MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 10

Let

$\displaystyle u = \overset{\infty}{\underset{0}{\int}} \dfrac{dx}{x^4 + 7x^2 + 1} \& v = \overset{\infty}{\underset{0}{\int}} \dfrac{x^2 dx}{x^4 + 7x^2 + 1}$ then:

0%
v > u
0%
6v = $\displaystyle \pi$
0%
$\displaystyle 3u+2v=5\pi /6$
0%
$\displaystyle u+v=\pi /3$

Explanation

$\displaystyle v=\int_{0}^{\infty }\frac{x^{2}dx}{x^{4}+7x^{2}+1}$ put

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

$\displaystyle v=-\int_{\infty }^{0}\frac{\frac{1}{t^{2}}\cdot \frac{1}{t^{2}}dt}{\frac{1}{t^{4}}+\frac{1}{t^{2}}+1}=\int_{0}^{\infty }\frac{dx}{x^{4}+7x^{2}+1}$
v=u hence

$\displaystyle 2u=\int_{0}^{\infty }\left ( \frac{x^{2}+1}{x^{4}+7x^{2}+1} \right )dx$

$\displaystyle =\int_{0}^{\infty }\left ( \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7} \right )dx=\int_{0}^{\infty }\frac{d\left ( x-\frac{1}{x} \right )}{\left ( x-\frac{1}{x} \right )^{2}+3^{2}}=\int_{0}^{\infty }\frac{dt}{t^{2}+9}$

$\displaystyle \frac{2}{3}\left [ \tan^{-1}\frac{t}{3} \right ]_{0}^{\infty }$

$\displaystyle 2u=\pi /3$

$\displaystyle I=\int _{8}^{15} \frac{dx}{(x-3)\sqrt{x+1} }$ then

$I$ equals

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

Explanation

Put

$\sqrt{x+1} =t$ or

$x + 1 =t^{2}$

$\displaystyle \therefore I=\int _{3}^{4} \frac{2t}{(t^{2}-4) t} dt=\frac{2}{(2)(2)}\log \left. \left | \frac{t-2}{t+2}\right |\right ]_{3}^{4}$

$\displaystyle =\frac{1}{2} \left[ \log \frac{1}{3}-\log \frac{1}{5}\right]$

$\displaystyle =\frac{1}{2}\log \frac{5}{3}$

37 If

$n > 1,$ and

$\displaystyle I=\int _{0}^{\infty} \frac{dx}{(x+\sqrt{1+x^{2}})^{n}}$ then

$I$ equals

0%
$\displaystyle \frac{n}{n^{2}-1}$
0%
$\displaystyle \frac{2n}{n^{2}-1}$
0%
$\displaystyle \frac{n}{2(n^{2}-1)}$
0%
$\displaystyle \sqrt{ n^{2}-1}$

Explanation

$x$ and

$\sqrt{1+x^{2}}$ are involved,
put

$\sqrt{1-x^{2}} = t - x$ or

$1 + x^{2}=t^{2} -2tx + x^{2}$

$\displaystyle \Rightarrow x= \frac{1}{2}\left(t-\frac{1}{t}\right)$

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

$\displaystyle \therefore I= \int_{0}^{\infty} \frac{1}{2} \left(1+\frac{1}{t^{2}}\right) \frac{dt}{t^{n}}$

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

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

$=\displaystyle 0+\frac{1}{2} \left( \frac{1}{n-1}+\frac{1}{n+1}\right) =\frac{n}{n^{2}-1}$

A function

$f$ is defined by

$\displaystyle f(x)=\frac{1}{2^{r-1}},\frac{1}{2r}<x\leq \frac{1}{2^{r-1}},r=1,2,3,.....$ then the value of

$\displaystyle \int _{0}^{1}f(x)dx$ is equal

0%
$\cfrac 13$
0%
$\cfrac 14$
0%
$\cfrac 23$
0%
$\cfrac 12$

Explanation

$\displaystyle \int_{0}^{1}f(x) dx =\sum_{r=1}^{\infty} \int_{2^{-r}}^{2^{-(r-1)}} \frac{1}{2^{r-1}}dx$

$\displaystyle =\sum_{r=1}^{\infty}\frac{1}{2^{r-1}}[2^{-(r-1)}-2^{-r}]$

$\displaystyle =\sum_{1}^{\infty}2^{-2(r-1)}- \sum _{1}^{\infty} 2^{-2r+1}$

$\displaystyle =(2^{2}-2)\sum_{1}^{\infty}2^{-2r}=2\cdot \frac{1}{4}\cdot \frac{1}{1-1/4}=\frac{2}{3}\cdot$

$\displaystyle \int_{0}^{\alpha}\frac{dx}{1-\cos \alpha \cos x}=\frac{A}{\sin \alpha}+B(a\neq 0)$
Then possible values of

$A$ and

$B$ are

0%
$\displaystyle A=\frac{\pi}{2},B=0$
0%
$\displaystyle A=\frac{\pi}{4},B=\frac{\pi}{4\sin \alpha}$
0%
$\displaystyle A=\frac{\pi}{6},B=\frac{\pi}{\sin \alpha}$
0%
$\displaystyle A=\pi ,B=\frac{\pi}{\sin \alpha}$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \alpha }{ \frac { dx }{ 1-\cos { \alpha } \cos { \alpha } } }$

