MCQ Questions for Class 12 Commerce Maths Integrals Quiz 5

$y= \int \sqrt{1+\sin2x} dx; y$ is equal to-

0%
$\sin x - \cos x+c$
0%
$\sin x+ \cos x +c$
0%
$\cos x - \sin x +c$
0%
None of these

Explanation

$\int { \sqrt { 1+\sin { 2x } } dx }$

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

$=\int { \sqrt { { \left( \cos { x } +\sin { x } \right) }^{ 2 } } dx }$

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

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

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

Hence, the answer is

$\sin x-\cos x+c.$

The value of

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

0%
$\dfrac 12$
0%
$\dfrac 34$
0%
$2$
0%
$None\ of\ these$

Explanation

$I=\displaystyle \int_0^{\pi/2}\sin x\cos xdx$

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

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

$I=\displaystyle \int _0^{\pi/2} t dt$

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

$I=\dfrac 12-0=\dfrac 12$

$\int _{ 0 }^{ \pi }{ x\ln { \left( \sin { x } \right) } } dx=$

0%
$\dfrac { \pi }{ 2 } \ln { 2 }$
0%
$\dfrac { -\pi^{2} }{ 2 } \ln { 2 }$
0%
$\dfrac { -\pi }{ 2 } \ln { 2 }$
0%
$-2p\ \ln { 2 }$

Explanation

$I = \int_0^\pi {x\ln \left( {\sin x} \right)\ln }$

$\Rightarrow I = \int_0^\pi {\left( {\pi - x} \right)\ln } \left[ {\sin \left( {\pi - x} \right)} \right]dx$

$\Rightarrow I = \int_0^\pi {\pi \ln \sin xdx - \int_0^\pi {x\ln \sin xdx} }$

$\Rightarrow I = \pi \int_0^\pi {\ln \sin xdx - I}$

$\Rightarrow 2I = \pi \int_0^\pi {\ln \sin xdx}$

$\Rightarrow 2I = 2\pi \int_0^\pi {\ln \sin xdx}$

$\Rightarrow I = \pi \int_0^\pi {\ln \sin xdx} - eq(1)$

$\Rightarrow I = \pi \int_0^{\pi /2} {\ln \sin \left( {\dfrac{\pi }{2} - x} \right)} dx$

$\Rightarrow I = \pi \int_0^{\pi /2} {\ln \cos x} dx - eq\left( 2 \right)$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln \dfrac{{2\sin a\cos x}}{2}} dx$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln {{\sin }^2}xdx - \pi \int_0^{\pi /2} {\ln 2dx} }$

$\Rightarrow 2I = \pi {\int_0^{\pi /2} {\ln {{\sin }^2}xdx - \pi \ln 2\left[ x \right]} _0}^{\pi /2}$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln {\mathop{\rm sintdt}\nolimits} = \pi \ln 2\left[ {\dfrac{\pi }{2}} \right]}$

$\Rightarrow 2I = \dfrac{\pi }{2}\int_0^\pi {\ln \sin xdx - \dfrac{{{\pi ^2}}}{2}{{\ln }^2}}$

$\Rightarrow 2I = \dfrac{\pi }{2} \times 2\int_0^{\pi /2} {\ln {\mathop{\rm sintdt}\nolimits} - \dfrac{{{\pi ^2}}}{2}{{\ln }^2}}$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln {\mathop{\rm sintdt}\nolimits} - \frac{{{\pi ^2}}}{2}{{\ln }^2}}$

$\Rightarrow 2I = \pi \times \frac{I}{\pi } - \dfrac{{{\pi ^2}}}{2}{\ln ^2}$

$\Rightarrow 2I - I = \dfrac{{ - {\pi ^2}}}{2}{\ln ^2}$

$\Rightarrow I = \dfrac{{ - {\pi ^2}}}{2}\ln 2$

$\displaystyle\int^{\pi/4}_0\sec^4xdx$ .

0%
$\dfrac{3}{4}$
0%
$\dfrac{-3}{4}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{5}{4}$

Explanation

We are given

$I=\displaystyle \int_{0}^{\pi/4}{\sec^4x dx}$

$I=\displaystyle \int_{0}^{\pi/4}{\sec^{2}.\sec^2x dx}$ (

$\therefore$ we know that

$\sec^2\theta=1+\tan^2\theta$ )

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

Let, take

$\tan x=t\Rightarrow \sec^2xdx=dt$

( differentiating both sides)

now, when

$x=0\Rightarrow t=\tan 0=0$

& when

$x=\pi/4\Rightarrow t=\tan \pi/4=1$

$I=\displaystyle \int_{0}^{1}{(1+t^2)dt}=\displaystyle \int_{0}^{1}{dt}+\displaystyle \int_{0}^{1}{t^2 dt}$

$=[t]_{0}^{1}+\left[\dfrac{t^3}{3}\right]_{0}^{1}$

$=(1-0)+(1/3-0)$

$I=1+\dfrac{1}{3}=\dfrac{4}{3}$

$\int_{0}^{2\pi}{ln(1+\cos{x})}dx=$

0%
$\pi\ln{2}$
0%
$-\pi\ln{2}$
0%
$-2\pi\ln{2}$
0%
$2\pi\ln{2}$

Explanation

$I = \int_0^{2\pi } {\ln \left( {1 + \cos x} \right)dx}$

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 + \cos x} \right)dx}$ ---(1) (

${\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx,} } }$ if

${f\left( {2a - x} \right) = f\left( x \right)}$ )

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 + \cos \left( {\pi - x} \right)} \right)dx}$

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 - \cos x} \right)dx}$ ---(2)

Adding (1) and (2) we get

$\Rightarrow 2I = 2\int_{}^{} {\ln \left[ {\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)} \right]dx}$

