sin{x}+\cos{x}\right)}+c$

$\therefore I=\dfrac{\left(-x\cos{x}+x\sin{x}\right)}{\left(x\sin{x}+\cos{x}\right)}+c$

Evaluate the given integral.
$\displaystyle \int { \sin ^{ -1 }{ \left( \cfrac { 2x }{ 1+{ x }^{ 2 } } \right) } } dx\quad \quad$

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

Let

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

$=\displaystyle\int{{\sin}^{-1}{\left(\dfrac{2\tan{\theta}}{1+{\tan}^{2}{\theta}}\right)}{\sec}^{2}{\theta}d\theta}$

We know that

$\sin{2\theta}=\dfrac{2\tan{\theta}}{1+{\tan}^{2}{\theta}}$

$=\displaystyle\int{{\sin}^{-1}{\sin{2\theta}}{\sec}^{2}{\theta}d\theta}$

$=2\displaystyle\int{\theta{\sec}^{2}{\theta}d\theta}$

Let

$u=\theta\Rightarrow\,du=d\theta$

$dv={\sec}^{2}{\theta}d\theta\Rightarrow\,v=\tan{\theta}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$=2\left[\theta\tan{\theta}-\displaystyle\int{\tan{\theta}d\theta}\right]$

$=2\left[\theta\tan{\theta}-\log{\sec{\theta}}\right]+c$

$=2\theta\tan{\theta}-3\log{\sec{\theta}}+c$

We know that

$x=\tan{\theta}\Rightarrow\,{x}^{2}={\tan}^{2}{\theta}$

$\Rightarrow\,1+{x}^{2}=1+{\tan}^{2}{\theta}={\sec}^{2}{\theta}$

$\Rightarrow\,\sec{\theta}=\sqrt{1+{x}^{2}}$

$=2x{\tan}^{-1}{x}-2\log{\left|\sqrt{1+{x}^{2}}\right|}+c$

$=2x{\tan}^{-1}{x}-2\log{\left|{\left(1+{x}^{2}\right)}^{\frac{1}{2}}\right|}+c$

$=2x{\tan}^{-1}{x}-\dfrac{2}{2}\log{\left|1+{x}^{2}\right|}+c$

$=2x{\tan}^{-1}{x}-\log{\left|1+{x}^{2}\right|}+c$

Evaluate the given integral.
$\int { \cos ^{ -1 }{ \left( 4{ x }^{ 3 }-3x \right) } } dx$

$I=\displaystyle\int{{\cos}^{-1}{\left(4{x}^{3}-3x\right)}dx}$

Let

$x=\cos{\theta}\Rightarrow\,dx=-\sin{\theta}d\theta$

$=\displaystyle\int{{\cos}^{-1}{\left(4{\cos}^{3}{\theta}-3\cos{\theta}\right)}-\sin{\theta}d\theta}$

$=\displaystyle\int{{\cos}^{-1}{\cos{3\theta}}-\sin{\theta}d\theta}$

$=-3\displaystyle\int{{\theta}\sin{\theta}d\theta}$

Integrating by parts, we have

$u=\theta\Rightarrow\,du=d\theta$

$dv=\sin{\theta}d\theta\Rightarrow\,v=-\cos{\theta}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$=-3\left[-\theta\cos{\theta}-\displaystyle\int{-\cos{\theta}d\theta}\right]$

$=3\theta\cos{\theta}-3\displaystyle\int{\cos{\theta}d\theta}$

$=3\theta\cos{\theta}-3\sin{\theta}+c$

$=3{\cos}^{-1}{x}\times x-3\sqrt{1-{x}^{2}}+c$ .......since

$x=\cos{\theta}\Rightarrow\,\sqrt{1-{x}^{2}}=\sin{\theta}$

$=3x{\cos}^{-1}{x}-3\sqrt{1-{x}^{2}}+c$

Evaluate the given integral.
$\int {x.{\sin}^{-1}{x}}$

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

Let

$u={\sin}^{-1}{x}\Rightarrow\,du=\dfrac{1}{\sqrt{1-{x}^{2}}}dx$

$dv=x\,dx\Rightarrow\,v=\dfrac{{x}^{2}}{2}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

