MCQ Questions for Class 12 Commerce Maths Integrals Quiz 3

Evaluate the following definite integral:

$\displaystyle \int_{0}^{1}$

$\sin \left(2 {t} {a} {n}^{-1}\sqrt{\dfrac{1- {x}}{1+ {x}}}\right) {d} {x}$

0%
$\pi$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

Consider, $I=\displaystyle \int_{0}^{1}\sin \left (2 tan^{-1}\sqrt{\dfrac{1-x}{1+x}} \right ) dx$

let, $x= \cos \theta$ $\Rightarrow$ $dx=-\sin \theta d \theta$

$\Rightarrow$ $x\rightarrow 0 \Rightarrow \cos \theta \rightarrow \dfrac{\pi}{2}$ and $x\rightarrow 1 \Rightarrow \cos \theta \rightarrow 0$

$\Rightarrow$ $I=\displaystyle \int_{\frac{\pi}{2}}^{0} \sin ( 2 tan^{-1}(tan(\dfrac{\theta}{2})) (-\sin \theta) d\theta$

$\Rightarrow$ $I=\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^2 \theta d \theta = \displaystyle \int_{0}^{\frac{\pi}{2}}\left [ \dfrac{1-\cos 2 \theta }{2} \right ] d\theta$

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

$\Rightarrow$ $I=\left[\left ( \dfrac{\pi}{4}-0 \right )+\dfrac{1}{2} \sin 2 \theta \right]_{0}^{\frac{\pi}{2}}$

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

The value of

$\displaystyle \int_{0}^{\pi /4}\displaystyle \frac{\sin^{ \frac{1}{2}}x}{\cos^{ \frac{5}{2}}x} dx$ is

0%
$0$
0%
$\displaystyle \frac{\pi}{4}$
0%
$1$
0%
$\dfrac{2}{3}$

Explanation

$\displaystyle\int_{0}^{\pi/4} \dfrac{\sin^{\frac{1}{2}}x}{\cos^{\frac{5}{2}}x} dx$

$=\displaystyle\int_{0}^{\pi/4} \dfrac{\tan^{\frac{1}{2}}x}{\cos^2{x}}dx$

$=\displaystyle\int_{0}^{\pi/4} \sec^2{x} \tan^{\frac{1}{2}}dx$

Now, let

$z=\tan{x}$

$\implies dz=\sec^2{x}dx$

And at

$x=0, z=0$ and at

$x=\dfrac{\pi}{4}, z=1$

Now, integration becomes

$\displaystyle\int_{0}^{1} \sqrt{z} dz$

$=\dfrac{2}{3} \left[z^{\frac{3}{2}} \right]_{0}^{1}$

$=\cfrac 23$

$\displaystyle \int_{0}^{\pi/2}(2\tan\frac{x}{2}+x\sec^{2}\frac{x}{2})dx=$

0%
$\pi$
0%
$\pi/2$
0%
2 $\pi/3$
0%
$\pi/6$

Explanation

$\int_{0}^{\dfrac{\pi}{2}}\left ( 2 tan \dfrac{x}{2} + x sec^2 \dfrac{x}{2} \right ) dx$
$=2 \int_{0}^{\dfrac{\pi}{2}}\left ( tan \dfrac{x}{2} + \dfrac{1}{2} x sec^2 \dfrac{x}{2} \right ) dx$
$=2 \int_{0}^{\dfrac{\pi}{2}}d\left ( x tan \dfrac{x}{2} \right )$
$=2 \left ( x tan \dfrac{x}{2} \right ) \int_{0}^{\dfrac{\pi}{2}}$
$=2 \left [ \left ( \dfrac{\pi}{2} 1 \right ) - 0 \right ]$
$=\pi$

$\int_{2}^{3}\dfrac{2-x}{\sqrt{5x-6-x^{2}}} dx =$

0%
$\pi /2$
0%
$-\pi /2$
0%
$-\pi /3$
0%
$\pi$

Explanation

Let

$I=\int { \cfrac { 2-x }{ \sqrt { 5x-6-{ x }^{ 2 } } } dx } =\int { \cfrac { 2-x }{ \sqrt { \cfrac { 1 }{ 4 } -\left( x-\cfrac { 5 }{ 2 } \right) ^{ 2 } } } dx }$
Substitute

$t=x-\cfrac { 5 }{ 2 } \Rightarrow dt=dx$

$I=\int { \cfrac { -t-\cfrac { 1 }{ 2 } }{ \sqrt { \cfrac { 1 }{ 4 } -{ t }^{ 2 } } } dt }$
Substitute

$t=\cfrac { sinu }{ 2 } \Rightarrow dt=\cfrac { cosu }{ 2 }du$

$I=\int { \left( -\cfrac { sinu }{ 2 } -\cfrac { 1 }{ 2 } \right) du } =\cfrac { cosu }{ 2 } -\cfrac { u }{ 2 } =\cfrac { cos\left( { sin }^{ -1 }2t \right) }{ 2 } -\cfrac { { sin }^{ -1 }2t }{ 2 } \\ =\cfrac { cos\left( { sin }^{ -1 }2\left( x-\cfrac { 5 }{ 2 } \right) \right) }{ 2 } -\cfrac { { sin }^{ -1 }2\left( x-\cfrac { 5 }{ 2 } \right) }{ 2 } =\cfrac { cos\left( { sin }^{ -1 }\left( 2x-5 \right) \right) }{ 2 } -\cfrac { { sin }^{ -1 }\left( 2x-5 \right) }{ 2 }$
Therefore

$\int _{ 2 }^{ 3 }{ Idx= } { \left[ \cfrac { cos\left( { sin }^{ -1 }\left( 2x-5 \right) \right) }{ 2 } -\cfrac { { sin }^{ -1 }\left( 2x-5 \right) }{ 2 } \right] }_{ 2 }^{ 3 }=0-\cfrac { \pi }{ 4 } -0-\cfrac { \pi }{ 4 } =-\cfrac { \pi }{ 2 }$

Evaluate the following definite integral:

$\displaystyle \int^{\pi /4}_{-\pi /4}\log(\cos x+\sin x)dx$

0%
$\pi$ log2
0%
$-\pi$ log2
0%
$-\displaystyle \frac{\pi}{4} \log 2$
0%
$\pi^{2}\log 2$

Explanation

Consider, $\displaystyle I=\int_{-\pi/4}^{\pi /4}log (cos x + sin x)dx$

$I=\displaystyle \int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}}log \left [ \sqrt{2} sin \left ( \dfrac{\pi}{4}+ x \right ) \right ] dx$

let, $\dfrac{\pi}{4}+x=t$ $\Rightarrow$ $dx=dt$

So, $x \rightarrow \dfrac{-\pi}{4} \Rightarrow t \rightarrow 0$ and $x \rightarrow \dfrac{\pi}{4} \Rightarrow t \rightarrow \dfrac{\pi}{2}$

$\Rightarrow$ $\displaystyle I=\int_{0}^{\dfrac{\pi}{2}} \log (\sqrt{2} \sin t)dt$

$\Rightarrow$ $\displaystyle I=\int_{0}^{\dfrac{\pi}{2}} \log (\sqrt{2}) + \int_{0}^{\dfrac{\pi}{2}} \log (\sin t)dt$

$\Rightarrow$ $\displaystyle I=\dfrac{\pi}{2}\log (\sqrt{2})-\dfrac{\pi}{2} \log (2)$

