MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 2

$\displaystyle \int_{0}^{1}\frac{1}{\sqrt{2+3x}}dx_{=}$

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

Explanation

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

Put

$3x+2=t$

$\Rightarrow dx=\displaystyle \frac{dt}{3}$

$\displaystyle I=\frac {1}{3}\int_{2}^{5}\frac{1}{\sqrt{t}}dt$

$\Rightarrow \displaystyle I =\frac{2}{3}[\sqrt{t}]_{2}^{5}$

$\Rightarrow I=\dfrac{2}{3} (\sqrt{5}-\sqrt{2})$

Evaluate the integral

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

0%
$\log 2$
0%
$\displaystyle \frac{1}{2} \log2$
0%
$2$
0%
$\log 4$

Explanation

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

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

$=\displaystyle \int_{0}^{1}\dfrac{x}{1+x^2}dx=\left [ \dfrac{1}{2} \log (1+x^2)+c \right ]_{0}^{1}$

$=\left ( \dfrac{1}{2} \log (2)+c \right ) - \left ( \dfrac{1}{2} \log (1)+c\right )$

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

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

The integral

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

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

Explanation

Let $\displaystyle I=\int_{0}^{1}\dfrac{x^3}{1+x^8} dx$

Now, $\displaystyle \int \dfrac{x^3}{1+x^8 }dx =\dfrac{1}{4}\int \dfrac{d(x^4)}{1+x^8}=\dfrac{1}{4}\tan^{-1}(x^4)+c$

$I=\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx=\left(\dfrac{1}{4}\tan^{-1}(x^4)+c\right)_{0}^{1}$

$=\left [\dfrac{1}{4} \tan^{-1}(1)+c\right ]-\left [\dfrac{1}{4} \tan^{-1}(0)+c\right ]$

$=\dfrac{1}{4} \tan^{-1}(1)$

$=\dfrac{1}{4} \cdot \dfrac{\pi}{4}=\dfrac{\pi}{16}$

Hence, $\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx =\dfrac{\pi}{16}$

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

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

Explanation

$\int_{0}^{1}\dfrac{1}{1+x}dx=\ln(1+x)+c$

$\int_{0}^{1}\dfrac{1}{1+x}dx =\left [ \ln (1+x)+ c \right ]\int_{0}^{1}$

$=(\ln(2)+c)-(\ln(1)+c)$

$=\ln2$

$\int_{0}^{1}\dfrac{1}{1+x}dx=ln2$

The integral

$\displaystyle \int_{0}^{1}\frac{(\mathrm{t}\mathrm{a}\mathrm{n}^{-1}{\mathrm{x})^{3}}}{1+\mathrm{x}^{2}}\mathrm{d}\mathrm{x}=$

0%
$\displaystyle \frac{\pi^{4}}{64}$
0%
$\displaystyle \frac{\pi^{4}}{256}$
0%
$\displaystyle \frac{\pi^{4}}{1024}$
0%
$\displaystyle \frac{\pi^{4}}{512}$

Explanation

$\displaystyle \int_{0}^{1}\dfrac{(\tan^{-1}x)^3}{1+x^2}dx$

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

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

$\displaystyle \int_{0}^{1}\dfrac{(\tan^{-1}x)^3}{1+x^2}\cdot dx=\left ( \dfrac{(\tan^{-1}x)^4}{4}+c \right )_{0}^{1}$

$=\left [ \dfrac{(\tan^{-1}1)^4}{4}+c \right ] - \left [ \dfrac{(\tan^{-1}0)^4}{4}+1 \right ]$

$=\left [ \dfrac{\left(\dfrac{\pi}{4}\right)^4}{4}+c \right ] - [0+c]$

$=\dfrac{\pi^4}{4^{5}}=\dfrac{\pi^4}{1024}$

The integral

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

0%
$e-1$
0%
${e}^{-1}-1$
0%
${e}^{-1}+1$
0%
${e}^{-2}-1$

Explanation

Consider, $I=\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{e^{\tan x}}{\cos ^2 x}dx$

$\displaystyle \int e^{\tan x}\cdot \sec ^2 x dx =\int e^{\tan x}\cdot d({\tan x})$

$=\int e^{\tan x} d(\tan x )=e^{\tan x}+c$

$\therefore \displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{e^{\tan x}}{\cos ^2 x}dx=\left ( e^{\tan x}+ c \right )_{0}^{\frac{\pi}{4}}$

$=\left ( e^{\tan \frac{\pi}{4}}+c\right ) -\left ( e^{\tan 0}+c \right )$

$=(e+c)-(1+c)$

$=e-1$

Hence, $\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{e^{\tan x}}{\cos ^2 x}dx=e-1$

$\displaystyle \int_{0}^{1/2}\mathrm{e}^{\mathrm{x}}\left[ { s }{ i }{ n }^{ -1 }{ x }+\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \right]$ dx

$=$

0%
$\displaystyle \frac{e^{4}}{4}$
0%
$\displaystyle \frac{\pi\sqrt{\mathrm{e}}}{6}$
0%
$\displaystyle \frac{\sqrt{\pi \mathrm{e}}}{4}$
0%
$\displaystyle \frac{\pi\sqrt{\mathrm{e}}}{2}$

Explanation

$\displaystyle \int_{0}^{\frac{1}{2}}e^x \left [ \sin^{-1}x +\dfrac{1}{\sqrt{1-x^2}} \right ] dx$

Consider,

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

It is in the form of

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

$=e^x f(x)+c$

$\therefore \displaystyle \int_{0}^{\frac{1}{2}} e^x\left ( \sin^{-1}x+\dfrac{1}{\sqrt{1-x^2}} \right ) dx=\left[e^x \sin^{-1} x+c\right]_{0}^{\frac{1}{2}}$

$=e^{\frac{1}{2}} \sin^{-1}\left(\dfrac{1}{2}\right) - ( e^{0} \sin^{-1}0)$

$=e^{\frac{1}{2}} \cdot \dfrac{\pi}{6}=\dfrac{\pi\sqrt{e}}{6}$

$\displaystyle \int_{0}^{k}\frac{\cos x}{1+\sin^{2}x}dx=\frac{\pi}{4}$ then

${k}=?$

0%
$\pi/6$
0%
$1$
0%
$\pi/4$
0%
$\pi/2$

Explanation

Let

$sin x = t$
Thus

$cos x dx = dt$

Substituting the values back in the integral and performing indefinite integration,