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

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

Let

$u={x}^{2}\Rightarrow\,du=2x\,dx$

$\Rightarrow\,dv=\dfrac{1}{\sqrt{1-{x}^{2}}}dx\Rightarrow\,v={\sin}^{-1}{x}$

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

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

Let

$t={\sin}^{-1}{x}\Rightarrow\,dt=\dfrac{dx}{\sqrt{1-{x}^{2}}}$

$I=\dfrac{1}{2}\displaystyle\int{t\,dt}$

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

$I=\dfrac{1}{4}{\left({\sin}^{-1}{x}\right)}^{2}+c$ .......where

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

Evaluate the given integral.
$\displaystyle \int { \tan ^{ -1 }{ \left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) } } dx\quad$

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

Let

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

$\Rightarrow\,I=\displaystyle\int{{\tan}^{-1}{\left(\dfrac{2\tan{\theta}}{1-{\tan}^{2}{\theta}}\right)}dx}$

$\Rightarrow\,I=\displaystyle\int{{\tan}^{-1}{\tan{2\theta}}{\sec}^{2}{\theta}d\theta}$

$\Rightarrow\,I=2\displaystyle\int{\theta{\sec}^{2}{\theta}d\theta}$

Let

$u=\theta\Rightarrow\,du=d\theta$

$dv={\sec}^{2}{\theta}d\theta\Rightarrow\,v=\tan{\theta}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$\Rightarrow\,I=2\left[\theta\tan{\theta}-\displaystyle\int{\tan{\theta}d\theta}\right]$

$\Rightarrow\,I=2\left[\theta\tan{\theta}-\log{\left|\sec{\theta}\right|}\right]+c$

We have

$x=\tan{\theta}\Rightarrow\,1+{x}^{2}=1+{\tan}^{2}{\theta}={\sec}^{2}{\theta}$

$\Rightarrow\,\sec{\theta}=\sqrt{1+{x}^{2}}$

$\Rightarrow\,I=2\left[x{\tan}^{-1}{x}-\log{\left|\sqrt{1+{x}^{2}}\right|}\right]+c$

$\Rightarrow\,I=2\left[x{\tan}^{-1}{x}-\dfrac{1}{2}\log{\left|1+{x}^{2}\right|}\right]+c$

$\therefore\,I=2x{\tan}^{-1}{x}-\log{\left|1+{x}^{2}\right|}+c$

Evaluate the given integral.
$\int { \sin ^{ -1 }{ \left( 3x-4{ x }^{ 3 } \right) } } dx$

$I=\displaystyle\int{{\sin}^{-1}{\left(3x-4{x}^{3}\right)}\,dx}$

Let

$x=\sin{\theta}\Rightarrow\,dx=\cos{\theta}d\theta$

$=\displaystyle\int{{\sin}^{-1}{\left(3\sin{\theta}-4{\sin}^{3}{\theta}\right)}\,\cos{\theta}d\theta}$

We know that

$\sin{3\theta}=3\sin{\theta}-4{\sin}^{3}{\theta}$

$=\displaystyle\int{{\sin}^{-1}{\sin{3\theta}}\,\cos{\theta}d\theta}$

$=3\displaystyle\int{{\theta}\cos{\theta}d\theta}$

Integrating by parts,we get

Let

$u=\theta\Rightarrow\,du=d\theta$

$dv=\cos{\theta}d\theta\Rightarrow\,v=\sin{\theta}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$=3\left[\theta\sin{\theta}-\displaystyle\int{\sin{\theta}d\theta}\right]$