$\displaystyle =\int _{ 0 }^{ \alpha }{ \frac { dx }{ \left( \cos ^{ 2 }{ \frac { x }{ 2 } } +\sin ^{ 2 }{ \frac { x }{ 2 } } \right) -\cos { \alpha } \left( \cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } \right) } }$

$\displaystyle =\int _{ 0 }^{ \alpha }{ \frac { dx }{ \left( 1-\cos { \alpha } \right) \cos ^{ 2 }{ \frac { x }{ 2 } } +\left( 1+\cos { \alpha } \right) \sin ^{ 2 }{ \frac { x }{ 2 } } } }$

$\displaystyle =\int _{ 0 }^{ \alpha }{ \frac { dx }{ 2\sin ^{ 2 }{ \frac { \alpha }{ 2 } } \cos ^{ 2 }{ \frac { x }{ 2 } } +2\cos ^{ 2 }{ \frac { \alpha }{ 2 } } \sin ^{ 2 }{ \frac { x }{ 2 } } } }$

$\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ \alpha }{ \frac { \sec ^{ 2 }{ \frac { \alpha }{ 2 } } \sec ^{ 2 }{ \frac { x }{ 2 } } }{ \tan ^{ 2 }{ \frac { \alpha }{ 2 } } +\tan ^{ 2 }{ \frac { x }{ 2 } } } dx }$
put

$\displaystyle \tan { \frac { x }{ 2 } } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt$

$\displaystyle \therefore I=\int _{ 0 }^{ \tan { \frac { \alpha }{ 2 } } }{ \frac { \sec ^{ 2 }{ \frac { \alpha }{ 2 } } }{ { t }^{ 2 }+\tan ^{ 2 }{ \frac { \alpha }{ 2 } } } dt }$

$\displaystyle =\sec ^{ 2 }{ \frac { \alpha }{ 2 } } \cot { \frac { \alpha }{ 2 } } \left[ \tan ^{ -1 }{ \left( \frac { t }{ \tan { \frac { \alpha }{ 2 } } } \right) } \right] _{ 0 }^{ \tan { \frac { \alpha }{ 2 } } }$

$\displaystyle =\frac { 2 }{ \sin { \alpha } } .\frac { \pi }{ 4 } =\frac { \pi }{ 2\sin { \alpha } }$

$\displaystyle A=\frac { \pi }{ 2 } ,B=0$ or

$\displaystyle A=\frac { \pi }{ 4 } ,B=\frac { \pi }{ 4\sin { \alpha } }$

$\displaystyle \int_{e^{e^{e}}}^{e^{e^{e^{e}}}}\frac{dx}{xlnx\cdot ln\left ( lnx \right )\cdot ln\left ( ln\left ( lnx \right ) \right )}$ equals

0%
1
0%
1/e
0%
e-1
0%
1+e

Explanation

Let

$\displaystyle I=\int _{ { e }_{ { e }_{ e } } }^{ { e }^{ { e }^{ { e }^{ e } } } }{ \frac { 1 }{ x\log { x } .\log { \left( \log { x } \right) . } \log { \left( \log { \left( \log { x } \right) } \right) } } } dx$

Substitute

$\displaystyle \log { \left( \log { x } \right) } =t\Rightarrow \frac { 1 }{ x\log { x } } dx=dt$

$\displaystyle I=\int _{ e }^{ { { e }^{ e } } }{ \frac { 1 }{ t\log { t } } dt } =\left[ \log { \log { t } } \right] _{ e }^{ { e }^{ e } }=\left[ 1-0 \right] =1$

$f\left( x \right) =\int _{ 0 }^{ 1 }{ \left( xf\left( t \right) +1 \right) dt,then\int _{ 0 }^{ 3 }{ f\left( x \right) dx=12 } }$
because
Statement-2: f(x) = 3x + 1

0%
Statements-1 is true, statements-2 is true and statements-2 is correct explanation for statement-1.
0%
Statements-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statements-1.
0%
Statements-1 is true, statements-2 is false.
0%
Statements-1 is false, statements-2 is true.

Value of

$\displaystyle \int_{0}^{\pi /2}\displaystyle \frac{\sin 4\Theta }{\sin \Theta }\: d\Theta$ is

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

Explanation

$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \frac { \sin { 4\Theta } }{ \sin { \Theta } } d\Theta } =\int _{ 0 }^{ \pi /2 }{ \sin { 4\Theta } \csc { \Theta } d\Theta }$

$=\int _{ 0 }^{ \pi /2 }{ \left( 2\cos ^{ 3 }{ \Theta } +2\cos { \Theta } -6\sin ^{ 2 }{ \Theta } \cos { \Theta } \right) d\Theta }$

$\displaystyle ={ \left[ -2\sin ^{ 3 }{ \Theta } +\frac { 10\sin { \Theta } }{ 3 } +\frac { 2 }{ 3 } \sin { \Theta } \cos ^{ 2 }{ \Theta } \right] }_{ 0 }^{ \pi /2 }=\frac { 4 }{ 3 }$

Evaluate

$\displaystyle \int_{0}^{1}\left ( tx+1-x \right )^{n}dx,$ where n is a positive integer and t is a parameter independent of x. Hence

$\displaystyle \int_{0}^{1}x^{k}\left ( 1-x \right )^{n-k}dx=\frac{P}{\left [ ^{n}C_{k}\left ( n+1 \right ) \right ]}for\:k=0,1,......n$ , then