$\Rightarrow I = \int_0^\pi {\ln \left( {1 - {{\cos }^2}x} \right)dx}$

$\Rightarrow I = \int_0^\pi {\ln {{\sin }^2}xdx}$

$\Rightarrow I = 2\int_0^\pi {\ln \sin xdx}$ ---(A)

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {\ln \sin xdx}$ ---(3) (

${\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx,} } }$

${f\left( {2a - x} \right) = f\left( x \right)}$ )

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {\ln \sin \left( {\frac{\pi }{2} - x} \right)dx}$

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {ln(cosx)dx}$ ---(4)

Adding (3) and (4)

$\Rightarrow 2I = 4\int_0^{\frac{\pi }{2}} {\ln \left( {\frac{{2\sin x\cos x}}{2}} \right)dx}$

$\Rightarrow I = 2\int_0^{\frac{\pi }{2}} {\ln \sin 2xdx - 2\int_0^{\frac{\pi }{2}} {\ln 2dx} }$

$\Rightarrow I = 2\int_0^{\frac{\pi }{2}} {\ln \sin 2xdx - 2\ln 2\left[ x \right]_0^{\frac{\pi }{2}}}$

putting

$2x=t$

$\Rightarrow 2dx = dt$

$\Rightarrow I = \int_0^\pi {\ln \sin tdt} - 2\ln 2\left[ {\frac{\pi }{2}} \right]$

$\Rightarrow I = \int_0^\pi {\ln \sin x dx} - \pi \ln 2$ (since

${\int_a^b {f\left( t \right)dt = \int_a^b {f\left( x \right)dx} } }$ )

$\Rightarrow I = \frac{I}{2} - \pi \ln 2$ (from (A))

$\Rightarrow \frac{I}{2} = - \pi \ln 2$

$\Rightarrow I = - 2\pi \ln 2$

Evaluate

$\displaystyle \int_\cfrac{\pi}{4}^\cfrac{\pi}{2}(\sqrt{\tan x}+\sqrt{\cot x})dx=$

0%
$\dfrac{\pi}{2\sqrt{2}}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{\sqrt{2}}$
0%
$\dfrac{\pi}{3}$

Explanation

$\int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx}$

$= \int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \cfrac{1}{{\sqrt {\tan x} }}} \right)dx}$

$= \int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\cfrac{{1 + \tan x}}{{\sqrt {\tan x} }}} \right)dx}$

putting

$\tan x = {t^2}$

$\Rightarrow {\sec ^2}xdx = 2tdt$

$\Rightarrow \left( {1 + {{\tan }^2}x} \right)dx = 2tdt$

$\Rightarrow \left( {1 + {t^4}} \right)dx = 2tdt$

$\Rightarrow dx = \cfrac{{2tdt}}{{1 + {t^4}}}$

$= \int_1^\infty {\cfrac{{\left( {1 + {t^2}} \right)}}{{t\left( {1 + {t^4}} \right)}} \times 2tdt}$

$= 2\int_1^\infty {\cfrac{{1 + {t^2}}}{{1 + {t^4}}}} dt$

$= 2\int_1^\infty {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{\left( {{t^2} + \cfrac{1}{{{t^2}}}} \right)}}dt}$

$= 2\int_1^\infty {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{\left( {{t^2} + \cfrac{1}{{{t^2}}} - 2 + 2} \right)}}dt}$

$= 2\int_1^\infty {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{{{\left( {t - \cfrac{1}{t}} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}dt}$

putting

$t - \cfrac{1}{t} = y$

$\ \Rightarrow \left( {1 + \cfrac{1}{{{t^2}}}} \right)dt = dy$

$= 2\int_0^\infty {\cfrac{{dy}}{{{y^2} + {{\left( {\sqrt 2 } \right)}^2}}}}$

$= 2 \times \cfrac{1}{{\sqrt 2 }}\left[ {{{\tan }^{ - 1}}\left( {\cfrac{y}{{\sqrt 2 }}} \right)} \right]_0^\infty$

$= \sqrt 2 \left[ {{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right]$

$= \sqrt 2 \left( {\cfrac{\pi }{2}} \right)$

$= \cfrac{\pi }{{\sqrt 2 }}$

The value of

$\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \log { \left( \dfrac { 4+3\sin { x } }{ 4+3\cos { x } } \right) dx } }$ is

0%
$2$
0%
$\dfrac{3}{4}$
0%
$0$
0%
$-2$

Explanation

As,

$\begin{array}{l}\int_0^a {f\left( x \right)} dx = \int_0^a {f\left( {x - a} \right)} dx\\So,\\I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin \left( {\frac{\pi }{2} - x} \right)}}{{4 + 3\cos \left( {\frac{\pi }{2} - x} \right)}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \end{array}$

Adding the above two integral,

$\begin{array}{l}I + I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} dx + \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} dx\\2I = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) + \log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\As,\\\log a + \log b = \log ab\\So,\\2I = \int_0^{\frac{\pi }{2}} {\log \left[ {\left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) \times \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\ = \int_0^{\frac{\pi }{2}} {\log \left( 1 \right)dx} \\ = \int_0^{\frac{\pi }{2}} {0dx} \\2I = 0\\I = 0\end{array}$

Thus Option C

$\displaystyle \int _{ 0 }^{ \pi /4 }{ \tan ^{ 2 }{ x } dx= }$

0%
$1-\dfrac {\pi}{4}$
0%
$1+\dfrac {\pi}{4}$
0%
$\dfrac {-\pi}{4}-1$
0%
$\dfrac {\pi}{4}-1$

Explanation

$tan^2x = sec^2x - 1$

$\Rightarrow$

$\int_0^{\pi/4} -1 dx$ +

$\int_0^{\pi/4} (sec^2x) dx$

$-\pi/4$ + [

$tanx$ }

$_0^{\pi/4}$

= 1 -

$\pi/4$

So, the correct option is 'A'.