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

$=$

${tan}^{-1}t + c$

Substituting the value of t,

$=$

${tan}^{-1}(sin x) + c$

Putting in the limits,

$=$

${tan}^{-1}(sin k) - {tan}^{-1} 0 = \dfrac{\pi}{4}$

$=$

${tan}^{-1}(sin k) - 0 = \dfrac{\pi}{4}$

$sin k = tan \dfrac{\pi}{4}$

$= sin k = 1$

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

$\displaystyle \int_{0}^{1}\tanh xdx=$

0%
$\log(\mathrm{e}+1/\mathrm{e})$
0%
$\log$ $(e-1/e)$
0%
$\log(\mathrm{e}/2+1/2\mathrm{e})$
0%
$\displaystyle \log(\frac{\mathrm{e}}{2}-\frac{1}{\mathrm{e}})$

Explanation

$\int_{0}^{1} tan h x dx$
$\int tan h x dx=\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}} dx =log (e^x+e^{-x})+c$
$\int_{0}^{1} tan h x dx = log (e^x+e^{-x})+c \int_{0}^{1}$
$=\left ( log (e+\dfrac{1}{e}) +c \right ) - \left ( log (2)+c \right )$
$=log (e+\dfrac{1}{e})-log 2$
$=log (\dfrac{e}{2}+\dfrac{1}{2e})$
$\int_{0}^{1}tan hx dx=log (\dfrac{e}{2}+\dfrac{1}{2e})$

Evaluate the integral

$\displaystyle \int_{1}^{e}\frac{(\ln x)^{3}}{x}dx$

0%
$e^{4}/4$
0%
$\dfrac{1}{4}$
0%
$\displaystyle \frac{1}{4}({e}^{4}-1)$
0%
$e^{4}-1$

Explanation

$\displaystyle \int_{1}^{e}\frac{(ln x)^3}{x}dx$

$\displaystyle \int_{1}^{e}(ln x)^3d(ln x)=\frac{(ln x)^4}{4}+c$

$\displaystyle \int_{1}^{e}(ln x)^3d(ln x)=\left ( \frac{(1)^4}{4}+c \right ) - \left ( \frac{0}{4}+ c \right )$

$=\dfrac{1}{4}$

Thus

$\displaystyle \int_{1}^{e}\frac{(ln x)^3}{x}dx=\frac{1}{4}$

$\displaystyle \int_{0}^{\pi}\frac{\tan x}{\sec x+\cos x}dx_{=}$

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

Explanation

$\int_{0}^{\pi}\dfrac{\tan x}{\sec x+\cos x}dx$

$\int \dfrac{\tan x}{\sec x+\cos x}dx=\int \dfrac{d (\sec x)}{1+\sec ^2 x}$

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

$\int_{0}^{\pi}\dfrac{\tan x}{\sec x+\cos x}dx=\left [ \tan (\sec x)+c \right ] |_{0}^{\pi}$

$=(\tan ^{-1}(\sec \pi)+c)- [ \tan ^{-1}(\sec 0) ]$

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

$\displaystyle \int_{0}^{\infty}(a^{-x}-b^{-x})dx=(\mathrm{a}>1,\mathrm{b}>1)$

0%
$\displaystyle \dfrac{1}{\log a}-\dfrac{1}{\log b}$
0%
$\log a-\log b$
0%
$\log a + \log b$
0%
$\displaystyle \dfrac{1}{\log a}+\dfrac{1}{\log b}$

Explanation

$\displaystyle I=\int_{0}^{\infty}(a^{-x}-b^{-x})dx$

$\displaystyle = -\frac{1}{\log a}(a^{-x})_{0}^{\infty}+\frac{1}{\log b}(b^{-x})_{0}^{\infty}$

$\displaystyle =\frac{1}{\log a}-\frac{1}{\log b}$

Evaluate:

$\displaystyle \int_{\sqrt{8}}^{\sqrt{15}}x\sqrt{1+x^{2}}.dx$

0%
$\dfrac{15}{8}$
0%
$\dfrac{37}{3}$
0%
$\dfrac{37}{6}$
0%
$\displaystyle \frac{37}{9}$

Explanation

$\displaystyle\int_{\sqrt{8}}^{\sqrt{15}} x\sqrt{1+x^2} dx$

Let

$z=1+x^2$

$\implies dz=2xdx$

$\implies xdx=\dfrac{dz}{2}$

For,

$x=\sqrt{8}, z=9$ and for

$x=\sqrt{15}, z=16$

Hence, integration becomes-

$\displaystyle\int_{9}^{16} \sqrt{z} \dfrac{dz}{2}$

$=\dfrac{1}{2} \left[\dfrac{z^\left(\tfrac{1}{2}+1\right)}{\dfrac{1}{2}+1}\right]_{9}^{16}$

$=\dfrac{1}{2}\times \dfrac{2}{3} \displaystyle\left[z^\left(\tfrac{3}{2}\right)\right]_{9}^{16}$

$= \dfrac{1}{3} \displaystyle\left[16^\left(\tfrac{3}{2}\right)-9^\left(\tfrac{3}{2}\right) \right]$

$=\dfrac{1}{3} [64-27]$

$=\dfrac{37}{3}$

$\displaystyle \int_{0}^{1}\mathrm{e}^{\mathrm{x}}(\mathrm{e}^{\mathrm{x}}+1)^{3}\mathrm{d}\mathrm{x}=$

0%
$\displaystyle \frac{e^{4}}{4}-4$
0%
$\displaystyle \frac{(e+1)^{4}}{4}-4$
0%
$\displaystyle \frac{(e+1)^{4}+16}{4}$
0%
$\displaystyle \frac{(\mathrm{e}+1)^{4}}{4}+4$