$P=$

0%
2
0%
1
0%
3
0%
None of these

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ 1 }{ { \left( tx+1-x \right) }^{ n } } dx=\int _{ 0 }^{ 1 }{ { \left\{ \left( t-1 \right) x+1 \right\} }^{ n } } dx=\left[ \frac { \left( t-1 \right) { x+1 }^{ n+1 } }{ \left( n+1 \right) \left( t-1 \right) } \right] _{ 0 }^{ 1 }$

$\displaystyle =\frac { 1 }{ n+1 } \left( \frac { { t }^{ n+1 }-1 }{ t-1 } \right) =\frac { 1 }{ n+1 } \left( 1+t+{ t }^{ 2 }+...{ t }^{ n } \right)$ ...(1)
Again

$\displaystyle\int _{ 0 }^{ 1 }{ { \left( tx+1-x \right) }^{ n } } dx=\int _{ 0 }^{ 1 }{ { \left[ \left( 1-x \right) +tx \right] }^{ n } } dx$

$\displaystyle=\int _{ 0 }^{ 1 }{ \left[ _{ }^{ n }{ { C }_{ 0 } }{ \left( 1-x \right) }^{ n }+^{ n }{ { C }_{ 1 } }{ \left( 1-x \right) }^{ n-1 }\left( tx \right) +^{ n }{ { C }_{ 2 } }{ \left( 1-x \right) }^{ n-2 }{ \left( tx \right) }^{ 2 }+...+^{ n }{ { C }_{ n } }{ \left( tx \right) }^{ n } \right] } dx$

$\displaystyle=\int _{ 0 }^{ 1 }{ \sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } } { \left( 1-x \right) }^{ n-r }{ \left( tx \right) }^{ r }dx=\sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } \left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-r }{ x }^{ r }dx } \right] { t }^{ r }$ ...(2)
From (1) and (2)

$\displaystyle \sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } \left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-r }.{ x }^{ r }dx } \right] { t }^{ r }=\frac { 1 }{ n+1 } \left( 1+t+...{ t }^{ n } \right)$
On equating coefficient of

${ t }^{ k }$ on both sides , we get

$\displaystyle ^{ n }{ { C }_{ k } }\left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-r },{ x }^{ r } } dx \right] \Rightarrow \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-k } } { x }^{ h }dx=\frac { 1 }{ { \left( n+1 \right) }^{ n }{ { C }_{ k } } }$

For x > 0, let f(x) =

$\displaystyle \int_{1}^{x}{ \frac{log t}{1 + t}} dt$ . Then f(x) + f

$\displaystyle \left( \frac{1}{x} \right)$ is equal to:

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

Explanation

$f(x)=\int _{ 1 }^{ x }{ \cfrac { \log { t } }{ 1+{ t } } }$

$--(1)$

$f\left( \cfrac { 1 }{ x } \right) =\int _{ 1 }^{ 1/x }{ \cfrac { \log { t } }{ 1+{ t } } }$

Put t=1/u

$f\left( \cfrac { 1 }{ x } \right) =\displaystyle\int _{ 1 }^{ x }{ \cfrac { -u.\log { u } }{ (1+{ u) } } \cfrac { -du }{ { u }^{ 2 } } }$

$f\left( \cfrac { 1 }{ x } \right) =\displaystyle \int _{ 1 }^{ x }{ \cfrac { \log { u } }{ u(1+{ u) } } du }$

$--(2)$

Add (1) and (2)

$=\displaystyle \int _{ 1 }^{ x }{ \cfrac { \log { u } }{ u } du }$

$z=logu\\ dz=\cfrac { du }{ u }$

$\int _{ 1 }^{ x }{ zdz }$

$={ \cfrac { 1 }{ 2 } (\log { x } })^{ 2 }$

The value of

$\displaystyle\int _{ 0 }^{ { 1 }/{ \sqrt { 2 } } }{ \dfrac { \sin ^{ -1 }{ x } }{ { \left( 1-{ x }^{ 2 } \right) }^{ { 3 }/{ 2 } } } dx }$ is

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

Explanation

Let

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

When

$x=0$ , then

$\sin { t } =0=\sin { 0 } \Rightarrow t=0$

When

$x=\dfrac { 1 }{ \sqrt { 2 } }$ , then

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

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

Put

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

$\therefore l=\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \dfrac { t }{ \cos ^{ 3 }{ t } } \cdot \cos { t } dt } =\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ t\sec ^{ 2 }{ t } dt }$

$={ \left[ t\cdot \tan { t } \right] }_{ 0 }^{ { \pi }/{ 4 } }-\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \tan { t } dt }$

$=\dfrac { \pi }{ 4 } -{ \left[ \log { \sec { t } } \right] }_{ 0 }^{ { \pi }/{ 4 } }$

$=\dfrac { \pi }{ 4 } -\log { \sqrt { 2 } } -0=\dfrac { \pi }{ 4 } -\log { { 2 }^{ { 1 }/{ 2 } } } =\dfrac { \pi }{ 4 } -\dfrac { 1 }{ 2 } \log { 2 }$