$\Rightarrow$ $\displaystyle I=\dfrac{\pi}{4}\log 2 -\dfrac{\pi}{2} \log (2)$

$\Rightarrow$ $\displaystyle I=-\dfrac{\pi}{4} \log 2$

Evaluate the integral

$\displaystyle \int_{0}^{a}\sqrt{\frac{a+x}{a-x}}dx$

0%
$\displaystyle \frac{a}{2}(\pi+2)$
0%
$\displaystyle \frac{a}{2}(\pi-2)$
0%
$\displaystyle \frac{a}{3}(\pi+2)$
0%
$\displaystyle \frac{a}{2}(\pi+3)$

Explanation

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

$x=a cos \theta =dx -a sin \theta d \theta$

$=\displaystyle \int_{\dfrac{\pi}{2}}^{a}\sqrt{\dfrac{a+ a cos \theta}{a- a cos \theta}}(-a sin \theta)d \theta$

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

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

$=a\displaystyle \int_{0}^{\dfrac{\pi}{2}}2 cos^2 \dfrac{\theta}{2} d \theta = a \displaystyle \int_{0}^{\dfrac{\pi}{2}}(cos \theta -1) d \theta$

$=a \left ( sin \theta\right )_{0}^{\frac{\pi}{2}} - (\theta)_{0}^{\frac{\pi}{2}}$

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

$=a\left (\dfrac{\pi}{2}+1 \right )$

$=\dfrac{a}{2}(\pi+2)$

$\displaystyle \int_{0}^{1}\sqrt{\frac{\mathrm{x}}{1-\mathrm{x}^{3}}}\mathrm{d}\mathrm{x}=$

0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{6}$
0%
$\displaystyle \frac{\pi}{2}$

Explanation

Let

$I=\int { \sqrt { \cfrac { x }{ 1-{ x }^{ 3 } } } dx }$
Substitute

$t={ x }^{ 3/2 }\Rightarrow dt=\cfrac { 3\sqrt { x } }{ 2 } dx$
We get

$I=\cfrac { 2 }{ 3 } \int { \cfrac { 1 }{ \sqrt { 1-{ t }^{ 2 } } } dt } =\cfrac { 2 }{ 3 } { sin }^{ -1 }t=\cfrac { 2 }{ 3 } { sin }^{ -1 }{ x }^{ 3/2 }$
Therefore

$\int _{ 0 }^{ 1 }{ Idx } =\int _{ 0 }^{ 1 }{ \cfrac { 2 }{ 3 } { sin }^{ -1 }{ x }^{ 3/2 } } =\cfrac { 2 }{ 3 } \left( \cfrac { \pi }{ 2 } -0 \right) =\cfrac { \pi }{ 3 }$

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

0%
9/2
0%
27/4
0%
116/15
0%
112/15

Explanation

Initially, we perform indefinite integration,
Let 1 + x = t
dx = dt
Putting the values back,

$\int (t-1){t}^{0.5} dt$
=

$\int {t}^{1.5} - {t}^{0.5} dt$
=

$\frac{2}{5} {t}^{2.5} - \frac{2}{3}{t}^{1.5}$
Putting the value of t back,

$\frac{2}{5} {(1+x)}^{2.5} - \frac{2}{3}{(1+x)}^{1.5}$
Now applying the limits to the expression,

$(\frac{2}{5} {(1+3)}^{2.5} - \frac{2}{3}{(1+3)}^{1.5}) - (\frac{2}{5} {(1+0)}^{2.5} - \frac{2}{3}{(1+0)}^{1.5})$
=

$\displaystyle \frac{64}{5} - \displaystyle \frac{16}{3} - \displaystyle \frac{2}{5} + \displaystyle \frac{2}{3}$

$\displaystyle \frac{62}{5} - \displaystyle \frac{14}{3}$
=

$\displaystyle \frac{116}{15}$

$\displaystyle \int_{\pi^{2}/16}^{\pi^{2}/4}\frac{\sin\sqrt{x}}{\sqrt{x}}dx=$

0%
$\sqrt{2}$
0%
$1/\sqrt{2}$
0%
2 $\sqrt{2}$
0%
${\pi}/2$

Explanation

Let

${x}^{0.5} = t$

$\dfrac{1}{2{x}^{0.5}} dx = dt$
Putting the values back in the given expression,

$\int 2 sin t dt$
Integrating the expression, indefinitely
=

$-2 cos t + c$
Putting back the value of t =

${x}^{0.5}$
=

$- 2cos({x}^{0.5}) + c$
Applying the limits,

$(2 cos (\dfrac{\pi}{4})) - (2 cos (\dfrac{\pi}{2}))$
=

${2}^{0.5}$

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

0%
log 2
0%
2 log 2
0%
3 log 3
0%
$\displaystyle \frac{1}{2}$ log2

Explanation

$\int_{0}^{1} \dfrac{dx}{(x+\sqrt{x})}$
$x=t^2$
$\int_{0}^{1}\dfrac{2t dt}{t^2+ t}=2\int_{0}^{1}\dfrac{dt}{t+1}=2 log (t+1)\int_{0}^{1}$
$=2 log (2)-2log (1)$
$=2 log 2$

$\displaystyle \int_{0}^{k}\frac{dx}{2+8x^{2}}=\frac{\pi}{16}$ then

$\mathrm{k}=$

0%
1
0%
1/2
0%
$\pi/2$
0%
2

Explanation

$\int_{0}^{k}\dfrac{dx}{2+ 8x^2}=\dfrac{\pi}{16}$
$\dfrac{1}{2}\int_{0}^{k}\dfrac{dx}{1+(2x)^2}=\dfrac{\pi}{16}$
$\dfrac{1}{2}(\dfrac{1}{2})tan^{-1}(2x)\int_{0}^{k}=\dfrac{\pi}{16}$
$\dfrac{1}{4}tan^{-1}(2k)=\dfrac{\pi}{16}$
$tan^{-1}(2k)=\dfrac{\pi}{4}$
$2k=1$
$k=\dfrac{1}{2}$

$\displaystyle {I}_{ {n}}=\int^{\pi/2 }_{\pi/4}( {T} {a} {n}\theta)^{- {n}}. {d}\theta$ for

$( {n}>1)$

then

$I_{n}+I_{n+2} = ?$

0%
$\displaystyle \frac{1}{\mathrm{n}+1}$
0%
$\displaystyle \frac{-1}{\mathrm{n}+1}$
0%
$\displaystyle \frac{1}{\mathrm{n}-1}$
0%
$\displaystyle \frac{-1}{\mathrm{n}-1}$

Explanation

$I_{n}=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}$

$I_{n}+I_{n+2}=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}[(cot\ \theta)^{n}+(cot\ \theta)^{n+2}]d\ \theta$

$=\displaystyle \int_{^{\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}cosec^{2}d\ \theta$

$=\displaystyle \int_{^{-\pi}/_{4}}^{^{\pi}/_{2}}(cot\ \theta)^{n}d(cot\ \theta)$

$-\dfrac{(cot\ \theta)^{n+1}}{n+1}|_{^{-\pi}/_{4}}^{^{\pi}/_{2}}$

$-\left [ 0-\dfrac{1}{n+1} \right ]$

$=\dfrac{1}{n+1}$

Evaluate the integral

$\displaystyle \int_{1}^{\sqrt[7]{2}}\frac{1}{x(2x^{7}+1)} dx$

0%
$\log \dfrac{6}{5}$
0%
$6 \log \dfrac{6}{5}$
0%
$\dfrac{1}{7} \log \dfrac{6}{5}$
0%
$\dfrac{1}{5} \log \dfrac{6}{5}$