Explanation

Let D =

$\int_{0}^{1}e^x(e^x+1)^3dx$
Consider

$I = \int e^x(e^x+1)^3dx=\int (e^x+1)^3e^x.dx=\frac{(e^x+1)^4}{4}+c$

$I = \int e^x(e^x+1)^3dx=\frac{(e^x+1)^4}{4}+c$

$D = \int_0^1 e^x(e^x+1)^3dx$

$= \cfrac{(e^1+1)^4}{4} - 4$
Thus,

$\int_{0}^{1}e^x(e^x+1)^3dx=\frac{(e+1)^4}{4}-4$

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

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

Explanation

$\int_{0}^{1}\dfrac{x dx}{(x^2+1)^2}$
$\int \dfrac{x dx}{(x^2+1)^2} =\dfrac{1}{2}\int \dfrac{dx^2}{(1+x^2)^2}=-\dfrac{1}{2}\left ( \dfrac{1}{1+x^2} \right )^{-1}$
$=-\dfrac{1}{2}(1+x^2)^{+c}$
$=\int_{0}^{1}\dfrac{x dx}{(x^2+1)^2}=\left [ -\dfrac{1}{2} (1+x^2)+c \right ] \int_{0}^{1}$
$=\left [ -\dfrac{1}{2} (2)+c \right ] - \left [ -\dfrac{1}{2}+c \right ]$
$=-\dfrac{1}{4}+\dfrac{1}{2} =\dfrac{1}{4}$
$\int_{0}^{1}\dfrac{x dx}{(x^2+1)^2}=\dfrac{1}{4}$

Evaluate:

$\displaystyle \int_{0}^{\dfrac{\pi}{2}}e^{\sin^2 x}\sin 2xdx$

0%
$e$
0%
$e+1$
0%
$e-1$
0%
$2{e}+1$

Explanation

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

Let

$z=\sin^2 x$

$\implies dz=2\sin{x} \cos{x}dx=\sin{2x}dx$

For

$x=0, z=0$ and for

$x=\dfrac{\pi}{2} ,z=1$ .

Hence, integration becomes-

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

$=\left[e^z\right]_{0}^{1}$

$=e-1$

$\displaystyle \int_{0}^{4}\sqrt{16-x^{2}}d_{X}=$

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

Explanation

$\int_{0}^{4}\sqrt{16-x^2} dx$

$\int \sqrt{16-x^2} dx =\int \sqrt{4^2-x^2}dx$

$=\frac{x}{2} \sqrt{16-x^2}+ \frac{16}{2} sin^{-1} (\frac{x}{14})+c$

$=\int_{0}^{4}\sqrt{16-x^2}dx \left [ \frac{x}{2} \sqrt{16-x^2}+ \frac{16}{2}sin^{-1}\left ( \frac{x}{4} \right ) +c \right ]\int_{0}^{4}$

$=\left ( \frac{4}{2}\sqrt{16-16}+ 8 sin^{-1}\left ( \frac{4}{4}\right ) +c\right )- (0+0+c)$

$=8\left ( \dfrac{\pi}{2} \right ) =4\pi$

$\int_{0}^{4}\sqrt{16-x^2}dx=4\pi$

Find

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}xdx}{(\sec x+\tan x)^{n}}$ , where

$(\mathrm{n}>2)$

0%
$\displaystyle \frac{1}{n^{2}-1}$
0%
$\displaystyle \frac{n}{n^{2}-1}$
0%
$\displaystyle \frac{n}{n^{2}+1}$
0%
$\displaystyle \frac{2}{n^{2}-1}$

Explanation

Let

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

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

$\displaystyle \Rightarrow \sec { x } dx=\dfrac { dt }{ t }$

$\displaystyle \therefore \sec { x } -\tan { x } =\dfrac { 1 }{ t } \Rightarrow \sec { x } =\dfrac { 1 }{ 2 } \left( t+\dfrac { 1 }{ t } \right)$

$\displaystyle \therefore I=\dfrac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \dfrac { \left( t-\dfrac { 1 }{ t } \right) \dfrac { dt }{ t } }{ { t }^{ n } } } =\dfrac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \left( \dfrac { 1 }{ { t }^{ n } } +\dfrac { 1 }{ { t }^{ +n-12 } } \right) } dt$

$\displaystyle =\dfrac { 1 }{ 2 } \left[ \dfrac { 1 }{ \left( 1-n \right) { t }^{ n-1 } } +\dfrac { 1 }{ \left( n+1 \right) { t }^{ n+1 } } \right] _{ 0 }^{ \infty }=\dfrac { n }{ { n }^{ 2 }-1 }$

$\displaystyle \int_{0}^{\pi/2}\frac{d_{X}}{4\cos^{2}x+9\sin^{2}x}=$

0%
$\displaystyle \frac{\pi}{12}$
0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{9}$
0%
$\displaystyle \frac{\pi}{6}$

Explanation

$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4 cos^2 x + 9 sec^2 x}$
$\int \dfrac{dx}{4 cos^2 x + 9 sec^2 x}=\int \dfrac{sec^2 x dx} {4+9 tan^2 x}=\int \dfrac{d(tan x)}{4+9 tan^2 x}$
$=\dfrac{1}{4} tan^{-1}\left ( \dfrac{3}{2} tan x \right ) \dfrac{2}{3} +c$
$=\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan x \right ) +c$
$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4 cos^2 x + 9 sec^2 x}=\left [ \dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan x \right ) +c \right ]\int_{0}^{\dfrac{\pi}{2}}$
$=\left (\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan \dfrac{\pi}{2} \right ) +c \right ]- \left ( \dfrac{1}{6}tan^{-1}(0)+c \right )$
$=\dfrac{1}{6}\times \dfrac{\pi}{2}=\dfrac{\pi}{12}$

Evaluate the following definite integral:

$\displaystyle \int_{0}^{\pi_/{2}}\frac{1}{4+5\cos x}dx=$

0%
$\displaystyle \frac{1}{5}\log 2$
0%
$\displaystyle \frac{1}{2}\log 2$
0%
$\displaystyle \frac{1}{3}\log 3$
0%
$\displaystyle \frac{1}{3}\log 2$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{\frac{\pi}{2}} { \cfrac { 1 }{ 4+5cosx } dx }$