Evaluate

$\displaystyle \int_{0}^{\pi /4}\frac{\cos x-\sin x}{10+\sin 2x}dx$

0%
$\displaystyle \frac{1}{3}\left (\tan^{-1} \frac{\sqrt{2}}{3}+\tan^{-1} \frac{1}{3} \right )$
0%
$\displaystyle \frac{1}{3}\left (\tan^{-1} \frac{\sqrt{1}}{3}-\cot^{-1} \frac{2}{3} \right )$
0%
$\displaystyle \frac{1}{3}\left ( \tan^{-1} \frac{\sqrt{2}}{3}-\tan^{-1} \frac{1}{3} \right )$
0%
$\displaystyle \frac{1}{3}\left ( \tan^{-1} \frac{\sqrt{1}}{3}-\cot^{-1} \frac{1}{3} \right )$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { \cos { x } -\sin { x } }{ 10+\sin { 2x } } } dx$

Put

$\\ \cos { x } +\sin { x } =t\Rightarrow \left( -\sin { x } +\cos { x } \right) dx=dt$

$\displaystyle I=\int _{ 1 }^{ \sqrt { 2 } }{ \frac { 1 }{ 10+\left( { t }^{ 2 }-1 \right) } dt } =\int _{ 1 }^{ \sqrt { 2 } }{ \frac { 1 }{ { t }^{ 2 }+9 } } dt$

$\displaystyle =\frac { 1 }{ 3 } \left[ \tan ^{ -1 }{ \frac { t }{ 3 } } \right] _{ 1 }^{ \sqrt { 2 } }=\frac { 1 }{ 3 } \left( \tan ^{ -1 }{ \frac { \sqrt { 2 } }{ 3 } -\tan ^{ -1 }{ \frac { 1 }{ 3 } } } \right)$

The value of

$\displaystyle \int_{3}^{4}\sqrt {(4 - x)(x - 3)}dx$ is

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

Explanation

Let

$\displaystyle I = \int_{3}^{4} \sqrt {(4 - x)(x - 3)}\ dx$

$\displaystyle = \int_{3}^{4} \sqrt {-x^{2} + 7x - 12}\ dx$

$\displaystyle = \int_{3}^{4} \sqrt {-\left (x^{2} - 7x + \dfrac {49}{4} - \dfrac {49}{4}\right ) - 12}\ dx$

$\displaystyle = \int_{3}^{4} \sqrt {- \left (x - \dfrac {7}{2}\right )^{2} + \dfrac {49}{4} - 12}\ dx$

$\displaystyle = \int_{3}^{4} \sqrt {\dfrac {1}{4} - \left (x - \dfrac {7}{2}\right )^{2}}\ dx$

Let

$t = x - \dfrac {7}{2}\rightarrow dt = dx$

$\therefore$ Upper limit

$= \dfrac {1}{2}$ and lower limit

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

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

$\displaystyle = 2\int_{0}^{1/2} \sqrt {\left (\dfrac {1}{2}\right )^{2} - t^{2}}dx$

$\displaystyle = 2\left [\dfrac {t}{2}\sqrt {\dfrac {1}{4} - t^{2}} + \dfrac {1}{8} \sin^{-1} 2t\right ]_{0}^{\frac12}$

$= 2\left [0 + \dfrac {1}{8}\times \dfrac {\pi}{2}\right ]$

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

The value of the integral

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

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

Explanation

Let

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

Put

$x=\sin\theta$

$\Rightarrow dx=\cos\theta \, d\theta$

When

$x=\cfrac {1}{3},\theta =\sin^{-1}\left (\cfrac {1}{3}\right )$ and when

$x=1, \theta =\cfrac {\pi}{2}$

$\Rightarrow \displaystyle I=\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta-\sin^3\theta)^{\frac {1}{3}}}{\sin^4\theta}cos \theta \,d\theta$

$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(1-\sin^2\theta)^{\frac {1}{3}}}{\sin^4\theta}\cos \theta \, d\theta$

$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(\cos\theta)^{\frac {2}{3}}}{\sin^4\theta}\cos \theta \, d\theta$

$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(\cos\theta)^{\frac {2}{3}}}{\sin^2\theta \sin^2\theta}\cos \theta d\theta$

$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\cos\theta)^{\frac {5}{3}}}{(\sin\theta)^{\frac {5}{3}}}cosec^2\theta d\theta$

$\displaystyle=\int_{sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}(\cot\theta)^{\frac {5}{3}}cosec^2\theta d\theta$

Put

$\cot \theta=t$

$\Rightarrow -cosec^2\theta \,d\theta=dt$
When

$\theta=\sin^{-1}\left (\dfrac {1}{3}\right ),t=2\sqrt 2$ and when

$\theta=\dfrac {\pi}{2}, t=0$

$\therefore I=-\int_{2\sqrt 2}^0(t)^{\frac {5}{3}}dt$

$\Rightarrow \displaystyle I=\int_0^{2\sqrt 2} (t)^{\frac {5}{3}}dt$

$\displaystyle =\left [\frac {3}{8}(t)^{\frac {8}{3}}\right ]_0^{2\sqrt 2}$

$\displaystyle =\frac {3}{8}[(2\sqrt 2)^{\frac {8}{3}}]$

$=\dfrac {3}{8}[(\sqrt 8)^{\frac {8}{3}}]$

$=\cfrac {3}{8}[(8)^{\frac {4}{3}}]$

$=\cfrac {3}{8}[16]$

$=3\times 2$

$=6$
Hence, the correct Answer is A.