Explanation

$\displaystyle \int_{1}^{\sqrt[7]{2}}\dfrac{1}{x(2x^{7}+1)}dx=\displaystyle \int_{1}^{\sqrt[7]{2}}\dfrac{1}{x^{8}(2+1/x^{7})dx}$

$1/x^{7}= 1\Rightarrow\dfrac{-7}{x^{8}}dx=dt$

$\Rightarrow \dfrac{dx}{x^{8}}-\dfrac{dx}{7}$

$\Rightarrow \dfrac{1}{-7}\displaystyle \int_{1}^{1/2}\dfrac{dt}{(2+t)}=\dfrac{1}{7}\displaystyle \int_{1/2}^{1}\left [ \log(2+t) \right ]_{1/2}^{1}$

$=\dfrac{1}{7} \left [ \log(3)-\log\dfrac{5}{2} \right ]$

$=\dfrac{1}{7} \log \dfrac{6}{5}$

Evaluate:

$\displaystyle \int_{0}^{\pi /4}\tan^{5}xdx$

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

Explanation

$\tan^5{x}=\tan^3{x}(\sec^2{x}-1)=\tan^3{x}\sec^2{x}-\tan^3{x}$

$=\tan^3{x}\sec^2{x}-\tan{x}(\sec^2{x}-1)$

$\implies \tan^5{x}=\tan^3{x}\sec^2{x}-\tan{x}\sec^2{x}+\tan{x}$

Hence, integration becomes:-

$\displaystyle\int_{0}^{\pi/4}\tan^3{x}\sec^2{x}dx-\int_{0}^{\pi/4}\tan{x}\sec^2{x}dx+\int_{0}^{\pi/4}\tan{x}dx$

Now, let

$z=\tan{x}\implies dz=\sec^2{x}dx$

And when,

$x=0,z=0$ and when

$x=\dfrac{\pi}{4}, z=1$

$\implies\displaystyle\int_{0}^{1}z^3dz-\int_{0}^{1}zdz+\int_{0}^{\pi/4}\tan{x}dx$

$=\dfrac{1}{4}\left[z^4\right]_{0}^{1}-\dfrac{1}{2}\left[z^2\right]_{0}^{1}+\left[\log{|\sec{x}|}\right]_{0}^{\pi/4}$

$=\dfrac{1}{4}-\dfrac{1}{2}+\log{\sqrt{2}}-\log{1}$

$=\dfrac{1}{2}\log{2}-\dfrac{1}{4}$

$\displaystyle \mathrm{U}_{\mathrm{n}}=\int^{\pi /4}_{0} \mathrm{t}\mathrm{a}\mathrm{n}^{\mathrm{n}}\theta \mathrm{d}\theta$ , then

$\mathrm{u}_{10}+\mathrm{u}_{12}$ is equal to:

0%
$\displaystyle \frac{1}{10}$
0%
$\displaystyle \frac{1}{12}$
0%
$\displaystyle \frac{1}{11}$
0%
$\displaystyle \frac{1}{22}$

Explanation

$U_n=\displaystyle\int_{0}^{\pi/4} \tan^n {\theta}d\theta$

$U_{10}+U_{12}=\displaystyle\int_{0}^{\pi/4} \tan^{10} {\theta} d\theta+ \int_{0}^{\pi/4} \tan^{12} {\theta}d\theta$

$=\displaystyle\int_{0}^{\pi/4} \tan^{10}{\theta} (1+\tan^2{\theta})d\theta$

$=\displaystyle\int_{0}^{\pi/4} \tan^{10}{\theta} \sec^2\theta d\theta$

Let,

$z=\tan\theta$

$\implies dz=\sec^2\theta d\theta$

And when

$\theta=0,z=0$ and when

$\theta=\dfrac{\pi}{4},z=1$

Now, integration becomes:-

$\displaystyle\int_{0}^{1} z^{10}dz=\dfrac{1}{11} \left[z^{11}\right]_{0}^{1}$

$=\dfrac{1}{11}$

Evaluate:

$\displaystyle \int_{0}^{1}\frac{x^{3}}{1+x^{8}} dx$

0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{8}$
0%
$\displaystyle \frac{\pi}{16}$
0%
$\displaystyle \frac{\pi}{2}$

Explanation

Let,

$z=x^4\implies dz=4x^3dx\implies x^3dx=\dfrac{dz}{4}$

When

$x=0,z=0$ and when

$x=1,z=1$

Hence,integration becomes:-

$\displaystyle\dfrac{1}{4}\int_{0}^{1} \dfrac{dz}{1+z^2}$

$=\dfrac{1}{4}\displaystyle\left[\tan^{-1}{z}\right]_{0}^{1}$

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

$=\dfrac{\pi}{16}$

Hence, answer is option-(C).

Evaluate:

$\int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ ({ \tan }^{ 4 }x+{ \tan }^{ 2 }x)dx }$

0%
$1$
0%
$1/2$
0%
$1/3$
0%
$1/4$

Explanation

$\int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ ({ \tan }^{ 4 }x+{ \tan }^{ 2 }x)dx } \\ = \int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ { \tan }^{ 2 }x({ \tan }^{ 2 }x+1)dx } \\ = \\ = \int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ { \tan }^{ 2 }x(\sec ^{ 2 }x)dx }$

We know that:

$\int { [f(x)]^{ n } } f'(x)dx=\dfrac { [f(x)]^{ n+1 } }{ n+1 } +C\\ = \int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ { \tan }^{ 2 }x\; \times \; \sec ^{ 2 }x\; dx } \\ = \left[\dfrac { { \tan }^{ 3 }x }{ 3 } \right]_{ 0 }^{ \tfrac { \pi }{ 4 } }\\ = \dfrac { 1 }{ 3 } [\tan ^{ 3 }(\tfrac { \pi }{ 4 }) -\tan ^{ 3 }0]\\ = \dfrac { 1 }{ 3 } [1-0]\\ = \dfrac { 1 }{ 3 }$

Evaluate:

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

0%
$\pi/8$
0%
$\pi/4$
0%
$\pi/2$
0%
$\pi$

Explanation

$\displaystyle\int_{1/2}^{1} \dfrac{1}{\sqrt{x-x^2}}dx$

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

we know that

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

Now, our integration becomes:-

$\displaystyle\left[\sin^{-1} \left(\dfrac{x-\dfrac{1}{2}}{\dfrac{1}{2}}\right)\right]_{1/2}^{1}$

$=\left[\sin^{-1} {1}-\sin^{-1} {0}\right]$

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

Hence,answer is option_(C).