Substitute

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

So,

$x \rightarrow 0 \ \Rightarrow t \rightarrow 0$ and

$x \rightarrow \dfrac{\pi}{2} \ \Rightarrow t \rightarrow {1}$

gives

$\cos x=\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } ,dx=\cfrac { 2dt }{ 1+{ t }^{ 2 } }$

$\Rightarrow$

$\displaystyle I=\int _{ 0 }^{ 1 } { \cfrac { 1 }{ 4+5\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } } \ \cfrac { 2dt }{ 1+{ t }^{ 2 } } } =2\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ 9-{ t }^{ 2 } } dt }$

$\Rightarrow$

$\displaystyle I =\cfrac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \cfrac { 1 }{ 1-\cfrac { { t }^{ 2 } }{ 9 } } dt }$

Substitute

$\cfrac { t }{ 3 } =u\Rightarrow dt=3du$

So,

$t \rightarrow 0 \ \Rightarrow u \rightarrow 0$ and

$t \rightarrow 1 \ \Rightarrow u \rightarrow \dfrac{1}{3}$

$\Rightarrow$ $\displaystyle I=\frac { 2 }{ 3 } \int _{ 0 }^{ \frac { 1 }{ 3 } }{ \cfrac { 1 }{ 1-{ u }^{ 2 } } du } =\cfrac { 2 }{ 3 } { \left[ \cfrac { 1 }{ 2 } \log { \cfrac { 1+u }{ 1-u } } \right] }_{ 0 }^{ \frac { 1 }{ 3 } }$

$\Rightarrow$

$\displaystyle I=\cfrac { 1 }{ 3 } \left( \log { 2 } -0 \right) =\cfrac { 1 }{ 3 } \log { 2 }$

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

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

Explanation

To find : $\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a} \cdot dx$
Consider, $\displaystyle \int \dfrac{x-a}{x+a} \cdot dx=\int 1 - 2a \int \dfrac{1}{(x+a)}dx$
$=x-2a \log (x+a)+c$
Then, $\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a} \cdot dx=\left[x-2a \log (x+a)+c\right]_{0}^{a}$
$=(a-2a \log (1a)+c)- [0-2a \log (a)+c]$
$=a-2a \log 2$
$\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a}\cdot dx= a-2a \log 2$

$\displaystyle \int_{0}^{60}\frac{dx}{2x+1}=\log a$ , then

$a=$

0%
$3$
0%
$11$
0%
$81$
0%
$40$

Explanation

Let

$I=\displaystyle \int_{0} ^{60} \dfrac{dx}{2x+1}$

$=\left [\dfrac{\log(2x+1)}{2}\right]_{0} ^{60}$

$=\dfrac{1}{2}[\log (121)-\log(1)]$

$=\dfrac{\log (121)}{2}$

$=\log(\sqrt{121})$

$=\log 11$
Thus

$\log 11=\log a$

$\Rightarrow a=11$

Evaluate the integral

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

0%
$\log 4$
0%
$\log (\dfrac{4}{e})$
0%
$1$
0%
$\log (\dfrac{e}{4})$

Explanation

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

$\displaystyle \int \dfrac{1-x}{1+x} dx = \displaystyle \int \dfrac{2}{1+x}dx -\displaystyle \int1 \cdot dx$

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

$\displaystyle \int \dfrac{1-x }{1+x}dx = 2 \log (1+x)- x+c$

$\displaystyle \int_{0}^{1}\dfrac{1-x}{1+x}dx= \left(2 \log (1+x)-x+c\right)_{0}^{1}$

$=2 \log (2)-1$

$=\log 4- \log \ e$

$=\log \left(\dfrac{4}{e}\right)$

$\displaystyle \int_{1}^{2}\frac{\mathrm{d}\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}=$

0%
$\displaystyle \log_{\mathrm{e}}(\frac{2+\sqrt{5}}{\sqrt{2}+1})$
0%
$\displaystyle \log_{\mathrm{e}}(\frac{\sqrt{2}+1}{2+\sqrt{5}})$
0%
$\displaystyle \log_{\mathrm{e}}(\frac{2-\sqrt{5}}{\sqrt{2}-1})$
0%
0

Explanation

$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log |x+\sqrt{x^2+a}|+c$
$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log |x+\sqrt{x^2+1}|+c$
$=\log |2+\sqrt{5}|-\log |1+\sqrt{2}|$
$=\log \left | \dfrac{2+\sqrt{5}}{1+\sqrt{2}} \right |$
$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log \left | \dfrac{2+\sqrt{5}}{1+\sqrt{2}} \right |$

$\int_{-1}^{1} \displaystyle \frac{d{x}}{1+x^{2}}=$

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

Explanation

$=\int_{-1}^{1}\dfrac{dx}{1+x^2}$
$\int \dfrac{dx}{1+x^2}=tan^{-1}x+c$
$\int_{-1}^{1}\dfrac{dx}{1+x^2}=(tan^{-1}x+c)\int_{-1}^{1}$
$=tan^{-1}1-tan^{-1}(-1)$
$=\dfrac{\pi}{4}-(\dfrac{-\pi}{4})$
$=\dfrac{\pi}{2}$
So,
$\int_{-1}^{1}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$

$\displaystyle \int_{0}^{a}\frac{x^{5}dx}{\sqrt{a^{2}-x^{2}}}=$

0%
$\displaystyle \frac{\mathrm{a}^{5}}{15}$
0%
$\displaystyle \frac{8\mathrm{a}^{5}}{15}$
0%
$\displaystyle \frac{8a}{15}$
0%
$\displaystyle \frac{11\mathrm{a}^{2}}{15}$

Explanation

$\int_{a}^{0}\dfrac{x^5dx}{\sqrt{a^2-x^2}}x=a sin \theta$
$\int_{o}^{a}\dfrac{a^5sin ^5\theta}{a cos \theta}a cos \theta d \theta$
$\int_{0}^{\dfrac{\pi}{2}}a^5 sin^5\theta d \theta=a^5\int_{0}^{\dfrac{\pi}{2}}sin ^5\theta d \theta$
$=a^5\left ( \dfrac{4}{5}\right )\left ( \dfrac{2}{3}\right )$
$=\dfrac{8a^5}{15}$