${ I }_{ m }=\overset { e }{ \underset { 1 }{ \int } } (lnx)^{ m }dx,$ where

$m\epsilon N,$ then

${ I }_{ 10 }+10{ I }_{ 9 }$ is equal to-

0%
${ e }^{ 10 }$
0%
$\frac { { e }^{ 10 } }{ 10 }$
0%
e
0%
e-1

If I =

$\overset { 2 }{ \underset { -3 }{ \int } }$ (|x + 1 | + |x + 2| + |x - 1|)dx, then I equal

0%
$\dfrac{31}{2}$
0%
$\dfrac{35}{2}$
0%
$\dfrac{47}{2}$
0%
None of these

$\displaystyle\int^{\pi /3}_0\frac{\cos \theta}{5-4\sin \theta}d\theta$ equal to.

0%
$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5}{5+2\sqrt{3}}\right)$
0%
$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5}{5-2\sqrt{3}}\right)$
0%
$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5+2\sqrt{3}}{5}\right)$
0%
$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5-2\sqrt{3}}{5}\right)$

Explanation

$I=\int _{ 0 }^{ { \pi }/{ 3 } }{ \cfrac { \cos { \theta } }{ 5-4\sin { \theta } } d\theta }$

Let

$5-4\sin { \theta =t }$

$\Rightarrow \cos { \theta } d\theta =-\cfrac { dt }{ 4 }$

$I=\int _{ 5 }^{ 5-2\sqrt { 3 } }{ \cfrac { -dt }{ 4t } }$

$I=\cfrac { 1 }{ 4 } \int _{ 5-2\sqrt { 3 } }^{ 5 }{ \cfrac { dt }{ t } }$

$I=\cfrac { 1 }{ 4 } { \left[ \log { t } \right] }^{ 5 }_{ 5-2\sqrt { 3 } }$

$I=\cfrac { 1 }{ 4 } { \left[ \log { 5 } - \log { (5-2\sqrt { 3 } ) } \right] }$

$I=\cfrac { 1 }{ 4 } { \left[ \log \left({ \cfrac { 5 }{ 5-2\sqrt { 3 } } }\right) \right] }$

$\therefore I=\cfrac { 1 }{ 4 } \log { \left( \cfrac { 5 }{ 5-2\sqrt { 3 } } \right) }$

The value of

$\int _{ 1/3 }^{ 3 }{ \cfrac { \tan { \left( { x }^{ 2 }+\cfrac { 1 }{ { x }^{ 2 } } \right) } }{ \sin { \left( x+\cfrac { 1 }{ x } \right) } } } \cfrac { dx }{ x }$ is

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

Let

$I_{1} =\displaystyle \int_{0}^{1}\dfrac {e^{x}dx}{1 + x}$ and

$I_{2} = \displaystyle \int_{0}^{1} \dfrac {x^{2}dx}{e^{x^{3}}(2 - x^{3})}$ , then

$\dfrac {I_{1}}{I_{2}}$ is

0%
$3/e$
0%
$e/3$
0%
$3e$
0%
$1/3e$

Explanation

Given :

${ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) } } \quad { I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ x^{ 3 } }(2-{ x }^{ 3 }) } }$

${ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) } } \\ { I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ x^{ 3 } }(2-{ x }^{ 3 }) } }$

Let :

$1-{ x }^{ 3 }=t\quad x\rightarrow 0\quad t\rightarrow 1$

$-3{ x }^{ 2 }=dt\quad x\rightarrow 1\quad t\rightarrow 0\\ { I }_{ 2 }=-\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ 3 } \cfrac { { e }^{ t-1 } }{ (1+t) } dt } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t } }{ (1+t) } dt } \\ \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) } } }{ \cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t } }{ (1+t) } dt } } =3e$

Hence the correct answer is

$3e$

The value of

$\int^3_{1/3}\dfrac{tan(x^2-\dfrac{1}{x^2})}{sin(x+\dfrac{1}{x}) }\dfrac{dx}{x}$ is

0%
0
0%
$\dfrac{3}{2}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{4}{3}$

$\int _{ 0 }^{ \pi /2 }{ \sin ^{ 8 }{ x } \cos ^{ 2 }{ x } dx }$ is equal to

0%
$\cfrac { \pi }{ 512 }$
0%
$\cfrac {3 \pi }{ 512 }$
0%
$\cfrac { 5\pi }{ 512 }$
0%
$\cfrac {7 \pi }{ 512 }$

The value of the integral

$\displaystyle \int_0^1 \dfrac{x^{\alpha}-1}{\log x}dx$ is

0%
$\log (\alpha+1)$
0%
$2\log (\alpha+1)$
0%
$3\log \alpha$
0%
none of these

Explanation

$I(\alpha)=\displaystyle\int_0^1\dfrac{x^{\alpha}-1}{\text{log }x}dx$

Differentiating with respect to

$\alpha,$ we get

$,$

$I'(\alpha)=\displaystyle\int_0^1x^\alpha dx=\dfrac{1}{\alpha +1}$

We need to solve this simple differential equation

$:I'(\alpha)=\dfrac{1}{\alpha +1}$

On solving, we get,

$I(\alpha)=\text{log}(\alpha +1)+C$

Now, from the initial equation, we can find that

$I(0)=0$

$I(0)=\text{log}(1)+C=0$

$\Rightarrow C=0$

So,

$I(\alpha)=\text{log}(\alpha +1)$

The value of the definite integral

$\displaystyle\int^{\pi/2}_{0}\left(\cos ^{10}x\cdot \sin 12x\right)dx$ , is equal to.