$f(x)=\begin{vmatrix}\sin x+\sin2x+\sin3x & \sin2x & \sin3x\\ 3+4\sin x & 3 & 4\sin x\\ 1+\sin x & \sin x & 1\end{vmatrix}$ , then the value of

$\displaystyle \int_{0}^{\frac{\pi}2}f(x)dx$ , is

0%
$3$
0%
$\dfrac23$
0%
$\dfrac13$
0%
$0$

Explanation

Using Determinant properties:

$C_1 \rightarrow C_1 - (C_2+C_3)$

We get,

$\begin{vmatrix} \sin\ x &\sin\ 2x &\sin\ 3x \\0 &3 &4\sin\ x \\0 &\sin\ x &1 \end{vmatrix}$

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

$=3\ \sin\ x-4\ \sin^{3}x$

$=\sin\ 3x$

$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin\ 3x\ dx=\left.\dfrac{-\cos\ 3x}{3}\right|_{0}^{\frac{\pi}{2}}=0+\dfrac{1}{3}$

$=\dfrac{1}{3}$

Evaluate:

$\displaystyle \int_{0}^{\pi /4}\tan^{6}xdx$

0%
$\displaystyle \frac{13}{15}-\frac{\pi}{4}$
0%
$\displaystyle \frac{13}{15}+\frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{4}-\frac{2}{3}$
0%
$\displaystyle \frac{13}{15}-4\pi$

Explanation

$\tan^6{x}=\tan^4{x}(\sec^2{x}-1)=\tan^4{x}\sec^2{x}-\tan^4{x}$

$\implies\tan^6{x}=\tan^4{x}\sec^2{x}-\tan^2{x}(\sec^2{x}-1)$

$\implies\tan^6{x}=\tan^4{x}\sec^2{x}-\tan^2{x}\sec^2{x}+\tan^2{x}$

$\implies\tan^6{x}=\tan^4{x}\sec^2{x}-\tan^2{x}\sec^2{x}+\sec^2{x}-1$

Now, let

$z=\tan{x}\implies dz=\sec^2{x}dx$ And when

$x=0, z=0$ and when

$x=\dfrac{\pi}{4}, z=1$

Hence,

$\displaystyle\int_{0}^{\pi/4}\tan^6{x}dx=\int_{0}^{\pi/4}\tan^4{x}\sec^2{x}dx-\int_{0}^{\pi/4}\tan^2{x}\sec^2{x}dx+\int_{0}^{\pi/4} \sec^2{x}dx-\int_{0}^{\pi/4} 1.dx$

$=\displaystyle\int_{0}^{1}z^4dz-\int_{0}^{1}z^2dz+\int_{0}^{1}dz-\int_{0}^{\pi/4}1.dx$

$=\left[\dfrac{z^5}{5}\right]_{0}^{1}-\left[\dfrac{z^3}{3}\right]_{0}^{1}+\left[z\right]_{0}^{1}-\left[x\right]_{0}^{\pi/4}$

$=\dfrac{1}{5}-\dfrac{1}{3}+1-\dfrac{\pi}{4}$

$=\dfrac{13}{15}-\dfrac{\pi}{4}$

$I_{\mathrm{n}}=\displaystyle \int_{0}^{\dfrac{\pi}{4}}$

$\tan^{n} xdx$ , then

$\displaystyle \lim_{n\rightarrow\infty}n[I_{n}+I_{n+2}]=$

0%
$\displaystyle \dfrac{1}{2}$
0%
$1$
0%
$\infty$
0%
$0$

Explanation

Given,

$I_{n}=\int_{0}^{\pi /4}\tan ^{n}x dx$

and we have to find

$\lim_{n\rightarrow \infty }n[I_{n}+I_{n+2}]$

Let us first solve the function and get the answer in terms of

$n$ and then apply limit .

Now

$n[I_{n}+I_{n+2}]$

$=n[\int_{0}^{\pi /4}\tan^{n}x dx$

$+\int_{0}^{\pi /4}\tan^{n+2}xdx]$

$=n[\int_{0}^{\pi/4}(\tan^{n}x+\tan^{n+2}x)dx]$

Since the limits of both the integrals is same it can be combined into one integral

$=n[\int_{0}^{\pi/4}(\tan^{n}x(1+\tan^{2}x)dx]$

Since

$1+\tan^{2}x=\sec^{2}x$

Therefore,

$=n[\int_{0}^{\pi /4}\tan^{n}x\sec^{2}x]$

Taking

$\tan x=t$

$dx=\sec^{2}xdt$

Upper limit becomes

$\tan(\pi/4)=1$ and lower limit becomes

$\tan(0)=0$

Using this the integrand becomes

$n\int_{0}^{1} t^{n}dt$

$=n(t^{n+1}/n+1)_{0}^{1}$

Substituting limits, we get

$\dfrac {n}{n+1}$

Now applying limit tending to infinity to this value

$\lim_{n\rightarrow \infty }n[I_{n}+I_{n+2}]$

$=\lim_{n \to \infty }\dfrac {n}{n+1}$

$=1$

The value of

$\displaystyle \int_{3}^{5}\frac{x^{2}}{x^{2}-4} dx$ is

0%
$2(1-\displaystyle \mathrm{l} \mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))$
0%
$2(1+\displaystyle \mathrm{l}\mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))$
0%
$2(1+4\log_{\mathrm{e}}3-4\log_{\mathrm{e}}7+4\log_{\mathrm{e}}5)$
0%
$2(1-\displaystyle \tan^{-1}(\frac{15}{7}))$

Explanation

$\displaystyle \int_{3}^{5}\dfrac{x^{2}}{x^{2}-4}dx$

$\displaystyle =\int_{3}^{5}1dx+4\int_{3}^{5}\dfrac{1}{x^{2}-4}dx=2+4.\dfrac{1} {2}log(\dfrac{x-2}{x+2})|_{3}^{5}$

$\displaystyle 2+2[log(3/7)-log(1/5)]$

$\displaystyle =2(1+log(\dfrac{15}{7}))$
Hence, option 'B' is correct.

$\displaystyle \int_{0}^{1}\frac{\tan^{-1}x}{x}dx=k\int_{0}^{\pi/2}\frac{x}{\sin x}dx$ , then the value of

$k$ is

0%
$1$
0%
$\displaystyle \frac{1}{4}$
0%
$4$
0%
${2}$

Explanation

Let

$I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { x }{ sinx } dx }$

Substitute

$tan\left( \cfrac { x }{ 2 } \right) =t\Rightarrow \cfrac { 1 }{ 2 } { sec }^{ 2 }\left( \cfrac { x }{ 2 } \right) dx=dt$

gives

$sinx=\cfrac { 2t }{ 1+{ t }^{ 2 } } ,dx=\cfrac { 2dt }{ 1+{ t }^{ 2 } }$

$I=\int _{ 0 }^{ 1 }{ \cfrac { 2{ tan }^{ -1 }t }{ \cfrac { 2t }{ 1+{ t }^{ 2 } } } \cfrac { 2dt }{ 1+{ t }^{ 2 } } } =2\int _{ 0 }^{ 1 }{ \cfrac { { tan }^{ -1 }t }{ t } dt } =2\int _{ 0 }^{ 1 }{ \cfrac { { tan }^{ -1 }x }{ x } dx } \\ \Rightarrow k=2$

The value of the integral

$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{4}} dx$ is

0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{\sqrt{2}}$
0%
$\displaystyle \frac{\pi}{2\sqrt{2}}$
0%
$\pi/4$

Explanation

$I=\int _{ 0 }^{ \infty }{ \cfrac { 1 }{ x^{ 4 }+1 } dx } =\int _{ 0 }^{ \infty }{ \left( \cfrac { \sqrt { 2 } x-2 }{ 4\left( -x^{ 2 }+\sqrt { 2 } x-1 \right) } +\cfrac { \sqrt { 2 } x+2 }{ 4\left( x^{ 2 }+\sqrt { 2 } x+1 \right) } \right) dx } \\ =\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x+2 }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } +\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x-2 }{ -x^{ 2 }+\sqrt { 2 } x-1 } dx } \\ =\cfrac { 1 }{ 4\sqrt { 2 } } \int { \cfrac { \sqrt { 2 } +2x }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } +\cfrac { 1 }{ 4 } \int { \cfrac { 1 }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } +\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x-2 }{ -x^{ 2 }+\sqrt { 2 } x-1 } dx }$
Let

${ I }_{ 1 }=\cfrac { 1 }{ 4\sqrt { 2 } } \int { \cfrac { \sqrt { 2 } +2x }{ x^{ 2 }+\sqrt { 2 } x+1 } dx }$
Substituting