Evaluate the integral

$\displaystyle \int_{1}^{2}\sqrt{(x-1)(2-x)}dx$

0%
$\dfrac {\pi }{8}$
0%
$\dfrac {\pi }{4}$
0%
$\dfrac {1}{8}$
0%
$\dfrac {1}{4}$

Explanation

$\displaystyle \int_{1}^{2}\sqrt{-x^2+3x-2}dx$

$=\displaystyle \int_{1}^{2}\sqrt{-x^2+3x-\dfrac{9}{4}+\dfrac{9}{4}-2} dx$

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

$=\left ( \dfrac{x}{2}\sqrt{(x-1)(2-x)+\dfrac{(\dfrac{1}{2})^2}{2}}\sin^{-1}\left ( \dfrac{x-\dfrac{3}{2}}{\dfrac{1}{2}} \right )+c \right )_{1}^{2}$

$=\left ( \dfrac{x}{2}\sqrt{(x-1)(2-x)}+\dfrac{1}{8}\sin^{-1}(2x-3)+c \right )_{1}^{2}$

$=\dfrac{1}{8}\sin^{-1}(1)-\dfrac{1}{8}\sin^{-1}(-1)$

$=\dfrac{1}{8}\left ( \dfrac{\pi}{2}+\dfrac{\pi}{8} \right )$

$=\dfrac{\pi}{8}$

Evaluate:

$\displaystyle \int_{0}^{\pi /8} \cos^{3}4x\ dx$

0%
$1/6$
0%
$1/5$
0%
$-1/3$
0%
$1/8$

Explanation

Given that

$\int_{0}^{\tfrac{\pi}{8}}\cos^3(4x)dx$

We can write

$\cos^3(4x)= \cos^2(4x)\cos(4x) =[1-\sin^2(4x)]\cos(4x)$

$\int_{0}^{\tfrac{\pi}{8}}[1-\sin^2(4x)]\cos(4x)$ ............1

Now let

$\sin(4x)= t$ ........2

when

$x=0, t = 0$ ; When

$x=\cfrac{\pi}{8}, t = 1$

On differentiating equation 2 we get,

$\cfrac{\cos(4x)}{4}dx=dt$ .........3

Substituting the values of equation 2 and 3 in equation 1.

$\int_{0}^{1}\cfrac{[1-t^2]}{4}dt$

$=\int_{0}^1 \cfrac{1}{4}dt - \int_{0}^1 \cfrac{t^2}{4}dt$

$=[\cfrac{t}{4}]_{0}^{1} - [\cfrac{t^3}{12}]_{0}^{1}$

$=\cfrac{1}{4} - \cfrac{1}{12}$

$=\cfrac{3-1}{12}$

$=\cfrac{2}{12}$

$=\cfrac{1}{6}$

$\displaystyle \int_{1}^{\infty}\left( \frac { 1 }{ 1+x^{ 2 } } \right) d{ x }=$

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

Explanation

$\int_{1}^{\infty }\left [ \dfrac{1}{1+x^2} \right ] dx$
$tan^{-1}x \int_{1}^{\infty }$
$\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$

$\displaystyle \int_{0}^{\pi/4}\frac{\sqrt{\tan x}}{sin x cos x}dx=$

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

Explanation

$\int_{0}^{\dfrac{\pi}{4}}\dfrac{sec^2 x \sqrt{tan x}}{tan x}dx$
dividing and multiplying by $sec^2 x$
$=\int_{0}^{\dfrac{\pi}{4}}\dfrac{d tan x}{\sqrt{tan x}}$
$=(2\sqrt{tanx}+c)\int_{0}^{\dfrac{\pi}{4}}$
$=2$

Evaluate:

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

0%
$\displaystyle \frac{1}{12}$
0%
$\displaystyle \frac{\pi}{24}$
0%
$\displaystyle \frac{\pi}{12}$
0%
$\displaystyle \frac{1}{24}$

Explanation

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

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

$=\displaystyle\int_{0}^{\pi/2} \sin^3{x}(1-\sin^2{x})\cos{x}dx$

$=\displaystyle\int_{0}^{\pi/2} \sin^3{x} \cos{x} dx- \displaystyle\int_{0}^{\pi/2} \sin^5{x} \cos{x}dx$

$=I_1-I_2$ (let)

Now let

$t=\sin{x}$

$\implies dt=\cos{x}dx$

And at

$x=0, t=0$ and at

$x=\dfrac{\pi}{2}, t=1$

$\displaystyle\int_{0}^{1} t^3dt-\displaystyle\int_{0}^{1} t^5 dt$

$=\left[\dfrac{t^4}{4}\right]_{0}^{1}- \left[\dfrac{t^6}{6}\right]_{0}^{1}$

$=\left(\dfrac{1}{4}-\dfrac{1}{6}\right)=\dfrac{1}{12}$

$\int_{0}^{\pi /2} \sin^{4}x.\cos^{2}xd_{X=}$

0%
$\displaystyle \frac{\pi}{32}$
0%
$\displaystyle \frac{\pi^{2}}{16}$
0%
$\displaystyle \frac{\pi}{15}$
0%
$\displaystyle \frac{\pi}{64}$

Explanation

$\int_{0}^{\dfrac{\pi}{2}} sin ^4 x cos^2 x dx$
$\int_{0}^{\dfrac{\pi}{2}}sin ^m x cos ^x   dx$
$I_{m, n}=\left (\dfrac{m-1}{m+n} \right )\left (\dfrac{m-3}{m+n-2} \right )\left (\dfrac{m-5}{m+n-4} \right ) ..... I_0, n    or   I_1,,,,,$
$I_{4,2}=\left (\dfrac{4-1}{6} \right )\left (\dfrac{4-3}{4} \right )I_{0,2}$
$\int_{0}^{\dfrac{\pi}{2}}cos^2x   dx=\dfrac{1}{2}\times \dfrac{\pi}{2}$
$\dfrac{3}{6}\left (\dfrac{1}{4} \right )\times \dfrac{\pi}{4}=\dfrac{\pi}{32}$