The value of

$\displaystyle \underset{0}{\overset{x}{\int}} \dfrac{(t - |t|)^2}{(1 + t^2)} dt$ is equal to

0%
$4(x - \tan^{-1} x), \, \ \text{if} \, x < 0$
0%
$0 \, if \, x > 0$
0%
$\ln ( 1 + x^3) \, \ \text{if} \, x > 0$
0%
$4(x + \tan^{-1} x ) \, \text{if} \, x < 0$

$\displaystyle \int _0^4\dfrac{2x+3}{x^2+3x+2}dx$

0%
$\log 3$
0%
$\log 5$
0%
$\log 15$
0%
$\log 2$

Explanation

Let

$x^2+3x+2=t\implies 2x+3=dt$

$x\to 0 \>to \>4\\t\to 2\to 30\\\displaystyle \int _2^{30}\dfrac 1tdt\\\left.\log t\right]_2^{30}\\\log 30-\log 2\\\\\log \dfrac {30}2\\\\log 15$

Evaluate

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

0%
$I=\sqrt{{{x}^{2}}-2}+C$
0%
$I=\sqrt{{{x}^{2}}+2}+C$
0%
$I=\sqrt{{{x}^{3}}+2}+C$
0%
$I=\sqrt{{{x}^{3}}-2}+C$

Explanation

Consider the given integral.

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

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

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

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

Therefore,

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

$I=\dfrac{1}{2}\left( 2\sqrt{t} \right)+C$

$I=\sqrt{t}+C$

On putting the value of $t$ , we get

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

Hence, this is the answer.

$\displaystyle \int \sqrt{\dfrac{a+x}{a-x}}dx$ is equal to-

0%
$a\sin^{-1} (x/a)-\sqrt{a^{2}-x^{2}}+c$
0%
$a\cos^{-1} (x/a)-\sqrt{a^{2}-x^{2}}+c$
0%
$a\sin^{-1} (x/a)-\sqrt{a^{2}+x^{2}}+c$
0%
$a\cos^{-1} (x/a)-\sqrt{a^{2}+x^{2}}+c$

Explanation

Consider the given integral.

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

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

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

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

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

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

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

Hence, this is the correct answer.

$\int ( 1+3x+3x^2+4x^3+........)dx (|x| <1)$ -

0%
$(1+x)^{-1}+c$
0%
$(1-x)^{-1}+c$
0%
$(1+x)^{-2}+c$
0%
None of these

The angle between the tangent lines to the graph of the function

$f(x) =\int_\limits{2}^x (2t -5)dt$ at the point where the graph cuts the

$x$ -axis is

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{2}$

$\int { P\left( x \right) { e }^{ kx }dx=Q\left( x \right) { e }^{ 4x }+C }$ , where

$P(x)$ is polynomial of degree

$n$ and

$Q(x)$ is polynomial of degree

$7$ . Then the value of

$n+7+k+\lim _{ x\rightarrow \infty }{ \dfrac { P\left( x \right) }{ Q\left( x \right) } }$ is:

0%
$18$
0%
$19$
0%
$20$
0%
$22$

For

$x > 0$ , let

$f(x)=\displaystyle \int_{1}^{x}\dfrac {\log t}{1+t} \ dt$ then, the value of

$f(x)+f(1/x)$ will be

0%
$\dfrac {1}{4}\log x^{2}$
0%
$\log x$
0%
$\dfrac {1}{4}(\log x)^{2}$
0%
$\dfrac {1}{2}(\log x)^{2}$

Evaluate using limit of sum:

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

0%
$26$
0%
$28$
0%
$30$
0%
$32$

Explanation

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

Let

$t=(x+1)\implies dt=dx$

$x \to 1\to 3$

$t\to 2 \to 4$

$\displaystyle \int _2^4 t^2 dt$

$\left.\dfrac {t^3}3\right]_2^4$

$\dfrac{4^3}{2}-\dfrac {2^3}{2}$

$32-4=28$

$\displaystyle \int { \dfrac { { 2 }^{ x } }{ \sqrt { 1-{ 4 }^{ x } } } dx=K\sin ^{ -1 }{ \left( { 2 }^{ x } \right) +C } }$ , then the value of

$K$ is equal to

0%
$\ell n 2$
0%
$\dfrac{1}{2}\ell 2$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{\ell n 2}$

Explanation

Consider the following integral.

$\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx$

Solution:

Let and differentiate w.r.t “x” we get.

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

$\dfrac{dt}{dx}={{2}^{x}}\ln 2$

$\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx\,\,\,\,\,\,\,\,......\left( 1 \right)\$

Put the value $dx\,\,$ of in eq . (1) we get.

$=\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx\,\,\,\,\,\,\,\,......\left( 1 \right)$

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

$=\int\limits_{{}}^{{}}{\dfrac{1}{\ln 2\sqrt{1-{{\left( {{t}^{2}} \right)}^{x}}}}}dx\,$

$=\dfrac{1}{\ln 2}{{\sin }^{-1}}x+C$

Hence, the value of $[K=\dfrac{1}{\ln 2}]$

Hence, this is the correct answer .

Solve

$\int\limits_l^e {\dfrac{{dx}}{{\ln \left( {{x^x} \cdot {e^x}} \right)}}}$

0%
$\ln 2$
0%
$2\ln 2$
0%
$-\ln 2$
0%
None of these

Explanation

Consider the given integral.