$=3\left[\theta\sin{\theta}+\cos{\theta}\right]+c$

$=3\theta\sin{\theta}+3\cos{\theta}+c$

We have

$x=\sin{\theta}\Rightarrow\,\theta={\sin}^{-1}{x}$ and

$\cos{\theta}=\sqrt{1-{\sin}^{2}{\theta}}=\sqrt{1-{x}^{2}}$

$=3x{\sin}^{-1}{x}+3\sqrt{1-{x}^{2}}+c$

Evaluate the given integral.
$\displaystyle \int { \tan ^{ -1 }{ \left( \cfrac { 3x-{ x }^{ 3 } }{ 1-3{ x }^{ 2 } } \right) } } dx$

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

Let

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

$=\displaystyle\int{{\tan}^{-1}{\left(\dfrac{3\tan{\theta}-{\tan}^{3}{\theta}}{1-3{\tan}^{2}{\theta}}\right)}{\sec}^{2}{\theta}d\theta}$

We know that

$\tan{3\theta}=\dfrac{3\tan{\theta}-{\tan}^{3}{\theta}}{1-3{\tan}^{2}{\theta}}$

$=\displaystyle\int{{\tan}^{-1}{\tan{3\theta}}{\sec}^{2}{\theta}d\theta}$

$=\displaystyle\int{{3\theta}{\sec}^{2}{\theta}d\theta}$

$=3\displaystyle\int{\theta{\sec}^{2}{\theta}d\theta}$

Let

$u=\theta\Rightarrow\,du=d\theta$

$dv={\sec}^{2}{\theta}d\theta\Rightarrow\,v=\tan{\theta}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula.

$=3\left[\theta\tan{\theta}-\displaystyle\int{\tan{\theta}d\theta}\right]$

$=3\left[\theta\tan{\theta}-\log{\sec{\theta}}\right]+c$

$=3\theta\tan{\theta}-3\log{\sec{\theta}}+c$

We know that

$x=\tan{\theta}\Rightarrow\,{x}^{2}={\tan}^{2}{\theta}$

$\Rightarrow\,1+{x}^{2}=1+{\tan}^{2}{\theta}={\sec}^{2}{\theta}$

$\Rightarrow\,\sec{\theta}=\sqrt{1+{x}^{2}}$

$=3x{\tan}^{-1}{x}-3\log{\left|\sqrt{1+{x}^{2}}\right|}+c$

$=3x{\tan}^{-1}{x}-3\log{\left|{\left(1+{x}^{2}\right)}^{\frac{1}{2}}\right|}+c$

$=3x{\tan}^{-1}{x}-\dfrac{3}{2}\log{\left|1+{x}^{2}\right|}+c$

Indefinite Integrals - Class 12 Commerce Applied Mathematics - Extra Questions

Integrate the following function with respect to xx.xsinhxx\sin { hx }

Solve :a∫0√x√x+√a−xdx\int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx}

The value of (∫x.e−xdx) \left ( \int x. e ^{-x} dx \right) is,

The integration of y=x22+cy=\dfrac{x^2}{2}+c . What does cc represent?

I=∫ex(2cos2x+2sinxcosxcos2x)dxI= \displaystyle\int{e^x}\left(\frac{2}{\cos^2x}+\frac{2\sin x\cos x}{\cos^2x}\right)dx

Integrate the function xsinxx \sin x

Evaluate ∫x2exdx\displaystyle \int x^{2}e^{x}dx.

Object: ∫cosec−1x,dx,x>1\displaystyle\int cosec^{-1}x, dx, x > 1.

The acceleration of a particle varies with time tt seconds according to the relation a = 6t + 6\ ms^{-2}a = 6t + 6\ ms^{-2}. Find velocity and position as functions of time. It is given that the particle starts from origin at t = 0t = 0 with velocity 2\ ms^{-1}2\ ms^{-1}.

Object: \displaystyle\int (ex)^x(2+log x)dx\displaystyle\int (ex)^x(2+log x)dx.