$t=x^{ 2 }+\sqrt { 2 } x+1\Rightarrow dt=\left( 2x+\sqrt { 2 } \right) dx$ , we get

${ I }_{ 1 }=\cfrac { 1 }{ 4\sqrt { 2 } } \int { \cfrac { 1 }{ t } dt } =\cfrac { \log { t } }{ 4\sqrt { 2 } } =\cfrac { \log { \left( x^{ 2 }+\sqrt { 2 } x+1 \right) } }{ 4\sqrt { 2 } }$
Now let

${ I }_{ 2 }=\cfrac { 1 }{ 4 } \int { \cfrac { 1 }{ x^{ 2 }+\sqrt { 2 } x+1 } dx }$
Substituting

$t=x+\cfrac { 1 }{ \sqrt { 2 } } \Rightarrow dt=dx$ , we get

${ I }_{ 2 }=\cfrac { 1 }{ 4 } \int { \cfrac { 1 }{ t^{ 2 }+\cfrac { 1 }{ 2 } } dt } =\cfrac { 1 }{ 4 } \int { \cfrac { 2 }{ 2t^{ 2 }+1 } } =tan^{ -1 }\left( \cfrac { \sqrt { 2 } x+1 }{ 2\sqrt { 2 } } \right)$
And let

${ I }_{ 3 }=\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x-2 }{ -x^{ 2 }+\sqrt { 2 } x-1 } dx }$

$=\cfrac { 1 }{ 4 } \int { \left( \cfrac { \sqrt { 2 } x-2 }{ \sqrt { 2 } \left( -x^{ 2 }+\sqrt { 2 } x-1 \right) } -\cfrac { 1 }{ -x^{ 2 }+\sqrt { 2 } x-1 } \right) dx } \\ =-\cfrac { log\left( -x^{ 2 }+\sqrt { 2 } x-1 \right) }{ 4\sqrt { 2 } } -\cfrac { tan\left( 1-\sqrt { 2 } x \right) }{ 2\sqrt { 2 } } \\$
Therefore resubstituting

${ { I }=I }_{ 1 }+{ I }_{ 2 }+{ I }_{ 3 }$

$\int _{ 0 }^{ \infty }{ \cfrac { 1 }{ x^{ 4 }+1 } dx } =\left[ \cfrac { \log { \left( x^{ 2 }+\sqrt { 2 } x+1 \right) } }{ 4\sqrt { 2 } } +tan^{ -1 }\left( \cfrac { \sqrt { 2 } x+1 }{ 2\sqrt { 2 } } \right) -\cfrac { log\left( -x^{ 2 }+\sqrt { 2 } x-1 \right) }{ 4\sqrt { 2 } } -\cfrac { tan\left( 1-\sqrt { 2 } x \right) }{ 2\sqrt { 2 } } \right] _{ 0 }^{ \infty }\\ =\cfrac { \pi }{ 2\sqrt { 2 } }$

The value of

$x$ such that

$\displaystyle \int_{\sqrt{2}}^{\mathrm{x}}\frac{1}{x\sqrt{x^{2}-1}}dx=\frac{\pi}{12}$ is

0%
$\mathrm{x}=3$
0%
$\mathrm{x}=4$
0%
$\mathrm{x}=1$
0%
$\mathrm{x}=2$

Explanation

Let

$x=\sec{\theta} \implies dx=\sec{\theta}\tan{\theta}d\theta$

When

$x=\sqrt{2}, \theta=\dfrac{\pi}{4}$ and when

$x=x, \theta=\sec^{-1} {x}$

Now,

$\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} \dfrac{1}{\sec{\theta} \sqrt{\sec^2{\theta} -1}}\sec{\theta}\tan{\theta}d\theta=\dfrac{\pi}{12}$

$\implies\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} \dfrac{1}{\sec{\theta} \tan{\theta}}\sec{\theta}\tan{\theta}d\theta=\dfrac{\pi}{12}$

$\implies\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} d\theta=\dfrac{\pi}{12}$

$\implies\sec^{-1} {x} - \dfrac{\pi}{4}=\dfrac{\pi}{4}$

$\implies\sec^{-1} {x}=\dfrac{\pi}{4}+\dfrac{\pi}{12}=\dfrac{\pi}{3}$

$\implies x=\sec{\dfrac{\pi}{3}}=2$

$\implies x=2$

Hence,answer is option-(D).

$\displaystyle \int e^{tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=$

0%
$\displaystyle x^{2}e^{\tan^{-1}x}+c$
0%
$\displaystyle x e^{\tan^{-1}x}+c$
0%
$\displaystyle e^{\tan^{-1}x}+c$
0%
$\displaystyle \frac{1}{2}e^{\tan^{-1}x}+c$

Explanation

Let

$I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx$

$=\displaystyle \int e^{\tan^{-1}x}\left[1+\frac {x}{1+x^2}\right]dx$

$=\displaystyle \int \left[e^{\tan^{-1}x}+\frac {xe^{\tan^{-1}x}}{1+x^2}\right]dx$
For

$f(x)=e^{\tan^{-1} x}$ , above integral

$I$ reduces in the form of

$\displaystyle \int [f(x)+x f^{'}(x)]dx$

But we know that,

$\dfrac{d}{dx} (xf(x)+c) = \displaystyle [f(x)+x f^{'}(x)]$

$\therefore \displaystyle \int [f(x)+x f^{'}(x)]dx=xf(x)+c$

$\therefore I = \displaystyle xe^{\tan^{-1}x}+c$

$\therefore I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=xe^{\tan ^{-1} x}+c$
Hence, option 'B' is correct.

$f(x)=\left\{\begin{array}{l}e^{\cos x}\sin x, for |x|\leq 2\\2 ; otherwise\end{array}\right.$ , then

$\displaystyle \int_{-2}^{3}f(x)dx$ is

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

$I=\displaystyle\int_{-2}^{3}f(x)dx$

$I=\displaystyle\int_{-2}^{2}e^{\cos{x}}\sin{x}dx+\int_{2}^{3}2dx$

consider,

$f(x)=e^{\cos{x}}\sin{x}$

so,

$f(-x)=e^{\cos{(-x)}}\sin{(-x)}$

$\Rightarrow$

$f(-x)=-e^{\cos{x}}\sin{x}=-f(x)$

$\Rightarrow$

$e^{\cos{x}}\sin{x}$ is an odd function

hence,

$\displaystyle\int_{-2}^{2}e^{\cos{x}}\sin{x}dx=0$

Thus,

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

$I=2\left[x\right]_{2}^{3}=2(3-2)$

$I=2$

Hence, answer is option-(C).

The value of the integral

$\displaystyle \int_{0}^{1}\frac{1}{(x^{2}+1)^{3/2}} dx$ is

0%
$\displaystyle \frac{3}{\sqrt {2}}$
0%
$\displaystyle \frac { 1 }{ \sqrt { 2 } }$
0%
$1$
0%
${\sqrt{2}}$

Explanation

Let

$I=\int { \cfrac { 1 }{ \left( x^{ 2 }+1 \right) ^{ \cfrac { 3 }{ 2 } } } dx }$
Substituting

$x=tant\Rightarrow dx=sec^{ 2 }tdt$ , we get

$I=\int { cot\left( t \right) dt } =sin\left( t \right) =sin\left( tan^{ -1 }x \right) =\cfrac { x }{ \sqrt { x^{ 2 }+1 } }$
Therefore,

$\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ \left( x^{ 2 }+1 \right) ^{ \cfrac { 3 }{ 2 } } } dx } =\left[ \cfrac { x }{ \sqrt { x^{ 2 }+1 } } \right] _{ 0 }^{ 1 }=\cfrac { 1 }{ \sqrt { 2 } }$

Area bounded by

$\mathrm{y}=\{\mathrm{x}\},\{.\}$ is fractional part of function and

$\mathrm{x}=\pm 1$ is in sq. units

0%
1
0%
2
0%
3
0%
4

Explanation

Area

$\displaystyle =2\int _{ 0 }^{ 1 }{ xdx } =2{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }=2.\frac { 1 }{ 2 } =1$

$\displaystyle \int^{x}_{\log 2} \dfrac{1}{\sqrt{e^{x}-1}}dx = \dfrac{\pi}{6}$

then

$x$ is equal to?