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

Explanation

Let

$I =\displaystyle \int_{0}^{\pi/2}(\cos^{10}x.\sin\, 12x) dx$

$\cos^{10}x.\sin\,12x = \cos^{10}x.\sin(11x+x) = \cos^{10}x.(\sin\,11x.\cos\,x + \cos\, 11x.\sin\,x)$

$= - \dfrac{1}{11}(-11\,\sin\,11x.\cos^{11}x - 11\,\cos\,11x.\cos^{10}x.\sin\,x)$

$= -\dfrac{1}{11} \times$

$\dfrac{d}{dx}(\cos\,11x.\cos^{11}x)$ ... (1)

Therefore,

$I = \left (-\dfrac {1}{11}\right)\displaystyle \int_{0}^{\pi/2} \dfrac{d}{dx}(\cos\,11x.\cos^{11}x) dx$

$I =- \dfrac {1}{11}\times \cos\,11x.\cos^{11}x]_{0}^{\pi/2}$

$I = -\dfrac {1}{11}\times (0-1)$

$I = \dfrac{1}{11}$

Correct answer is option B.

$\displaystyle I=\int _{ { -\pi }/{ 6 } }^{ { \pi }/{ 6 } }{ \dfrac { \pi +4{ x }^{ 5 } }{ 1-\sin { \left( \left| x \right| +\dfrac { \pi }{ 6 } \right) } } } dx$ , then I equals to

0%
$4\pi$
0%
$2\pi +\dfrac { 1 }{ \sqrt { 3 } }$
0%
$2\pi -\dfrac { 1 }{ \sqrt { 3 } }$
0%
$4\pi +\sqrt { 3 } -\dfrac { 1 }{ \sqrt { 3 } }$

If,

$\int^{\frac{\pi}{2}} _0 \frac{sin^2x}{(1 + cosx^2)} dx = 0$

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

${I}_{n}=\int_{1}^{e}{\left(logx\right)^{n}dx}$ and

${I}_{n}=A+{BI}_{n-1}$ then
A=........., B=............

0%
$e,-n$
0%
$1/e,n$
0%
$-e,n$
0%
$-e-n$

$\displaystyle \int _{ 0 }^{ x }{ \cfrac { \sin { x } }{ 1+\cos ^{ 2 }{ x } } } dx=\pi \cfrac { \cos { \alpha } }{ 1-\sin ^{ 2 }{ \alpha } }$

0%
for no value of $\alpha$
0%
for exactly two values of $\alpha$ in $\left( 0,\pi \right)$
0%
for atleast one $\alpha$ in $\left( \pi /2,\pi \right)$
0%
for exactly one $\alpha$ in $\left( 0,\pi /2 \right)$

What is

$\int _{ 0 }^{ 1 }{ x{ \left( 1-x \right) }^{ 9 }dx }$ equal to?

0%
$\dfrac{1}{240}$
0%
$\dfrac{1}{110}$
0%
$\dfrac{1}{132}$
0%
$\dfrac{1}{148}$

Explanation

We need to find

$\int_{0}^{1}x(1-x)^9dx$

$\int_{0}^{1}x(1-x)^9dx=\int_{0}^{1}(1-x)x^9dx$

$=\int_{0}^{1}(x^9-x^{10})dx$

$=\left[\dfrac{x^{10}}{10}-\dfrac{x^{11}}{11}\right]_{0}^{1}$

$=\dfrac{1}{10}-\dfrac{1}{11}$

$=\dfrac{11-10}{110}$

$=\dfrac{1}{110}$

$\therefore \int_{0}^{1}x(1-x)^9dx=\dfrac{1}{110}$

The value of

$\int _{ 0 }^{ 14 }{ \dfrac { \left[ { x }^{ 2 } \right] }{ \left[ { x }^{ 2 }-28x+196 \right] +\left[ { x }^{ 2 } \right] } }$
where [x] is the integral part of real x, is

0%
14
0%
0
0%
7
0%
49

$I_1 =\int^1_0 2x^2 dx, I_2 =\int^1_0 2^{x3} dx, I_3 =\int^2_1 2x^{x2} dx$ then

0%
$I_1 > I_2$
0%
$I_2 > I_1$
0%
$I_3 > I_4$
0%
$I_3 = I_4$

$\displaystyle E = \int_{R}^{\infty}\dfrac{GMm}{x^2}$ dx , (where

$G$ ,

$M$ ,

$m$ are constants ) equal to

0%
$-\dfrac{GMm}{R^2}$
0%
$+\dfrac{GMm}{R^2}$
0%
$-\dfrac{GMm}{R}$
0%
$+\dfrac{GMm}{R}$

$\int _{ 0 }^{ 4036 }{ \dfrac { { 2 }^{ x } }{ { 2 }^{ x }+{ 1 }^{ 4036-x } } } dx=............$

0%
2018
0%
4035
0%
2017
0%
-2015

$I_{1} = \int_{0}^{1} 2^{x^{3}} dx, I_{2} = \int_{0}^{1}2^{x^{2}}dx, I_{3} = \int_{1}^{2}2^{x^{2}}dx$ and