$I=\int_{1}^{e}{\dfrac{dx}{\ln \left( {{x}^{x}}\cdot {{e}^{x}} \right)}}$

$I=\int_{1}^{e}{\dfrac{dx}{\ln {{x}^{x}}+\ln {{e}^{x}}}}$

$I=\int_{1}^{e}{\dfrac{dx}{x\ln x+x\ln e}}$

$I=\int_{1}^{e}{\dfrac{dx}{x\ln x+x}}$

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

Let $t=\ln x+1$

$\dfrac{dt}{dx}=\dfrac{1}{x}+0$

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

Therefore,

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

$I=\left[ \ln \left( t \right) \right]_{1}^{2}$

$I=\ln \left( 2 \right)-\ln \left( 1 \right)$

$I=\ln 2$

Hence, this is the answer.

$\displaystyle \overset{\ln \pi}{\underset{\ln\pi - \ln2}{\int}} \dfrac{e^x}{1 - \cos \left(\tfrac{2}{3}e^x\right)}dx$ is equal to

0%
$\sqrt{3}$
0%
$-\sqrt{3}$
0%
$\dfrac{1}{\sqrt{3}}$
0%
$-\dfrac{1}{\sqrt{3}}$

Explanation

Now,

$\displaystyle \overset{\ln \pi}{\underset{\ln\pi - \ln2}{\int}} \dfrac{e^x}{1 - \cos \left(\dfrac{2}{3}e^x\right)}dx$

$=\displaystyle \overset{\ln \pi}{\underset{\ln\tfrac{\pi}{2}}{\int}} \dfrac{e^x}{1 - \cos \left(\dfrac{2}{3}e^x\right)}dx$

Put

$\dfrac{2e^x}{3}=z$ .

or,

$e^x dx=\dfrac{3}{2}dz$ , again when

$x=\ln{\pi}$ then

$z=\dfrac{2}{3}{\pi}$ and

$x=\ln\frac{\pi}{2}$ then

$z=\dfrac{\pi}{3}$ .

So the given integration becomes,

$=\displaystyle \overset{\tfrac{2\pi}{3} }{\underset{\tfrac{\pi}{3}}{\int}} \frac{\dfrac{3}{2}}{1 - \cos \left(z\right)}dz$

$=\dfrac{3}{4}\displaystyle \overset{ \tfrac{2\pi}{3}}{\underset{\tfrac{\pi}{3}}{\int}}\cos ec^2 \dfrac{z}{2}dz$

$=\dfrac{3}{4}\times 2\left[-\cot \dfrac{z}{2}\right]_{\tfrac{\pi}{3}}^{\tfrac{2\pi}{3}}$

$=\dfrac{3}{2}\times\left[\sqrt3-\dfrac{1}{\sqrt 3}\right]$

$=\sqrt 3$ .

$\displaystyle \int_{0}^{a} \dfrac{dx}{\sqrt{x+a}+\sqrt{x}}=\displaystyle \int_{0}^{\pi/8} \dfrac{2 \tan \theta}{\sin 2 \theta} d\theta$ , then value of

$a$ is equal to

$(a > 0)$

0%
$\dfrac{3}{4}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{3\pi}{4}$
0%
$\dfrac{9}{16}$

Explanation

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

$\\=(\dfrac{1}{a})\left[\int_{0}^{a}\sqrt{x+a}dx-\int_{0}^{a}\sqrt{x}dx\right]\\=(\dfrac{1}{a})\left[(\dfrac{(x+a)^{(\frac{3}{2})}}{(\dfrac{3}{2})})\right]_{0}^{a}-(\dfrac{1}{a})\left[(\dfrac{x^{(\frac{3}{2})}}{(\dfrac{3}{2})})\right]_{0}^{a}$

$\\=(\dfrac{1}{a})\left[(2a)^{(\frac{3}{2})}-a^{(\frac{3}{2})}\right]-(\dfrac{2}{3a})\left[a^{(\frac{3}{2})}-0\right]\\=(\dfrac{2}{3a})\>a^{(\frac{3}{2})}[2\sqrt{2}-1-1]\\=(\dfrac{2}{3a})\>a^{(\frac{3}{2})}(2\sqrt{2}-2)\\=(\dfrac{4}{3a})(\sqrt{2}-1)\>a^{(\frac{3}{2})}$

$\\=(\dfrac{4}{3})(\sqrt{2}-1)\>a^{(\frac{1}{2})}$

$\\RHS=\int_{0}^{(\frac{\pi}{8})}(\dfrac{2tan\theta}{sin2\theta})d\theta\\=\int_{0}^{(\frac{\pi}{8})}(\dfrac{2sin\theta}{cos\theta\times2sin\theta\>cos\theta})d\theta\\=\int_{0}^{(\frac{\pi}{8})}sec^2\theta\>d\theta=[tan\theta]_{0}^{\pi/8}\\=tan(\dfrac{\pi}{8})=(\sqrt{2}-1)\\=as\>LHS=RHS\\\therefore\>(\dfrac{4}{3})(\sqrt{2}-1)a^{(\frac{1}{2})}=(\sqrt{2}-1)\\\therefore\>a=(\dfrac{9}{16})$

$\int {{e^x}\left[ {{\mathop{\rm tanx}\nolimits} - log\left( {\cos x} \right)} \right]} dx =$

0%
${e^x}\log \left( {\sec x} \right) + c$
0%
${e^x}\log \left( {co\sec x} \right) + c$
0%
${e^x}\log \left( {\cos x} \right) + c$
0%
${e^x}\log \left( {\sin x} \right) + c$

Explanation

$\displaystyle\int e^x\left[\tan x-log (\cos x)\right]dx$

$=\displaystyle\int e^x\left[\tan x+log (\sec x)\right]d$

$=\displaystyle\int \dfrac{d}{dx}\left\{e^xlog(\sec x)\right\}dx$

$=e^xlog (\sec x)+c$ .