0%
$e^{2}$
0%
$1/e$
0%
$\log 4$
0%
$\log 2$

Explanation

Let

$\displaystyle I=\int { \cfrac { 1 }{ \sqrt { e^{ x }-1 } } dx }$
Substituting

$t=e^{ x }\Rightarrow dt=e^{ x }dx$ , we get

$\displaystyle I=\int { \cfrac { 1 }{ t\sqrt { t-1 } } dt }$
Again substituting

$u=t-1\Rightarrow du=dt$ , we get

$\displaystyle I=\int { \cfrac { 1 }{ \sqrt { u } \left( u+1 \right) } du }$
now substituting

$s=\sqrt { u } \Rightarrow ds=\cfrac { 1 }{ 2\sqrt { u } } du$ , we get

$\displaystyle I=2\int { \cfrac { 1 }{ s^{ 2 }+1 } ds } \\ =2tan^{ -1 }s=2tan^{ -1 }\sqrt { u } =2tan^{ -1 }\sqrt { t-1 } =2tan^{ -1 }\sqrt { e^{ x }-1 }$
Therefore,

$\displaystyle\int _{ \log { 2 } }^{ x }{ 2tan^{ -1 }\left( \sqrt { e^{ x }-1 } \right) } =\cfrac { \pi }{ 6 } \Rightarrow \left[ 2tan^{ -1 }\left( \sqrt { e^{ x }-1 } \right) \right] _{ \log { 2 } }^{ x }=\cfrac { \pi }{ 6 }$

$\therefore x=\log{4}$

Let

$\displaystyle \frac{d}{dx}F\left ( x \right )=\frac{e^{\sin x}}{x},x> 0.$ If

$\displaystyle \int_{1}^{4}\frac{2e^{\sin x^{2}}}{x}dx=F\left ( k \right )-F\left ( 1 \right )$ then one of the possible values of

$\displaystyle k$ is

0%
$4$
0%
$\displaystyle -4$
0%
$16$
0%
none of these

Explanation

$\int _{ 1 }^{ 4 }{ \cfrac { 2{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =F\left( k \right) -F\left( 1 \right)$
Substitute

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

$\int _{ 1 }^{ 16 }{ 2\cfrac { { e }^{ \sin { t } } }{ t } \cfrac { dt }{ 2 } } =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow \int _{ 1 }^{ 16 }{ \cfrac { { e }^{ \sin { t } } }{ t } dt } =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow { \left[ F\left( t \right) \right] }_{ 1 }^{ 16 }=F\left( k \right) -F\left( 1 \right) \\ \Rightarrow F\left( 16 \right) -F\left( 1 \right) =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow k=16$

The value of

$\displaystyle \int_{1}^{2}\left [ f\left \{ g\left ( x \right ) \right \} \right ]^{-1}.{f}'\left \{ g\left ( x \right ) \right \}.{g}'\left ( x \right )dx,$ where

$\displaystyle g\left ( 1 \right )=g\left ( 2 \right ),$ is equal to?

0%
$1$
0%
$2$
0%
$0$
0%
none of these

Explanation

Let

$I=\int _{ 1 }^{ 2 } \left[ f\left\{ g\left( x \right) \right\} \right] ^{ -1 }.{ f }'\left\{ g\left( x \right) \right\} .{ g }'\left( x \right) dx$

Substitute

$\left[ f\left\{ g\left( x \right) \right\} \right] =t\Rightarrow { f }'\left\{ g\left( x \right) \right\} .{ g }'\left( x \right) dx=dt$

$\displaystyle \therefore I=\int _{ f\left( g\left( 1 \right) \right) }^{ f\left( g\left( 2 \right) \right) }{ tdt } ={ \left[ \frac { { t }^{ 2 } }{ 2 } \right] }_{ f\left( g\left( 1 \right) \right) }^{ f\left( g\left( 2 \right) \right) }=0\\ \because g\left( 1 \right) =g\left( 2 \right)$

The value of integral

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

0%
$0$
0%
$\log 7$
0%
$5\log 13$
0%
$none\ of\ these$

Explanation

$\displaystyle I=\int_{0}^{\infty }\frac{x\log xdx}{(1+x^2)^2}$
Let

$x=\frac{1}{t}$

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

$\displaystyle I=-\int_{0}^{\infty }\frac{t\log t}{(1+t^2)^2}dt=-I$
or

$I=0$

Let

$\displaystyle \frac{{df(x)}}{{dx}} = \frac{{{e^{\sin x}}}}{x}, x>0$ . If

$\displaystyle \int_1^4 {\frac{{3{e^{\sin {x^3}}}}}{x}dx = f(k) - f(1)}$ then one of the possible values of k is

0%
$16$
0%
$63$
0%
$64$
0%
$15$

Explanation

$\displaystyle \dfrac{df(x)}{dx}=\displaystyle \dfrac{e^{sin x}}{x} , x > 0, \int_1^4 \dfrac{3e^{sin x^3}}{x}dx=f(k)-f(1)$
substitute $x^3=p$
$dp=3x^2 dx$
limit of p will be rom $1 \rightarrow 64$ as of x was $1 \rightarrow 4$
So, $\displaystyle \int_1^{64} \dfrac{3e^{sin p}}{x} \cdot \dfrac{dp}{3x^2}=\int_1^{64} \dfrac{e^{sin p}}{p}dp=\int_1^{64}df(p)=f(64)-f(1)$

Let

$f(0) = 0$ and

$\displaystyle \int_{0}^{2}{f}'(2t)e^{f(2t)} \:dt=5$ .

Then the value of

$f (4)$ is?

0%
$\log 2$
0%
$\log 7$
0%
$\log 11$
0%
$\log 13$

Explanation

We have

$\displaystyle \int_{0}^{2 }{f}'(2t)e^{f(2t)}dt=5$

Substitute

$e^{f(2t)}=y$

$\therefore 2{f}'(2t)e^{f(2t)}dt=dy$

$\therefore \dfrac{1}{2} \displaystyle \int_{e^{f(0)}}^{e^{f(4)}} dy=5$

$\displaystyle \int_{e^{f(0)}}^{e^{f(4)}} dy=10$

$e^{f(4)}-e^{f(0)}=10$

$e^{f(4)}=10+1=11$

$f(4)=\log 11$

$\displaystyle \int_{\sin \theta }^{\cos \theta }f(x \tan \theta )dx\left ( where \theta \neq \frac{n\pi}{2} ,n\epsilon I\right )$ is equal to

0%
$\displaystyle -\int_{1}^{\tan }f(x\sin \theta )dx$
0%
$\displaystyle -\tan \theta \int_{\sin \theta }^{\cos \theta }f(x)dx$
0%
$\displaystyle \sin \theta \int_{1}^{\tan }f(x\cos \theta )dx$
0%
$\displaystyle \frac{1}{\tan \theta }\int_{\sin \theta }^{\sin \theta \tan \theta}f(x)dx$