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

Evaluate:\displaystyle\int x^n\log_ex\ dx\displaystyle\int x^n\log_ex\ dx

Integrate with respect to xx:x \sin^{2}xx \sin^{2}x

\int {{e^x}.\,{a^x}} dx =\int {{e^x}.\,{a^x}} dx =

Integrate the followingi) \log{x}\log{x}ii) {\cos}^{-1}(\sqrt{x}){\cos}^{-1}(\sqrt{x})

Evaluate \int {\dfrac{1}{{\sqrt 3 \sin x + \cos x}}dx} \int {\dfrac{1}{{\sqrt 3 \sin x + \cos x}}dx}

Evaluate: \int \dfrac{e^x}{\sqrt{5-4e^x-e^{2x}}}dx\int \dfrac{e^x}{\sqrt{5-4e^x-e^{2x}}}dx

Solve: \displaystyle \int {e^x}{x^2} - {e^{2x}}dx\displaystyle \int {e^x}{x^2} - {e^{2x}}dx

Integrate: \displaystyle \int {e^x}\left[ {\tan x + \log \,\sec x} \right] dx\displaystyle \int {e^x}\left[ {\tan x + \log \,\sec x} \right] dx

Solve \int {x{{\sin }^2}xdx} \int {x{{\sin }^2}xdx}

Evaluate \int {\dfrac{{dx}}{{\sqrt {x + 1} \, - \sqrt x }}} \int {\dfrac{{dx}}{{\sqrt {x + 1} \, - \sqrt x }}}

\int {\dfrac{{x\,{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx} \int {\dfrac{{x\,{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx}

y=\displaystyle\int \dfrac{\cos x}{1+\sin x}dxy=\displaystyle\int \dfrac{\cos x}{1+\sin x}dx.

Using integration, find the area of the triangle PQRPQR, whose vertices are at P (2 , 5), Q (4 , 7)P (2 , 5), Q (4 , 7) and R (6 , 2)R (6 , 2).

Evaluate: \displaystyle\int \log \left( {\log x} \right) + {\left( {\log x} \right)^{ - 2}}= ?\displaystyle\int \log \left( {\log x} \right) + {\left( {\log x} \right)^{ - 2}}= ?

Evaluate: \displaystyle\int e^x(x^2+2x) \ dx\displaystyle\int e^x(x^2+2x) \ dx

\int { \left( x+1 \right) \log { x } } dx\int { \left( x+1 \right) \log { x } } dx

\int {{e^{2x}}\left( {\sin \left( {ax + b} \right)} \right)} \,dx\int {{e^{2x}}\left( {\sin \left( {ax + b} \right)} \right)} \,dx

\int { x\tan ^{ 2 }{ x } } dx\int { x\tan ^{ 2 }{ x } } dx

\int {\left( {\frac{1}{{\left( {\ell nx} \right)}} - \frac{1}{{{{(\ell nx)}^2}}}} \right)\,dx} \int {\left( {\frac{1}{{\left( {\ell nx} \right)}} - \frac{1}{{{{(\ell nx)}^2}}}} \right)\,dx}

\int { { x }^{ 2 } } \sin ^{ -1 }{ x } dx\int { { x }^{ 2 } } \sin ^{ -1 }{ x } dx

Solve \displaystyle\int {x{{\tan }^{ - 1}}x\,dx} \displaystyle\int {x{{\tan }^{ - 1}}x\,dx}

\int { \sin ^{ -1 }{ \left( 3x-4{ x }^{ 3 } \right) } } dx\int { \sin ^{ -1 }{ \left( 3x-4{ x }^{ 3 } \right) } } dx

Evaluate:\int_{} {x{e^{2x}}dx} \int_{} {x{e^{2x}}dx}

Find \int \left(1-\frac{1}{x^2}\right)e^{(x+\frac{1}{x}) }dx\int \left(1-\frac{1}{x^2}\right)e^{(x+\frac{1}{x}) }dx on I where I= (0, \infty ).I= (0, \infty ).