$\displaystyle \int \dfrac{dx}{\sqrt{\sin^3 x \cos^5 x}} = a \sqrt{\cot x } + b \sqrt {\tan^3x} + c$ where c is an arbitrary constant of integration then the values of

$'a'$ and

$'b'$ are respectively :

0%
$-2$ & $\dfrac{2}{3}$
0%
$2$ & $-\dfrac{2}{3}$
0%
$2$ & $\dfrac{2}{3}$
0%
None of these

Explanation

$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3x\cdot \cos^5x}}=a\sqrt{\cot x}+b\sqrt{\tan^3x}+c$ .......(1).

Now,

$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3x\cdot \cos^5x}}$

$=\displaystyle\int \dfrac{cosec^2 xdx}{\sqrt{\cot x.\cos^4 x}}$ [ Dividing the numerator and the denominator by

$\sin^2 x$ ]

$=\displaystyle\int \dfrac{cosec^2 x.\sec^2 xdx}{\sqrt{\cot x}}$

$=\displaystyle\int \dfrac{cosec^2 x.(1+\tan^2 x)dx}{\sqrt{\cot x}}$

$=\displaystyle\int \dfrac{cosec^2 x.dx}{\sqrt{\cot x}}$

$+\displaystyle\int \dfrac{cosec^2 x.\tan^2 x\ dx}{\sqrt{\cot x}}$

$=\displaystyle\int \dfrac{cosec^2 x.dx}{\sqrt{\cot x}}$

$+\displaystyle\int \dfrac{\sec^2 x\ dx}{\sqrt{\cot x}}$

$=-\displaystyle\int \dfrac{d(\cot x)}{\sqrt{\cot x}}$

$+\displaystyle\int {\sqrt{\tan x}d(\tan x)}$

$=-{2}\sqrt{\cot x}+\dfrac{2}{3}\sqrt{\tan^3x}+c$ .

Comparing this with (1) we get,

$a=-2$ and

$b=\dfrac{2}{3}$ .

$\int (1 + 2x + 3x^2 + 4x^3 + ...)dx =$

0%
$(1 + x)^{-1} + c$
0%
$(1 - x)^{-1} + c$
0%
$(1 - x)^{-1} - 1 + c$
0%
None of these

Explanation

$\displaystyle \int(1+2x+3x^2+4x^3+..)dx$

$=\displaystyle \int (1-x)^2dx\quad [(1-x)^2=1+2x+3x^2]$

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

$=(1-x)^{-1}+c$

$\therefore \boxed{\displaystyle \int(1+2x+3x^2+.....)dx}=(1-x)^{-1}+c$

Evaluate:

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

0%
$0$
0%
$1$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{4}$

Evaluate

$\displaystyle\int^{\pi}_0\dfrac{x}{1+\sin x}dx$ .

0%
$x\tan x - ln|\cos x| - x\sec x -ln|\sec x - \tan x| + C$
0%
$x\tan x + ln|\cos x| - x\sec x -ln|\sec x - \tan x| + C$
0%
$\tan x + ln|\cos x| - x\sec x -ln|\sec x - \tan x| + C$
0%
None of these

Explanation

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

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

$\displaystyle \int (x \sec^2x - x \tan x \sec x )dx$

Applying byparts,we get

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

Evaluate

$\displaystyle\int^{\pi/3}_{\pi/6}\dfrac{dx}{1+\sqrt{\tan x}}$ .

0%
$\cfrac{\pi}{12}$
0%
$\cfrac{7\pi}{12}$
0%
$\cfrac{5\pi}{12}$
0%
None of these

Explanation

$I=\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\tan x}}$

$\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\tan(\cfrac{\pi}{3}+\cfrac{\pi}{6} - x)}}$

$\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\cot x}}$

$\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}}$

Adding another I in the above equantion, we get

$2I = \displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}} + \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\tan x}}$

$2I = \displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} dx$

$2I = \cfrac{\pi}{6}$

$I = \cfrac{\pi}{12}$

$\displaystyle \int (x^2-x+5)\, dx$

0%
$\dfrac {x^3}3-\dfrac{x^2}2+5x+c$
0%
$\dfrac {x^3}3+\dfrac{x^2}2+5x+c$
0%
$\dfrac {x^2}2-\dfrac{x}2+5x+c$
0%
$\dfrac {x^4}4-\dfrac{x^4}3+5$

Explanation

Given

$\displaystyle \int x^2-x+5 \, dx$

$\implies \displaystyle \int x^2 dx-\int x dx+\int 5 dx$

$\implies \dfrac{x^3}3-\dfrac{x^2}2+5x+c$

$\displaystyle \int {{x^3}{e^{{x^2}}}dx = }$

0%
$\dfrac{1}{2}\left( {{x^2} + 1} \right){e^{{x^2}}} + c$
0%
$\left( {{x^2} + 1} \right){e^{{x^2}}} + c$
0%
$\dfrac{1}{2}\left( {{x^2} - 1} \right){e^{{x^2}}} + c$
0%
$\left( {{x^2} - 1} \right){e^{{x^2}}} + c$

Explanation

$\int x ^ { 3 } e ^ { x ^ { 2 } } d x$

put

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

$\dfrac { 1 } { 2 } \int t e ^ { t } d t = \dfrac { 1 } { 2 } \left[ \left( t e ^ { t } \right) - \int e ^ { t } d t \right]$ {integration by parts}

$= \dfrac { 1 } { 2 } \left[ t e ^ { t } - e ^ { t } \right]+c$

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

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

$\int \sqrt {1 + \sin x}dx =$

0%
$\dfrac {1}{2}\left (\sin \dfrac {x}{2} + \cos \dfrac {x}{2}\right ) + c$
0%
$\dfrac {1}{2}\left (\sin \dfrac {x}{2} - \cos \dfrac {x}{2}\right ) + c$
0%
$2\sqrt {1 + \sin x} + c$
0%
$-2\sqrt {1 - \sin x} + c$

Explanation

$\displaystyle\int \sqrt{1+\sin x} dx$

$=\displaystyle\int \dfrac{\sqrt{1+\sin x}.\sqrt{1-\sin x}}{\sqrt{1-\sin x}}\ dx$

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

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

Let

$u=\sqrt{1-\sin x}$

$du=\dfrac{1}{\sqrt{1-\sin x}}(-\cos x)dx$

$-2du=\dfrac{\cos x}{2\sqrt{1-\sin x}}dx$

$=\displaystyle\int -2du=-2u+c$

$=-2\sqrt{1-\sin x}+c$

$\displaystyle\int x^2e^{x^3}\cos \left(e^{x^3}\right)dx$ is equal to?