Evaluate the integral

$\displaystyle \int_{1}^{3}\frac{d_{X}}{\sqrt{(x-1)(3-x)}}$

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

Explanation

$\int_{1}^{3}\dfrac{dx}{\sqrt{(x-1)(3-x)}}$
$\int_{1}^{3}\dfrac{dx}{\sqrt{-3+4x-x^2}}=\int_{1}^{3}\dfrac{dx}{\sqrt{1-(x-2)^2}}$
$=\int_{1}^{3}\dfrac{dx}{\sqrt{1-(x-2)^2}}=\left [ sin^{-1}\left (\dfrac{x-2}{1}\right )+c\right ]_1^3$
$=sin^{-1}(x-2)_{1}^{3}$
$=sin^{-1}(1)-sin^{-1}(-1)$
$=\dfrac{\pi}{2}-\left ( \dfrac{-\pi}{2}\right )$
$=\pi$

$\displaystyle \int_{0}^{\infty}\frac{dx}{(x+\sqrt{x^{2}+1})^{5}}=$

0%
1/24
0%
1/5
0%
5/24
0%
5/36

Explanation

$\int_{0}^{\infty}\dfrac{dx}{(x+\sqrt{x^2+1})^5}=\int_{0}^{\infty}\dfrac{(x-\sqrt{x^2+1})^5}{-1}dx$
$=\int_{0}^{\infty}(\sqrt{x^2+1}-1)^5 dx$
$x=Sin h \theta$
$\int_{0}^{\infty}(cos h \theta- sin h \theta)^5 cos h \theta d\theta$
$\int_{0}^{\infty}(e^{-x})^5\left (\dfrac{e^x+e^{-x}}{2} \right ) d\theta$
$\int_{0}^{+\infty}\dfrac{e^{-4x}+e^{-6x}}{2} dx=\dfrac{1}{2}\left [ \left (\dfrac{-1}{4}\right )[0-1]+\left (\dfrac{-1}{6} \right )[0-1] \right ]$
$=\dfrac{1}{2}\left ( \dfrac{1}{4}+\dfrac{1}{6}\right )=\dfrac{1}{2}\left (\dfrac{1+4}{24}\right )$
$=\dfrac{5}{24}$

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

0%
$\pi /2$
0%
$\pi /4$
0%
$\pi /6$
0%
$\pi /8$

Explanation

$\int_{0}^{1}\sqrt{x(1-x)}dx=\int_{0}^{1}\sqrt{x-x^2}dx$

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

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

$=\frac{x}{2}\sqrt{x-x^2}+\frac{1}{8}sin^{-1}(2x-1)+c$

$\int_{0}^{6}\sqrt{x(1-x)}dx=(\frac{x}{2})\sqrt{x(1-x)}+\frac{1}{8}sin^{-1}(2x-1)+c)\int_{0}^{1}$

$=\frac{1}{8}[sin^{-1}(1)-sin^{-1}(-1)]$

$=\frac{1}{8}[\frac{\pi}{2}-(\frac{-\pi}{2})]$

$=\frac{1}{8}(\pi)$

$\int_{0}^{1}\sqrt{x(1-x)}dx=\frac{\pi}{8}$

Evaluate the integral

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

0%
$\displaystyle \frac{5\pi}{4}$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\pi$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

$\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{sin 2x}{cos^4 x + sin^4 x}dx=\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{d sin^2 x}{(1-sin^2 x)^2+sin^4x}$

$=\displaystyle \int_{\pi}^{\dfrac{3\pi}{4}}\dfrac{d sin^2 x}{(2 sin^2 x-2 sin^2 x +\dfrac{1}{2}+\dfrac{1}{2})}$

$sin^2 x=t$ $\displaystyle \int_{0}^{\dfrac{1}{2}}\dfrac{dt}{(\sqrt{2}t-\dfrac{1}{\sqrt{2}})^2+ (\dfrac{1}{\sqrt{2}})^2}=\dfrac{1}{\sqrt{2}}\dfrac{1}{(\dfrac{1}{\sqrt{2}})}tan^{-1}\left [ \dfrac{\left (\sqrt{2}t- \dfrac{1}{\sqrt{2}} \right )}{\dfrac{1}{\sqrt{2}}} \right ]_{0}^{\dfrac{1}{2}}$

$=tan^{-1}(2t-1)_{0}^{{1}/{2}}$

$=tan^{-1}(0)-tan^{-1}(-1)$

$=0-\left ( \dfrac{-\pi}{4} \right )$

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

Evaluate the integral

$\displaystyle \int_{0}^{1} cos ^{-1}\left(\displaystyle \frac{1- {x}^{2}}{1+ {x}^{2}}\right) {d} {x}$

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

Explanation

$\displaystyle \int_{0}^{1} cos^{-1}\left ( \dfrac{1-x^2}{1+x^2} \right )$

$x=tan \theta$

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

$\displaystyle \int_{0}^{\dfrac{\pi}{4}}cos^{-1}(cos 2\theta)cos^2\theta d\theta$

$\displaystyle \int_{0}^{\dfrac{\pi}{4}}2 \theta sec^2 \theta d \theta =2\displaystyle \int_{0}^{\dfrac{\pi}{4}}\theta sec^2 \theta d \theta$

$2 [ \theta tan \theta + log |cos \theta| ]_{0}^{\dfrac{\pi}{4}}$

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

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

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

Evaluate:

$\displaystyle \int_{1/3}^{1}\frac{(x-x^{3})^{1/3}}{x^{4}}dx$ .

0%
$3$
0%
$0$
0%
$6$
0%
$4$

Explanation

$\displaystyle \int_{\frac {1}{3}}^{1} \dfrac {(x – x^{3})^{\frac {1}{3}}}{x^{4}} dx = \int_{\frac {1}{3}}^{1} \frac {(x^{3})^{\frac {1}{3}} \left (\dfrac {1}{x^{2}} – 1\right )^{\frac {1}{3}}}{x^{4}} dx$

$\displaystyle = \int_{\dfrac {1}{3}}^{1} \dfrac {\left (\dfrac {1}{x^{2}} – 1\right )^{\frac {1}{3}}}{x^{3}} dx$ (let $\dfrac {1}{x^{2}} – 1 = t$ )

$\displaystyle = \int_{8}^{0} \dfrac{t^{\frac{1}{3}}}{-2} dt$ $\therefore \dfrac {-2}{x^{3}} dx = dt$

$= -\dfrac {1}{2}\left (\dfrac {t^{4/3}}{4/3}\right )_{8}^{0} = 6$ .