Evaluate \displaystyle \int \frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}} dx\displaystyle \int \frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}} dx

Solve \int { \sin ^{ -1 }{ \left( \dfrac { 2x }{ 1+{ x }^{ 2 } } \right) } dx } \int { \sin ^{ -1 }{ \left( \dfrac { 2x }{ 1+{ x }^{ 2 } } \right) } dx }

\int {{x^3}\log xdx = } \int {{x^3}\log xdx = }

Integrate with respect to xx:x\ ln\ xx\ ln\ x

\displaystyle \int \dfrac{(t^{2}+1)^{2}}{t^6+1} dt\displaystyle \int \dfrac{(t^{2}+1)^{2}}{t^6+1} dt

Integrate the function w.r.t x : x{\left( {\log x} \right)^2}x{\left( {\log x} \right)^2}

Solve: \int\limits_0^{\pi /4} {\left( {x\cos x - \cos 3x} \right)dx} \int\limits_0^{\pi /4} {\left( {x\cos x - \cos 3x} \right)dx}

Integrate with respect to xx:\ell n\ x\ell n\ x

Simplify:-\int {x\ell n\sqrt x dx} \int {x\ell n\sqrt x dx}

Integrate with respect to xx:x\sin^{2}xx\sin^{2}x

Integrate with respect to x:(i) \dfrac{x}{nx}\dfrac{x}{nx}(ii) x\sin^2xx\sin^2x

Solve : \int \, 4 \, x^3 \sqrt{5 - x^2 } \, dx\int \, 4 \, x^3 \sqrt{5 - x^2 } \, dx

Solve : \displaystyle \int \, \dfrac{dx}{x - \sqrt{x}}\displaystyle \int \, \dfrac{dx}{x - \sqrt{x}}

Solve : \displaystyle \int \dfrac{2x}{x^2+5}dx\displaystyle \int \dfrac{2x}{x^2+5}dx

Solve: \int 3^x.e^x dx =\int 3^x.e^x dx =?

Solve \int e^{ax}\sin bxdx\int e^{ax}\sin bxdx.

\displaystyle\int cosec^3x dx\displaystyle\int cosec^3x dx.

Evaluate:\displaystyle\int \dfrac{1}{24+1}dx=?\displaystyle\int \dfrac{1}{24+1}dx=?

Solve: \int tan (x)dx=?\int tan (x)dx=?

\displaystyle \int \dfrac {3ax}{ b^{2}+c^{2}x^{2}} dx\displaystyle \int \dfrac {3ax}{ b^{2}+c^{2}x^{2}} dx

Solve:\displaystyle\int\dfrac{(\ln x )^n}{x}dx\displaystyle\int\dfrac{(\ln x )^n}{x}dx

\int e^{\ 10x} dx\int e^{\ 10x} dx

Evaluate: - 2 \int \cos 2 \theta \sin 2\theta d \theta- 2 \int \cos 2 \theta \sin 2\theta d \theta

Solve:\sqrt{\sin x}dx\sqrt{\sin x}dx

Prove that \int e ^ { x } \left( \tan ^ { - 1 } x + \frac { 1 } { 1 + x ^ { 2 } } \right) d x\int e ^ { x } \left( \tan ^ { - 1 } x + \frac { 1 } { 1 + x ^ { 2 } } \right) d x

I= \int { \left( { 1-x } \right) \left( { 2+3x } \right) \left( { 5-4x } \right) dx } \int { \left( { 1-x } \right) \left( { 2+3x } \right) \left( { 5-4x } \right) dx }

\int {\tan x}dx\int {\tan x}dx

I = \int {\left( {3\sin x - 4\cos x + 5{{\sec }^2}x - 2\cos e{c^2}x} \right)dx} I = \int {\left( {3\sin x - 4\cos x + 5{{\sec }^2}x - 2\cos e{c^2}x} \right)dx}

Evaluate: \displaystyle \int{\dfrac{dx}{32-2x^{2}}}\displaystyle \int{\dfrac{dx}{32-2x^{2}}}