0%
$\sin\left(e^{x^3}\right)+C$
0%
$3\sin \left(e^{x^3}\right)+C$
0%
$\dfrac{1}{3}\sin\left(e^{x^3}\right)+C$
0%
$e^x\sin\left(e^{x^3}\right)+C$

Explanation

Let

$e^{x^3} = m$

$3x^2e^{x^3} dx = dm$

$\displaystyle \cfrac{1}{3} \int \cos m dm$

$\cfrac{1}{3} \sin e^{x^3}+C$

$\displaystyle \underset{0}{\overset{\pi}{\int}} \dfrac{x \, dx}{1 + \sin \, x} =$

0%
$\dfrac{\pi}{6}$
0%
$\pi$
0%
$\dfrac{\pi}{3}$
0%
none of these

Explanation

$\int \dfrac{x}{1+\sin x}dx$

$\int \dfrac{x \sec x}{\tan x+\sec x}dx$

$=-\dfrac{x}{\tan x+\sec x}-\int \dfrac{1}{-\tan x-\sec x}dx$

$=-\dfrac{x}{\tan x+\sec x}+\int \dfrac{1}{\tan x+\sec x}dx$

$=-\dfrac{x}{\tan x+\sec x}+\int \dfrac{\cos x}{1+\sin x}dx$

--------------------------------------------------------

substitute

$u=\sin x\rightarrow du=\cos x dx$

$=\int \dfrac{1}{u+1}du$

$=\ln (u+1)$

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

-----------------------------------------------------------------

$=-\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)+C$

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

$=\left [ -\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)\right ]_0^{\pi}$

$=\pi -0$

$=\pi$

The integral of

$\displaystyle\int e^{\sin x}(x\cos x-\sec x\tan x)dx$ is?

0%
$se^{\sin x}-e^{\sin x}\sec x+c$
0%
$(x+\sec x)e^{\sin x}+c$
0%
$e^{\sin x}\cos x+c$
0%
$e^{\sin x}(\cos x-\sec x)+c$

$\int\dfrac{e^x+e^{-x}+(e^x-e^{-x})sin x}{1+cos x}dx=$

0%
$(e^x+e^{-x}) tan (x/2)+C$
0%
$(e^x-e^{-x}) cot (x/2)+C$
0%
$(e^x-e^{-x}) tan (x/2)+C$
0%
$(e^x-e^{-x}) cosec (x/2)+C$

What is the value of

$\int_{0}^{a}\dfrac{x-a}{x+a}\ dx$ ?

0%
$a+2a\log 2$
0%
$a-2a\log 2$
0%
$2a\log 2-a$
0%
$2a\log 2$

Explanation

$I= _{0}^{a}\int \dfrac{x-a}{x+a} dx$

Let

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

When

$x= 0, t=a$

$x=a, t= 2a$

Substituting,

$I= _{a}^{2a} \int \dfrac{t-2a}{t} dt$

$= [t-2a \log |t|]^{2a}_{a}$

$=2a-2a \log 2a- a+2a \log a$

$=a-2a\log 2$

$\therefore$ Option

$B$ is correct.

$\int \dfrac {dx}{(x^{2} + 4x + 5)^{2}}$ is equal to

0%
$\dfrac {1}{2}\left [\tan^{-1}(x + 1) + \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
0%
$\dfrac {1}{2}\left [\tan^{-1}(x + 2) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
0%
$\dfrac {1}{2}\left [\tan^{-1}(x + 1) - \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$
0%
$\dfrac {1}{2}\left [\tan^{-1}(x + 2) + \dfrac {x + 2}{x^{2} + 4x + 5}\right ] + c$

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

0%
$\dfrac {(\ln (1+x))^2}2$
0%
$- \pi \ln (1+x)$
0%
$\frac{\pi }{2}\ln (1+x)$
0%
$- \frac{\pi }{2}\ln (1+x)$

Explanation

Given

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

let

$1+x=t$

$\implies dx=dt$

$\implies \displaystyle \int \dfrac {\ln t}{t}dt$

Let

$\ln t=p$

$\implies \dfrac 1tdt =dp$

$\implies \displaystyle \int pdp$

$\implies \dfrac{p^2}{2}$

$\implies \dfrac {(\ln t)^2}2$

$\implies \dfrac {(\ln (1+x))^2}2$

The value of

$\int\limits_0^{\pi /2} {\dfrac{{x\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx\,is}$ is

0%
$\dfrac{{{\pi ^2}}}{4}$
0%
$\dfrac{{{\pi ^2}}}{8}$
0%
$\dfrac{{{\pi ^2}}}{{16}}$
0%
$\dfrac{{3{\pi ^2}}}{{16}}$

Explanation

$\displaystyle I = \int_{0}^{\pi/2} \frac{x sinx \, cosx}{sin^4x + cos^4 x}dx$

using

$\displaystyle \int_{0}^{a} f(n)dx = \int_{0}^{a} f(a-x) dx,$ we get

$\displaystyle I= \int_{0}^{\pi/2} \frac{(\pi/2 -\pi) sin (\pi /2 -\pi) cos (\pi/2 - \pi)}{sin^4 (\pi/2 -\pi)+ cos^4 (\pi /2 -\pi)}$