$I_{4} = \int_{1}^{2}2^{x^{3}}dx$ , then

0%
$I_{1} > I_{2}$
0%
$I_{2} > I_{1}$
0%
$I_{3} > I_{4}$
0%
$I_{1} > I_{3}$

Explanation

Given :-

$I_{1} = \int_{0}^{1} 2^{x^{3}} dx, I_{2} = \int_{0}^{1}2^{x^{3}}dx, I_{3} = \int_{1}^{2}2^{x^{2}}dx$ and

$I_{4} = \int_{1}^{2}2^{x^{3}}dx$ ,

As we know

$x^2 < x^3$ for

$x\in(1,2)$ , So

$2^{x^{3}}$ will also be grater than

$2^{x^{2}}$ for

$x\in (0,1)$ and

$x\in (1,2)$

$\therefore I_{1}>I_2$

hence

$I_{3} = \int_{1}^{2}2^{x^{2}}dx$ <

$I_{4} = \int_{1}^{2}2^{x^{3}}dx$

And also

$2^{x^{2}}$ when

$x \in (1,2)$ is grater than

$2^{x^{3}}$ for

$x\in (0,1)$ hence

$I_{1} = \int_{0}^{1}2^{x^{3}}dx$ <

$I_{3} = \int_{1}^{2}2^{x^{2}}dx$

Hence,

$I_{1}>I_2$ is correct.

A function

$f(x)$ which satisfies the relation

$f(x) = e^{x} + \int_{0}^{1} e^{x}f(t)dt$ , then

$f(x)$ is

0%
$\dfrac {e^{x}}{2 - e}$
0%
$(e - 2)e^{x}$
0%
$2e^{x}$
0%
$\dfrac {e^{x}}{2}$

Value of the definite integral

$\displaystyle \int _{ 0 }^{ 1 }{ \cot ^{ -1 }{ \left( 1-x+{ x }^{ 2 } \right) } dx }$ is:

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

$I=\int _{ 0 }^{ \pi }{ \frac { x\sin { x } }{ 1+{ cos }^{ 2 }x } dx }$ , then the value of

$\sin { \sqrt { I } }$ , is

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

$if\,{l_n} = \int\limits_0^1 {{x^m}} {\left( {\ln x} \right)^n}dx,\,and\,{l_n} = \frac{1}{{m + 1}}{l_{n - 1}}$ then k is equal to

0%
n
0%
-n
0%
m-1
0%
None of these

$For x>0$ ,let

$f\left( x \right) =\int _{ 1 }^{ 2 }{ \dfrac { \log { t } }{ 1+t } } dt$ ,then

$f\left( x \right) +f\left( \dfrac { 1 }{ x } \right)$ is equal to:

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

Evaluate the integral,

$\int _{ 0 }^{ 1 }{ \cos { \left( 2\cot ^{ -1 }{ \sqrt { \dfrac { 1-x }{ 1+x } } } \right) } } dx=$

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

$\int _{ \pi /4 }^{ \pi /2 }{ \cos { \theta } \csc ^{ 2 }{ \theta } d\theta = }$

0%
$\sqrt {2}-1$
0%
$1-\sqrt {2}$
0%
$\sqrt {2}+1$
0%
$None\ of\ these$

Explanation

$\int _{ \pi /4 }^{ \pi /2 }{ \cfrac { \cos { \theta } }{ \sin ^{ 2 }{ \theta } } }$

$\sin { \theta } =t$

$\cos { \theta d\theta } =dt$

$\int _{ \pi /4 }^{ \pi /2 }{ { t }^{ -2 }dt } ={ \left( \cfrac { t-1 }{ -1 } \right) }_{ \pi /4 }^{ \pi /2 }={ -\left( \cfrac { 1 }{ t } \right) }_{ \pi /4 }^{ \pi /2 }={ \left( +\cfrac { 1 }{ \sin { \theta } } \right) }_{ \pi /4 }^{ \pi /2 }$

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

$\displaystyle \int_2^ 3\frac{(x+2)^2}{2x^2- 10x +53}dx$ is equal to

0%
$2$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$\dfrac{5}{2}$

Solve

$\displaystyle \int _{ 0 }^{ 1/2 }{ \dfrac { x\sin ^{ -1 }{ x } }{ \sqrt { 1-{ x }^{ 2 } } } dx= }$

0%
$\dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } \pi }{ 12 }$
0%
$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 3 } \pi }{ 12 }$
0%
$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 2 } \pi }{ 12 }$
0%
$None\ of\ these$

For n >2,

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

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

Explanation

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

substitute

$\sec x+\tan x = t\rightarrow \sec x dx = \dfrac{dt}{t}\Rightarrow x(0,\dfrac{\pi}{2})\rightarrow t(1,\infty )$

$\sec x = \dfrac{1}{2}\left ( t+\dfrac{1}{t} \right )$

$\int_{0}^{\frac{\pi}{2}} \dfrac{\sec ^2xdx}{(\sec x+\tan x)^n}=\int_{1}^{\infty } \dfrac{\frac{1}{2}\left ( t+\frac{1}{t} \right )\frac{dt}{t}}{t^n}$

$=\dfrac{1}{2}\int_{1}^{\infty }\left ( \dfrac{1}{t}+\dfrac{1}{t^{n+2}} \right )dt$