\int {\cot xdx}\int {\cot xdx}

\int {cosec\;x dx}\int {cosec\;x dx}

\int {\sec xdx}\int {\sec xdx}

I = \int {{x^9}dx} I = \int {{x^9}dx}

Evaluate: \int {\frac{{\tan x}}{{{{\left( {\cos x} \right)}^2}}}dx} \int {\frac{{\tan x}}{{{{\left( {\cos x} \right)}^2}}}dx}

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

Find the value of -\int xdx\int xdx

I = \int {\sqrt[3]{x}dx} I = \int {\sqrt[3]{x}dx}

Solve:\displaystyle \int \dfrac{\log{x}}{x^{2}}\ dx\displaystyle \int \dfrac{\log{x}}{x^{2}}\ dx

Solve\int \dfrac{\cos (x+a)}{\cos (x-a)} dx\int \dfrac{\cos (x+a)}{\cos (x-a)} dx

Solve:\displaystyle \int \dfrac{x\tan^{-1}{x^{2}}}{(1+x^{4})}\ dx\displaystyle \int \dfrac{x\tan^{-1}{x^{2}}}{(1+x^{4})}\ dx

Write the formula for integration by parts.

Find \displaystyle \int \dfrac {1}{9x^{2}+6x+5}dx\displaystyle \int \dfrac {1}{9x^{2}+6x+5}dx

\int {{x^9}dx} \int {{x^9}dx}

Evaluate \displaystyle \int{\cos^{2}x\, dx}\displaystyle \int{\cos^{2}x\, dx}.

Integrate the following function with respect to $x$ .
$x\sin { hx }$

Solve :

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

The value of $\left ( \int x. e ^{-x} dx \right)$ is,

The integration of $y=\dfrac{x^2}{2}+c$ . What does $c$ represent?

$I= \displaystyle\int{e^x}\left(\frac{2}{\cos^2x}+\frac{2\sin x\cos x}{\cos^2x}\right)dx$

Integrate the function $x \sin x$

Evaluate $\displaystyle \int x^{2}e^{x}dx$ .

Object: $\displaystyle\int cosec^{-1}x, dx, x > 1$ .

The acceleration of a particle varies with time $t$ seconds according to the relation $a = 6t + 6\ ms^{-2}$ . Find velocity and position as functions of time. It is given that the particle starts from origin at $t = 0$ with velocity $2\ ms^{-1}$ .

Object: $\displaystyle\int (ex)^x(2+log x)dx$ .

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

Evaluate:
$\displaystyle\int x^n\log_ex\ dx$

Integrate with respect to $x$ :
$x \sin^{2}x$

$\int {{e^x}.\,{a^x}} dx =$

Integrate the following
i) $\log{x}$
ii) ${\cos}^{-1}(\sqrt{x})$

Evaluate $\int {\dfrac{1}{{\sqrt 3 \sin x + \cos x}}dx}$

Evaluate:
$\int \dfrac{e^x}{\sqrt{5-4e^x-e^{2x}}}dx$

Solve: $\displaystyle \int {e^x}{x^2} - {e^{2x}}dx$

Integrate: $\displaystyle \int {e^x}\left[ {\tan x + \log \,\sec x} \right] dx$

Solve $\int {x{{\sin }^2}xdx}$

Evaluate $\int {\dfrac{{dx}}{{\sqrt {x + 1} \, - \sqrt x }}}$

$\int {\dfrac{{x\,{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx}$

$y=\displaystyle\int \dfrac{\cos x}{1+\sin x}dx$ .

Using integration, find the area of the triangle $PQR$ , whose vertices are at $P (2 , 5), Q (4 , 7)$ and $R (6 , 2)$ .