$\displaystyle = \int_{0}^{\pi/2} \frac{(\pi/2-\pi)cosx sinx}{cos^4x + sin^4 x} dx$

$\displaystyle I = \int_{0}^{\pi/2} \frac{cos sinx}{cos^4 x+ sin^4 x}dx - \int_{0}^{\pi/2} \frac{+xcosx \, sinx}{cos^4x+sin^4x}dx$

$\displaystyle \therefore 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{sinx cosx}{sin^4 x + cos^4x} dx - I$

$\displaystyle \therefore I = \dfrac{\pi}{2}\int_{0}^{1} \frac{dt/2}{t^2+(1-t)^2}$

$x = \pi / 2,$ then t = 1

and

$x = 0 ,$ then

$t =0$

$\displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{t^2+t^2+1-2t}$

$\displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{2t^2-2t+1}$

$\displaystyle = \dfrac{\pi}{16} \int_{0}^{1} \dfrac{dt}{(t-\dfrac{1}{2})^2 + (\dfrac{1}{2})^2}$

$\displaystyle = \dfrac{\pi}{16} \times \frac{1}{1/2} [ tan^{-1} (\frac{t-1/2}{1/2})]$

$\displaystyle \because \int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1}x/a$

$\displaystyle = [tan^{-1} 2(1-1/2)-tan^{-1}2(0-1/2)]$

$\displaystyle = \dfrac{\pi}{8} [tan^{-1}1-tan^{-1}(-1)]$

$\displaystyle \because tan^{-1}(-1) = -tan^{-1}1 = -\pi/4$

$\displaystyle I = \frac{\pi}{8} [ \frac{\pi}{4}+\frac{\pi}{4}] = \frac{\pi^2}{16}$

1067940_1138807_ans_ded56a648ed94f06a86b8d7916aeab9b.JPG

Find the value:

$I=\int_{1}^{2}{\frac{xdx}{(x+1)(x+2)}}$

0%
$\log \dfrac{32}{27}$
0%
$\log \dfrac{27}{20}$
0%
$\log \dfrac{27}{16}$
0%
$\log \dfrac{32}{9}$

$\displaystyle \int_{1/e}^{e}{|\ln x|dx}$ equals

0%
$e^{-1}-1$
0%
$2\left (1-\dfrac {1}{e}\right)$
0%
$1-\dfrac {1}{e}$
0%
$e-1$

Explanation

We know,

By parts integrals

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$\displaystyle \int lnx \cdot 1dx = lnx\int dx-\int \frac{1}{x}\int dx\,dx$

$\displaystyle = xlnx-x+c$

$\displaystyle \int _{1/e}^{e}\ |lnx|dx = \int_{1/e}^{1}|lnx|dx+\int _{1}^{e} |lnx|dx$

$\displaystyle =\int _{1/e}^{-1}lnxdx+\int_{1}^{e} lnxdx$

$\displaystyle = -[x(lnx-1)]_{1/e}^{1}+[x|lnx-1|]_{1}^{e}$

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

$\displaystyle = 2(1-\frac{1}{e})$

$\therefore$ Option B is correct

$y=(x+\sqrt{x^{2}-a^{2}})^{n}$ then

$(x^{2}-a^{2})(\dfrac{dy}{dx})^{2}$ =

0%
$n^{2}y$
0%
$-n^{2}y$
0%
$ny^{2}$
0%
$n^{2}y^{2}$

Explanation

$y = (x+\sqrt{x^2-a^2})^n$

${y}' = n(x+\sqrt{x^2-a^2})^{n-1} (1+\dfrac{1(2x)}{2\sqrt{x^2-a^2}})$

$= n \dfrac{y}{x+\sqrt{x^2-a^2}}(1+\dfrac{x}{\sqrt{x^2-a^2}})$

$= n \dfrac{y}{(x+\sqrt{x^{2}-a^2})}+\dfrac{(\sqrt{x^2+a^2+x})}{\sqrt{x^2-a^2}}$

$({y}')^2 = (\dfrac{ny}{x^2-a^2})^2$

$({y}')^2 = \dfrac{n^{2}y^2}{x^2-a^2}$

$\therefore (x^{2}-a^2) ({y}') = (x^2-a^2)\dfrac{x^2y^2}{(x^2-a^2)}$

$= n^{2}y^2$ option is D is correct.

1151063_1144362_ans_7dd05037f1194946a155a55fbb0dcae0.jpg

$\int _{ 0 }^{ \pi /4 }{ x.\sec ^{ 2 }{ x } dx=? }$

0%
$\frac {\pi}{4}+\log {\sqrt {2}}$
0%
$\frac {\pi}{4}-\log {\sqrt {2}}$
0%
$1+\log {\sqrt {2}}$
0%
$1-\frac{1}{2}\log {2}$

$\int\limits_0^\pi {\frac{{{x^2}{{\cos }^4}x\sin x}}{{2\pi x - {\pi ^2}}}dx} =$

0%
$\frac{2}{5}$
0%
$\frac{1}{5}$
0%
$\frac{{2\pi }}{5}$
0%
$\frac{\pi }{5}$

$\displaystyle \int\limits_0^\frac \pi 2 x\sqrt {1-x^2} dx =$

0%
$\dfrac{\pi }{4}$
0%
$\dfrac{\pi }{3}$
0%
$\dfrac{\pi }{6}$
0%
$\dfrac{\pi}{8}$

Explanation

$\displaystyle\int_{0}^{\frac{\pi}{2}}{x\sqrt{1-{x}^{2}}dx}$

Let

$x=\sin{\theta}\Rightarrow\,dx=\cos{\theta}d\theta$

$=\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin{\theta}\sqrt{1-{\sin}^{2}{\theta}}\cos{\theta}d\theta}$

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

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

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

$=\dfrac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{2}}{2\sin{\theta}\cos{\theta}\cos{\theta}d\theta}$