$=\left [ \dfrac{t^{-n+1}}{-n+1} +t^{-n+1}+\dfrac{t^{-(n+1)}(n+2)}{-(n+1)}\right ]_1^\infty$

$=-\dfrac{1}{2} \left [\dfrac{1}{-n+1}+1-\dfrac{n+2}{n+1} \right ]$

$=-\dfrac{1}{2} \left [ \dfrac{n^2+2n-1+1-n^2}{1-n^2} \right ]$

$=\dfrac{n}{n^2-1}$

Let

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

$J=\displaystyle\int _{ 0 }^{ 1 }{ \sqrt { \dfrac { 1-\sqrt { x } }{ 1+\sqrt { x } } } dx }$ , then the correct statement is

0%
$I+J=4$
0%
$I-J=\pi$
0%
$I=\dfrac{2+\pi}{2}$
0%
$J=\dfrac{4-\pi}{2}$

$\int _{ -\pi /4 }^{ \pi /4 }{ ln\sqrt { 1+\sin { 2x } } dx }$ has the value equal to:

0%
$-\dfrac {\pi}{4}l\ n2$
0%
$-\dfrac {\pi}{2}l\ n2$
0%
$-\dfrac {\pi}{8}l\ n2$
0%
$-\dfrac {\pi}{16}l\ n2$

$\int _{ -1 }^{ 1 }{ x } \tan { ^{ -1 } } xdx$

0%
$\left( \frac { \pi }{ 2 } -1 \right)$
0%
$\left( \frac { \pi }{ 2 } +1 \right)$
0%
$(\pi-1)$
0%
$None$

$\int _{ 0 }^{ 2\pi }{ \sqrt { 1+\sin { \dfrac { x }{ 2 } } } dx } =$

0%
$0$
0%
$2$
0%
$8$
0%
$4$

$\displaystyle\int^{1}_0\sqrt{x(1-x)}dx=$ .

0%
$\dfrac{\pi}{8}$
0%
$\dfrac{3\pi}{8}$
0%
$\dfrac{5\pi}{4}$
0%
$\dfrac{\pi}{2}$

Explanation

We are given,

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

let

$x=\sin^2 \theta \ \Rightarrow \ dx=2\sin \theta . \cos \theta \ d\theta =\sin 2\theta \ d\theta$

(differentiating both side) & (chain rule)

now when

$x=0, \sin^2 \theta =0\ \Rightarrow \theta =0$

& when

$x=1, \sin^2 \theta =1\ \Rightarrow \theta =\dfrac {\pi}{2}$

$I=\displaystyle \int_{0}^{\pi /2}\sqrt {\sin^2 \theta (1-\sin^2 \theta)} (\sin 2\theta \ d\theta)$

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

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

(here

$\sin \theta . \cos \theta$ are positive in

$[0, \pi /2$ ])

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

$=\dfrac {1}{2} \displaystyle \int_{0}^{\pi /2} (2\sin \theta \cos \theta ) \sin 2\theta \ d\theta$ (multiply & divide the integration by

$2$ to get

$2\cos \theta \sin \theta =\sin 2\theta$ )

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

$I=\dfrac {1}{2} \displaystyle \int_{0}^{\pi /2} \left (\dfrac {1-\cos 4\theta}{2}\right) d\theta$

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

$I=\dfrac {1}{2}\left [\left (\dfrac {\theta}{2}\right)_0^{\pi /2} -\left (\dfrac {\sin 4\theta}{4}\right)_0^{\pi /2} \right]$

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

$=\dfrac {1}{4}\left (\dfrac {\pi}{4}-0 \right)$

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

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

${ I }_{ 2 }=\int _{ 1 }^{ 1/x }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } } dt$ for x > 0, then

0%
${ I }_{ 1 }={ I }_{ 2 }$
0%
${ I }_{ 1 }>{ I }_{ 2 }$
0%
${ I }_{ 2 }>{ I }_{ 1 }$
0%
None of these.

$\displaystyle \overset{\pi/2}{\underset{0}{\int}} \dfrac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$ equals-

0%
$\pi/ab$
0%
$2\pi/ab$
0%
$ab/ \pi$
0%
$\pi/2ab$

Explanation

$I=\displaystyle\int^{\pi/2}_0\dfrac{dx}{a^2\cos^2x+b^2\sin^2x}$

$=\displaystyle\int^{\pi/2}_0\dfrac{\sec^2xdx}{a^2+b^2\tan^2x}$

Let

$\tan x=t$

$x=0, \tan x=0$

$\sec^2xdx=dt$

$x=\pi/2, \tan x=\infty$

$\Rightarrow \displaystyle\int^{\infty}_0\dfrac{dt}{a^2+b^2t^2}=\dfrac{1}{b^2}\displaystyle\int^{\infty}_0\dfrac{dt}{\left(\dfrac{a^2}{b^2}+t^2\right)}$

$\Rightarrow \dfrac{1}{b^2}\times \dfrac{1}{(a/b)}\left[\tan^{-1}\left(\dfrac{t}{(a/b)}\right)\right]^{\infty}_0$

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

$=\dfrac{1}{ab}[\pi/2]$

$\Rightarrow \dfrac{\pi}{2ab}$ .

1055448_1061298_ans_3c2304d12cfd410fba4d1226386b7727.png

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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