Explanation

Let

$I=\int _{ sin\theta }^{ cos\theta }{ f\left( xtan\theta \right) dx }$
Substitute

$xtan\theta =zsin\theta \Rightarrow dx=cos\theta dz$
Therefore

$x=sin\theta \Rightarrow z=tan\theta$ and

$x=cos\theta \Rightarrow z=1$
Gives

$I=\int _{ tan\theta }^{ 1 }{ f\left( zsin\theta \right) dz } =-\int _{ 1 }^{ tan\theta }{ f\left( zsin\theta \right) dz } =-\int _{ 1 }^{ tan\theta }{ f\left( xsin\theta \right) dx }$

The possible negative values of real number 'a' such that

$\overset { 0 }{ \underset { a }{ \int } } (9^{-2t}-2.9^{-t})dt\geq 0$ is

0%
-2013
0%
$-\frac{2013}{2}$
0%
-2
0%
-1

$I_1 = \displaystyle \int_x^1 \frac{1}{1 + t^2} dt$ and

$I_2 \displaystyle = \int_1^{1 / x}\frac{1}{1+ t^2}dt$ for

$x > 0$ , then

0%
$I_1 = I_2$
0%
$I_1 > I_2$
0%
$I_2 > I_1$
0%
none of these

Explanation

${ I }_{ 1 }=\int _{ x }^{ 1 }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } dt }$
Substitute

$t=\dfrac { 1 }{ u } \Rightarrow dt=-\dfrac { 1 }{ u^{ 2 } } du$
We get

$I_{ 1 }=\int _{ 1/x }^{ 1 } \dfrac { 1 }{ 1+\dfrac { 1 }{ u^{ 2 } } } \left( -\dfrac { 1 }{ u^{ 2 } } \right) du=-\int _{ 1/x }^{ 1 } \dfrac { du }{ u^{ 2 }+1 } =\int _{ 1 }^{ 1/x } \dfrac { 1 }{ 1+u^{ 2 } } du=I_{ 2 }$

Suppose that F(x) is an anti-derivative of

$\displaystyle f(x)=\frac{\sin x}{x}$ , where

$x>0$ .

Then

$\displaystyle \int_{1}^{3}\dfrac{\sin2x}{x} \:dx$ can be expressed as?

0%
$F(6)-F(2)$
0%
$\dfrac{1}{2}(F(6)-F(2))$
0%
$\dfrac{1}{2}(F(3)-F(1))$
0%
$2(F(6)-F(2))$

Explanation

Let

$I=\displaystyle \int _{ 1 }^{ 3 }{ \cfrac { sin2x }{ x } dx }$

Substitute

$2x=t\Rightarrow 2dx=dt$

Gives

$I=\displaystyle \int _{ 2 }^{ 6 }{ \cfrac { sint }{ t } dt }$

As antiderivative of

$f\left( x \right) =\cfrac { sinx }{ x }$ is

$F\left( x \right)$

$I={ \left[ F\left( x \right) \right] }_{ 2 }^{ 6 }=F\left( 6 \right) -F\left( 2 \right)$

$\displaystyle f\left ( x \right )=\int_{-1}^{1}\frac{\sin x}{1+t^{2}}dt$ then

$\displaystyle {f}'\left ( \frac{\pi }{3} \right )$ is

0%
nonexistent
0%
$\displaystyle \pi /4$
0%
$\displaystyle \pi \sqrt{3/4}$
0%
none of these

Explanation

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

$\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t } \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\sin { x } \left[ \frac { \pi }{ 2 } \right]$

$\displaystyle \therefore f'\left( x \right) =\frac { \pi }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi }{ 3 } \right) =\frac { \pi }{ 2 } \frac { 1 }{ 2 } =\frac { \pi }{ 4 }$

The value of

$\displaystyle \int_{0}^{\pi /4}\frac{\sec x}{\left ( \sec x+\tan x \right )^{2}}dx$ is

0%
$\displaystyle 1+\sqrt{2}$
0%
$\displaystyle -\left ( 1+\sqrt{2} \right )$
0%
$\displaystyle -\sqrt{2}$
0%
none of these

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \frac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } } dx$

Multiply numerator and denominator by

$\cos ^{ 2 }{ x }$ , we get

$\displaystyle I=\int _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }{ \dfrac { 1 }{ { \left( u+1 \right) }^{ 2 } } } du=\left[ \dfrac { -1 }{ u+1 } \right] _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }$

$\displaystyle =-\dfrac { 1 }{ \dfrac { 1 }{ \sqrt { 2 } } +1 }$

$=\dfrac { -\sqrt { 2 } }{ 1+\sqrt { 2 } }$

$\displaystyle \int_{\pi /4}^{3\pi /4}\frac{dx}{1+\cos x}$ is equal to

0%
$2$
0%
$\displaystyle -2$
0%
$1 / 2$
0%
$\displaystyle -1/2$

Explanation

Let

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

Substitute

$\displaystyle t=\tan { \dfrac { x }{ 2 } } \Rightarrow dt=\dfrac { 1 }{ 2 } dxsec^{ 2 }x$

$\displaystyle I=\int { \dfrac { 2 }{ \left( { u }^{ 2 }+1 \right) \left( \dfrac { 1-{ u }^{ 2 } }{ { u }^{ 2 }+1 } +1 \right) } } du=\int { du } =u=\tan { \dfrac { x }{ 2 } }$

$\displaystyle \therefore \int _{ \dfrac { \pi }{ 4 } }^{ \dfrac { 3\pi }{ 4 } }{ \dfrac { dx }{ 1+\cos { x } } } =\left[ \tan { \dfrac { x }{ 2 } } \right] _{ \dfrac { \pi }{ 4 } }^{ \dfrac { 3\pi }{ 4 } }=2$

The solution for x of the equation

$\displaystyle \int_{\sqrt{2}}^{x}\frac{dt}{t\sqrt{t^{2}-1}}=\frac{\pi }{2}$ is

0%
$\displaystyle \frac{\sqrt{3}}{2}$
0%
$-\sqrt{2}$
0%
$2$
0%
$\pi$

Explanation

As we know,

$\int{\dfrac{dt}{t\sqrt{t^2-1}}}= \left[\sec^{-1}t\right]+C$

$\therefore \displaystyle \int_{\sqrt 2}^{x}{\dfrac{dt}{t\sqrt{t^2-1}}}=\dfrac{\pi}{2}$

$\Rightarrow \left[\sec^{-1}t\right]_{\sqrt 2}^{x}=\dfrac{\pi}{2}$

$\Rightarrow [\sec^{-1}x-\sec^{-1}\sqrt{2}]=\dfrac{\pi}{2}$ .