$=\dfrac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{2}}{\sin{2\theta}\cos{\theta}d\theta}$

$=\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{2\sin{2\theta}\cos{\theta}d\theta}$

$=\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}{\left(\sin{3\theta}+\sin{\theta}\right)d\theta}$

$=\dfrac{1}{4}\left[\dfrac{-\cos{3\theta}}{3}-\cos{\theta}\right]_{0}^{\frac{\pi}{2}}$

$=\dfrac{-1}{4}\left[\dfrac{\cos{3\theta}}{3}+\cos{\theta}\right]_{0}^{\frac{\pi}{2}}$

$=\dfrac{-1}{4}\left[\dfrac{\cos{3\dfrac{\pi}{2}}}{3}-\dfrac{\cos{0}}{3}+\cos{\dfrac{\pi}{2}}-\cos{0}\right]$

$=-\dfrac{\pi}{4}\left[0-\dfrac{1}{3}+0-1\right]$

$=-\dfrac{\pi}{4}\left[-\dfrac{1+3}{3}\right]$

$=\dfrac{\pi}{4}\left[\dfrac{4}{3}\right]=\dfrac{\pi}{3}$

$\int\limits_0^{\dfrac{ \pi }{4}} {\sin x.{{\sec }^3}xdx}$

0%
$\dfrac{1}{2}$
0%
$\sqrt 2$
0%
$2$
0%
$\dfrac{1}{3}$

Explanation

Expalantion:-

$\int_{0}^{\frac{\Pi }{4}}sinx sec^{3}x dx =\int_{0}^{\frac{\Pi }{4}}\dfrac{sinx}{cosx}sec^{2}xdx$

$=\int_{0}^{\frac{\Pi }{4}}tanxsec^{2}xdx$

$Put tanx =y \Rightarrow sec^{2}xdx=dy$

$\Rightarrow \int_{0}^{\frac{\Pi }{4}}ydy=\left ( \dfrac{y^{2}}{2} \right )_{0}^{\frac{\Pi }{4}}$

$=(\dfrac{tan^{2}x}{2})_{0}^{\frac{\Pi }{4}}=\dfrac{1}{2}$

$\int \dfrac {\sin^{-1}x-\cos^{-1}x}{\sin^{-1}x+\cos^{-1}x}dx=$

0%
$\dfrac {4}{\pi}[x \sin^{-1}+\sqrt {1-x^{2}}]-x+c$
0%
$\log [\sin^{-1}+\sqrt {1-x^{2}}]-x+c$
0%
$\dfrac {4}{\pi}[x \sin^{-1}+\sqrt {1-x^{2}}]+c$
0%
$\dfrac {2}{\pi}[x \sin^{-1}x-x \cos^{-1}x+2\sqrt {1-x^{2}}]+c$

Explanation

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

$\sin^{-1}x+\cos^{-1}x=\pi /2$

$\dfrac{2}{\pi}\displaystyle \int{\sin^{-1}x}\displaystyle \int{\cos^{-1}x}dx$

$\Rightarrow \dfrac{2}{\pi}\left[x\sin^{-1}x+\sqrt{1-x^2}-x\cos^{-1}x+\sqrt{1-x^2}\right]+c$

$=\dfrac{2}{\pi}[x\sin^{-1}x-x\cos^{-1}x+2\sqrt{1-x^2}]+c$

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

0%
$x\ \tan \dfrac {x}{2}+c$
0%
$x\ \cot \dfrac {x}{2}$
0%
$x\ \sin\dfrac {x}{2}+c$
0%
$x\ \cos \dfrac {x}{2}$

Explanation

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

$\displaystyle \int \dfrac {x+2\sin \left (\dfrac {x}{2}\right) \cos \dfrac {x}{2}}{2\cos^2x/2}dx =\displaystyle \int \dfrac {xdx}{2\cos^2 (x/2)}+\displaystyle \int tan \dfrac {x}{2}dx$

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

integrate first integral by parts

$U=x$

$du=dx$

$du=\dfrac {1}{2}\sec^2 x/2 dx$

$u=\tan x/2$

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

$=x\tan (x/2)+c$

Solve

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

0%
$\dfrac{(x-1)}{2}(x-\dfrac{cos2x}{2})+C$
0%
$\dfrac{(x-1)}{2}(x-\dfrac{sin2x}{2})+C$
0%
$\dfrac{(x+1)}{2}(x-\dfrac{sin2x}{2})+C$
0%
$None\ of\ these$

Explanation

Solving By parts, We get

$\int xsin^2xdx$

$\implies \dfrac{x}{2}\int(1-cos2x)dx-\dfrac{1}{2}\int1.(1-cos2x)dx$

$\implies \dfrac{(x-1)}{2}\int(1-cos2x)dx$

$\implies \dfrac{(x-1)}{2}(x-\dfrac{sin2x}{2})+C$

Solve

$\int^{3}_{2}\dfrac {x}{x^{2}-1}dx$

0%
$\dfrac{1}{2}ln(\dfrac{5}{3})$
0%
$\dfrac{1}{2}ln(\dfrac{8}{3})$
0%
$\dfrac{1}{2}ln(\dfrac{4}{3})$
0%
None of these

Explanation

Let,

$x^2-1=t$

$\implies 2x.dx=dt$

Substituting values , we get

$\implies \int \dfrac{dt}{2t}$

$\implies \dfrac{1}{2}lnt$

$\implies \dfrac{1}{2}[ln(x^2-1)]^3_2$

$\implies \dfrac{1}{2}ln(\dfrac{8}{3})$

$\int _{ 0 }^{ \pi /6 }{ \dfrac { \sin { x } }{ \cos ^{ 3 }{ x } } dx } =$

0%
$2/3$
0%
$1/3$
0%
$2$
0%
$1/6$

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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