Evaluate: $\displaystyle\int \log \left( {\log x} \right) + {\left( {\log x} \right)^{ - 2}}= ?$

Evaluate: $\displaystyle\int e^x(x^2+2x) \ dx$

$\int { \left( x+1 \right) \log { x } } dx$

$\int {{e^{2x}}\left( {\sin \left( {ax + b} \right)} \right)} \,dx$

$\int { x\tan ^{ 2 }{ x } } dx$

$\int {\left( {\frac{1}{{\left( {\ell nx} \right)}} - \frac{1}{{{{(\ell nx)}^2}}}} \right)\,dx}$

$\int { { x }^{ 2 } } \sin ^{ -1 }{ x } dx$

Solve $\displaystyle\int {x{{\tan }^{ - 1}}x\,dx}$

$\int { \sin ^{ -1 }{ \left( 3x-4{ x }^{ 3 } \right) } } dx$

Evaluate:
$\int_{} {x{e^{2x}}dx}$

Find $\int \left(1-\frac{1}{x^2}\right)e^{(x+\frac{1}{x}) }dx$ on I where $I= (0, \infty ).$

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

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

$\int {{x^3}\log xdx = }$

Integrate with respect to $x$ :
$x\ ln\ x$

$\displaystyle \int \dfrac{(t^{2}+1)^{2}}{t^6+1} dt$

Integrate the function w.r.t x : $x{\left( {\log x} \right)^2}$

Solve: $\int\limits_0^{\pi /4} {\left( {x\cos x - \cos 3x} \right)dx}$

Integrate with respect to $x$ :
$\ell n\ x$

Simplify:-
$\int {x\ell n\sqrt x dx}$

Integrate with respect to $x$ :
$x\sin^{2}x$

Integrate with respect to x:
(i) $\dfrac{x}{nx}$
(ii) $x\sin^2x$

Solve : $\int \, 4 \, x^3 \sqrt{5 - x^2 } \, dx$

Solve : $\displaystyle \int \, \dfrac{dx}{x - \sqrt{x}}$

Solve : $\displaystyle \int \dfrac{2x}{x^2+5}dx$

Solve: $\int 3^x.e^x dx =$ ?

Solve $\int e^{ax}\sin bxdx$ .

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

Evaluate:
$\displaystyle\int \dfrac{1}{24+1}dx=?$

Solve: $\int tan (x)dx=?$

$\displaystyle \int \dfrac {3ax}{ b^{2}+c^{2}x^{2}} dx$

Solve: $\displaystyle\int\dfrac{(\ln x )^n}{x}dx$

$\int e^{\ 10x} dx$

Evaluate: $- 2 \int \cos 2 \theta \sin 2\theta d \theta$

Solve: $\sqrt{\sin x}dx$

Prove that
$\int e ^ { x } \left( \tan ^ { - 1 } x + \frac { 1 } { 1 + x ^ { 2 } } \right) d x$

I= $\int { \left( { 1-x } \right) \left( { 2+3x } \right) \left( { 5-4x } \right) dx }$

$\int {\tan x}dx$

$I = \int {\left( {3\sin x - 4\cos x + 5{{\sec }^2}x - 2\cos e{c^2}x} \right)dx}$

Evaluate: $\displaystyle \int{\dfrac{dx}{32-2x^{2}}}$

$\int {\cot xdx}$

$\int {cosec\;x dx}$

$\int {\sec xdx}$

$I = \int {{x^9}dx}$

Evaluate: $\int {\frac{{\tan x}}{{{{\left( {\cos x} \right)}^2}}}dx}$

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

Find the value of -
$\int xdx$

$I = \int {\sqrt[3]{x}dx}$

Solve:
$\displaystyle \int \dfrac{\log{x}}{x^{2}}\ dx$

Solve
$\int \dfrac{\cos (x+a)}{\cos (x-a)} dx$

Solve:
$\displaystyle \int \dfrac{x\tan^{-1}{x^{2}}}{(1+x^{4})}\ dx$

Find $\displaystyle \int \dfrac {1}{9x^{2}+6x+5}dx$

$\int {{x^9}dx}$

Evaluate $\displaystyle \int{\cos^{2}x\, dx}$ .

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

Evaluate:
$\displaystyle\int{\dfrac{1}{4+9x^{2}}}dx$