$\displaystyle \int_{\log 2}^{t}\frac{d_{X}}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$ , then

$\mathrm{t}=$

0%
$\log 8$
0%
4
0%
log 4
0%
log 2

Explanation

Let

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

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

$I=\int { \cfrac { 1 }{ u\sqrt { u-1 } } du }$
Substitute

$s=u-1\Rightarrow ds=du$ , we get

$I=\int { \cfrac { ds }{ \sqrt { s } \left( s+1 \right) } }$
Substitute

$p=\sqrt { s } \Rightarrow dp=\cfrac { 1 }{ 2\sqrt { s } } ds$ , we get

$I=2\int { \cfrac { 1 }{ { p }^{ 2 }+1 } } =2{ tan }^{ -1 }p=2{ tan }^{ -1 }\sqrt { s } =2{ tan }^{ -1 }\sqrt { u-1 }$

$=2{ tan }^{ -1 }\sqrt { { e }^{ x }-1 }$
Therefore,

$\int _{ \log { 2 } }^{ t }{ \cfrac { dx }{ \sqrt { { e }^{ x }-1 } } } =\cfrac { \pi }{ 6 } \\ \Rightarrow { \left[ 2{ tan }^{ -1 }\sqrt { { e }^{ x }-1 } \right] }_{ \log { 2 } }^{ t }=\cfrac { \pi }{ 6 } \Rightarrow 2{ tan }^{ -1 }\sqrt { { e }^{ t }-1 } -2{ tan }^{ -1 }1=\cfrac { \pi }{ 6 } \\ \Rightarrow 2{ tan }^{ -1 }\sqrt { { e }^{ t }-1 } -\cfrac { \pi }{ 2 } =\cfrac { \pi }{ 6 } \Rightarrow { tan }^{ -1 }\sqrt { { e }^{ t }-1 } =\cfrac { \pi }{ 3 } \\ \Rightarrow { e }^{ t }-1=3\Rightarrow t=\log { 4 }$

Evaluate:

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

$\left(2 \cot^{-1}\sqrt{\displaystyle \frac{1- {x}}{1+ {x}}}\right)dx$

0%
$\dfrac{-1}{2}$
0%
$\dfrac{1}{2}$
0%
$0$
0%
$1$

Explanation

Let,

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

And,

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

At,

$x=0, \theta=\dfrac{\pi}{2}$ and at

$x=1,\theta=0$

Now,

$\sqrt{\dfrac{1-x}{1+x}}=\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}$

$=\sqrt{\dfrac{2\sin^2\dfrac{\theta}{2}}{2\cos^2\dfrac{\theta}{2}}}$

$=\tan\dfrac{\theta}{2}=\cot\left(\dfrac{\pi}{2}-\dfrac{\theta}{2}\right)$

$\implies 2\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}=(\pi-\theta)$

Hence, integration becomes:-

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

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

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

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

$=\dfrac{-1}{4}(1+1)=\dfrac{-1}{2}$

Answer-(A)

$\displaystyle \int_{1}^{2}\frac{dx}{x^{2}-2x+4}=$

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

Explanation

$\int_{1}^{2}\dfrac{dx}{(x-1)^2+(\sqrt{3})^2}=\left [ \dfrac{1}{\sqrt{3}} tan^{-1} \left ( \dfrac{x-1}{\sqrt{3}}\right ) +c \right ] \int_{1}^{2}$
$=\dfrac{1}{\sqrt{3}}tan^{-1}\left ( \dfrac{1}{\sqrt{3}} \right )$
$=\dfrac{1}{\sqrt{3}}\cdot \dfrac{\pi}{6}$
$=\dfrac{\pi}{6\sqrt{3}}$

Evaluate the integral

$I=\displaystyle \int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \cfrac { { sin }^{ -1 }x }{ { \left( 1-{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } } } dx }$

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

Explanation

$I=\displaystyle \int _{ 0 }^{ \cfrac { 1 }{ \sqrt { 2 } } }{ \cfrac { { sin }^{ -1 }x }{ { \left( 1-{ x }^{ 2 } \right) }^{ \cfrac { 3 }{ 2 } } } dx }$

Substituting

$x=sint\Rightarrow dx=costdt$

$I=\displaystyle \int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ \cfrac { tcostdt }{ cost-{ sin }^{ 2 }tcost } } =\displaystyle \int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ t{ sec }^{ 2 }t } dt\\ ={ \left[ ttant \right] }_{ 0 }^{ \cfrac { \pi }{ 4 } }-\displaystyle \int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ tant } dt=\cfrac { \pi }{ 4 } -0-{ \left[ -\log { \left( cosx \right) } \right] }_{ 0 }^{ \cfrac { \pi }{ 4 } }\\ =\cfrac { \pi }{ 4 } +\log { \cfrac { 1 }{ \sqrt { 2 } } } =\cfrac { \pi }{ 4 } -\cfrac { 1 }{ 2 } log2$

$0<\mathrm{a}<\mathrm{c},\ 0<\mathrm{b}<\mathrm{c}$ then

$\displaystyle \int_{0}^{\infty}\frac{a^{x}-b^{x}}{c^{x}}dx=$

0%
$\displaystyle \log\frac{b}{c}-\log\frac{a}{c}$
0%
$\displaystyle \frac{\log a-\log b}{\log c}$
0%
$\displaystyle \frac{1}{\log b/c}-\frac{1}{\log a/c}$
0%
$l\displaystyle \mathrm{o}\mathrm{g}\frac{\mathrm{a}}{\mathrm{c}}-l\mathrm{o}\mathrm{g}\frac{\mathrm{b}}{\mathrm{c}}$