$\Rightarrow \sec^{-1}x=\sec^{-1}\sqrt 2+\dfrac{\pi}{2}$

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

$=\dfrac{3\pi}{4}$

$\Rightarrow x=\sec\left(\dfrac{3\pi}{4}\right)=\dfrac{1}{\cos \left(\dfrac{3\pi}{4}\right)}=\dfrac{1}{\cos(\pi-\dfrac{\pi}4)}=\dfrac{1}{-\cos \dfrac {\pi}4}$

$=\dfrac{1}{\dfrac {-1}{\sqrt 2}}$

$=-\sqrt 2$

$\therefore \boxed{x=-\sqrt 2}\rightarrow (B)$

$\displaystyle \int_{0}^{1}\frac{2^{x+1}-3^{x-1}}{6^{x}}dx$

0%
$\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{6}\log _{2}e.$
0%
$\displaystyle -\frac{4}{3}\log _{3}e+\frac{1}{6}\log _{2}e.$
0%
$\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{3}\log _{2}e.$
0%
$\displaystyle -\frac{2}{3}\log _{3}e+\frac{1}{6}\log _{2}e.$

Explanation

$\displaystyle \int _{ 0 }^{ 1 } \frac { 2^{ x+1 }-3^{ x-1 } }{ 6^{ x } } dx=\int _{ 0 }^{ 1 } \left( 2.3^{ -x }-\frac { 1 }{ 3 } 2^{ -x } \right) dx$

$\displaystyle =\left[ -2\frac { 3^{ -x } }{ \log 3 } +\frac { 1 }{ 3 } \frac { 2^{ -x } }{ \log 2 } \right] ^{ 1 }$

$\displaystyle =-\frac { 2 }{ \log 3 } \left( \frac { 1 }{ 3 } -1 \right) +\frac { 1 }{ 3\log 2 } \left( \frac { 1 }{ 2 } -1 \right)$

$\displaystyle =\frac { 4 }{ 3 } \log _{ 3 } e-\frac { 1 }{ 6 } \log _{ 2 } e$
Hence, option 'A' is correct.

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

0%
$\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}+\frac{8}{5}\sqrt{2}$
0%
$\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}$
0%
$\displaystyle \sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}$
0%
$\displaystyle \frac{3}{2}\sqrt{\left ( \frac{3}{2} \right )}-\frac{8}{5}\sqrt{2}$

Explanation

Let

$\displaystyle I=\int _{ 1 }^{ 2 } \left( x+\frac { 1 }{ x } \right) ^{ 3/2 }\frac { x^{ 2 }-1 }{ x^{ 2 } } dx$

Put

$\displaystyle x+\frac { 1 }{ x } =t\Rightarrow \left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx\Rightarrow dt$
Therefore

$\displaystyle I=\int _{ 2 }^{ 5/2 } t^{ 3/2 }dt=\frac { 2 }{ 5 } t^{ 5/2 }=\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 5/2 }-2^{ 5/2 } \right]$

$\displaystyle =\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 2 }\sqrt { \left( \frac { 5 }{ 2 } \right) } -2^{ 2 }\sqrt { 2 } \right] =\frac { 5 }{ 2 } \sqrt { \left( \frac { 5 }{ 2 } \right) } -\frac { 8 }{ 5 } \sqrt { 2 }$
Hence, option 'B' is correct.

The value of

$\displaystyle \int_{1}^{2}\frac{\cos \left ( \log x \right )}{x}dx$ is equal to

0%
$2\sin \left ( \log 2 \right )$
0%
$\sin \left ( \log 2 \right )$
0%
$\displaystyle\sin \log \left ( \frac{1}{2} \right )$
0%
None of these

Explanation

$\displaystyle \int _{ 1 }^{ 2 } \dfrac { \cos \left( \log x \right) }{ x } dx$

Substitute

$\log x=t$

$\displaystyle \frac { 1 }{ x } dx=dt$

$\displaystyle I=\displaystyle \int _{ \log { 1 } }^{ \log { 2 } } \cos { t } dt$

$\displaystyle I={ [\sin { t } ] }_{ \log { 1 } }^{ \log { 2 } }$

$\displaystyle I=\sin { (\log { 2) } }$

$\displaystyle f\left ( \frac{1}{x} \right )+x^{2}f\left ( x \right )=0$ for

$x> 0,$

and

$\displaystyle I=\int_{1/x}^{x}f\left ( z \right )dz, \frac{1}{2}\leq x\leq 2$

then

$\displaystyle I$ is?

0%
$\displaystyle f\left ( 2 \right )-f\left ( 1/2 \right )$
0%
$\displaystyle f\left ( 1/2 \right )-f\left ( 2 \right )$
0%
$0$
0%
None of these

Explanation

Substitute

$\displaystyle z=\dfrac { 1 }{ t }$

$\displaystyle I=\int _{ x }^{ \dfrac { 1 }{ t } }{ f\left( \dfrac { 1 }{ t } \right) } .\left( \dfrac { -1 }{ { t }^{ 2 } } \right) dt=\int _{ x }^{ \dfrac { 1 }{ x } }{ \left( -{ t }^{ 2 }.f\left( t \right) \right) } \left( \dfrac { -1 }{ { t }^{ 2 } } \right)$

$\displaystyle =\int _{ x }^{ \dfrac { 1 }{ x } }{ f\left( t \right) } dt=-\int _{ \dfrac { 1 }{ x } }^{ x }{ f\left( t \right) } dt=-I\\ \therefore 2I=0$

$\displaystyle I_{1}= \int_{-4}^{-5}e^{\left ( x +5 \right )^{2}}dx$ and

$\displaystyle I_{2}= 3\int_{{1}/{3}}^{{2}/{3}}e^{\left ( 3x -2 \right )^{2}}dx$

then

$I_{1}+I_{2}$ equals?

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle -\frac{1}{3}$
0%
$0$
0%
None of these

Explanation

$I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx$

$\displaystyle x=\frac { -y-3 }{ 3 }$

$\displaystyle dx=-\frac { dy }{ 3 }$

$\displaystyle \Rightarrow I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx=-\int _{ -4 }^{ -5 } e^{ \left( y+5 \right) ^{ 2 } }dx=-I_{ 1 }$

$\displaystyle \Rightarrow I_{ 1 }+I_{ 2 }=0$

$f\left ( a+x \right )= f\left ( x \right )$ , then

$\forall$

$a> 0, n\epsilon N$ the value of

$\displaystyle \int_{0}^{n a}f\left ( x \right )dx$ equals ?

0%
$\displaystyle \left ( n-1 \right )\int_{0}^{a}f\left ( x \right )dx$
0%
$\displaystyle \left ( 1-n \right )\int_{0}^{a}f\left ( x \right )dx$
0%
$\displaystyle n\int_{0}^{a}f\left ( x \right )dx$
0%
None of the above

Explanation

$I=\displaystyle \int _{ 0}^{ na }{ f\left( x \right) dx }$

Substitute

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

$I=\displaystyle \int _{ 0 }^{ \left( n-1 \right) a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+x \right) dx } =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( x \right) dx }$

$f\left ( x \right )= A\sin \left ( \dfrac {\pi x}{2} \right )\: +\: B,f{}'\left ( \dfrac 12 \right )= \sqrt{2}$ and

$\displaystyle \int_{0}^{1}f\left ( x \right )dx= \displaystyle \frac{2A}{\pi }$ ,

then the constants

$A$ and

$B$ are

0%
$A=\dfrac {\pi}{2}, B=\dfrac {\pi}{2}$
0%
$A=\dfrac {2}{\pi}, B=3\pi$
0%
$A=0, B=-4\pi$
0%
$A=\dfrac {4}{\pi}, B=0$

Explanation

$f(x)=A\sin{(\pi x/2)}+B$

$\implies f^{'}(x)=\dfrac{A\pi}{2}\cos{(\pi x/2)}$

$\implies f^{'}(1/2)=\dfrac{A\pi}{2}\cos{(\pi/4)=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}}$

$\implies\sqrt{2}=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}$

Hence,

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

$\displaystyle\int_{0}^{1} f(x)dx=\left[-\dfrac{2A}{\pi}\cos{(\pi x/2)}+Bx\right]_{0}^{1}$

$\implies\dfrac{2A}{\pi}=\dfrac{2A}{\pi}+B$

Hence,

$B=0$

Hence, answer is option-(D).

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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