Explanation

$\int_{0}^{\infty}\left ( \dfrac{a^x}{c^x}-\dfrac{b^x}{c^x} \right ) dx$
$=\int_{0}^{\infty}\left ( \dfrac{a}{c} \right )^xdx-\int_{0}^{\infty}\left ( \dfrac{b}{c}\right )^x dx$
$=\dfrac{(\dfrac{a}{c})^x}{log (\dfrac{a}{c})}\int_{0}^{\infty}-\dfrac{(\dfrac{b}{c})^x}{log (\dfrac{b}{c})}\int_{0}^{\infty}$
$=\left [ 1-\dfrac{1}{log(\dfrac{a}{c})} \right ] - \left [ 1-\dfrac{1}{log(\dfrac{b}{c})}\right ]$
$=\dfrac{1}{log(\dfrac{b}{c})}-\dfrac{1}{log(\dfrac{a}{c})}$

$\displaystyle \int_{0}^{1}\frac{xe^{x}}{(x+1)^{2}}dx=$

0%
$\displaystyle \frac{e}{2}$
0%
$\displaystyle \frac{e-1}{2}$
0%
$\displaystyle \frac{e}{2}-1$
0%
$\displaystyle \frac{e-3}{2}$

Explanation

$\displaystyle \int_{0}^{1}e^x\left ( \dfrac{1}{(x=+1)}-\dfrac{1}{(x+1)^2}\right ) dx$

It is in the form of

$\displaystyle \int e^x \left [ f(x)+f^1 (x) \right ] dx=e^x f(x)+c$

$=\left [ \dfrac{e^x}{(1+x)}+c \right ] _{0}^{1}$

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

The solution of the equation

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

0%
1
0%
1/2
0%
2
0%
-2

Explanation

$\int_{\sqrt{2}}^{x}\dfrac{dx}{x\sqrt{x^2-1}}=\dfrac{\pi}{2}$
$=\left ( sin^{-1}x+c \right ) \int_{\sqrt{2}}^{x}=\dfrac{\pi}{2}$
$\left ( sec^{-1}x - sec^{-1}\sqrt{2} \right ) =\dfrac{\pi}{2}$
$sec^{-1}x- \dfrac{\pi}{4}=\dfrac{\pi}{12}$
$sec^{-1}x =\dfrac{\pi}{12}+\dfrac{3\pi}{12}$
$sec^{-1}=\dfrac{\pi}{3}$
$x=sec \dfrac{\pi}{3}$
$=2$

Evaluate the integral

$\displaystyle \int_{0}^{1}\frac{ {d} {x}}{ {x}^{2}+2 {x} {c} {o} {s}\alpha+1}$

0%
$\sin \alpha$
0%
$\tan^{-1}$ $(\sin \alpha)$
0%
$\dfrac{\alpha}{(2\sin \alpha)}$
0%
$\alpha$ $(\sin \alpha)$

Explanation

$\int_{0}^{1}\dfrac{dx}{x^2+x cos_\alpha+1}$
$\int_{0}^{1}\dfrac{dx}{(x+cos \alpha)^2}+ sin^2 \alpha$
$\left ( \dfrac{1}{sin \alpha} \left [ tan^{-1} \left ( \dfrac{x+ cos \alpha}{sin \alpha} \right ) \right ]+c \right )_{0}^{1}$
$=\dfrac{1}{sin \alpha} \left [ tan^{-1} \left ( \dfrac{1+cos \alpha}{sin \alpha} \right ) \right ]-\dfrac{1}{sin \alpha}tan^{-1}\left ( \dfrac{cos \alpha}{sin \alpha} \right )$
$=\dfrac{1}{sin \alpha}\left [ tan^{-1} (cot \dfrac{\alpha}{2})-tan^{-1}(cot \alpha) \right ]$
$=\dfrac{1}{sin \alpha} \left ( \alpha -\dfrac{\alpha}{2} \right ) =\dfrac{\alpha}{2} sin \alpha$

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

0%
$\displaystyle \frac{\pi}{\sqrt{5}}$
0%
$\displaystyle \frac{\pi}{2\sqrt{5}}$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{3\sqrt{5}}$

Explanation

$\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4 sin^2 x}dx =\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4 cos^2 x}dx$
$\int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{1+4 cos^2x}dx =\int_{0}^{\dfrac{\pi}{4}} \dfrac{sec^2x}{(sec^2x + 4)}dx$
$=\int_{0}^{\dfrac{\pi}{4}} \dfrac{d(tan x)}{5 + tan^2 x} dx= \left [ \dfrac{1}{\sqrt{5}} tan^{-1}\left ( \dfrac{tan x}{\sqrt{5}} \right ) + c \right ]_{0}^{\dfrac{\pi}{2}}$
$=\dfrac{1}{\sqrt{5}}(\dfrac{\pi}{2})$
$=\dfrac{\pi}{2\sqrt{5}}$

$\displaystyle \int_{0}^{16}\frac{dx}{\sqrt{x+9}-\sqrt{x}}=$

0%
10
0%
12
0%
14
0%
16

Explanation

$\int_{0}^{16}\dfrac{dx}{\sqrt{x+9}-\sqrt{x}}$
$\int_{0}^{16}\dfrac{\sqrt{x}+9+\sqrt{x}}{9}dx$
$=\dfrac{1}{19}\left [ \int_{0}^{16} \sqrt{x+9}dx + \int_{0}^{16}\sqrt{x}dx \right ]$
$=\dfrac{1}{19}\times \dfrac{2}{3}\left [ (x+a)^{\dfrac{3}{2}}\int_{0}^{16}+ x^{\dfrac{3}{2}} \int_{0}^{16} \right ]$
$=\dfrac{2}{27}\left [ (25)^{\dfrac{3}{2}} - (9)^{\dfrac{3}{2}}+ (16)^{\dfrac{3}{2}} \right ]$
$=\dfrac{2}{27}\left [ 125-27+64 \right ]$
$=\dfrac{2}{27}\left [ 189-27 \right ] =\dfrac{2}{27}\left [ 162 \right ]$
$=12$

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

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,

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

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

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

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

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

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

Explanation

Let

$I=\int { \cfrac { \sqrt { x } }{ 1+x } dx }$
Substitute

$t=\sqrt { x } \Rightarrow dt=\cfrac { 1 }{ 2\sqrt { x } } ds$

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

$\int _{ 0 }^{ 1 }{ Idx } ={ \left[ 2\sqrt { x } -2{ tan }^{ -1 }\sqrt { x } \right] }_{ 0 }^{ 1 }=2-\cfrac { \pi }{ 2 }$

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.