MCQ Questions for Class 12 Commerce Maths Integrals Quiz 9

$\displaystyle \frac{\pi }{4}< \alpha < \displaystyle \frac{\pi }{2}$ , value of

$\displaystyle \int_{-\pi /2}^{\pi /2}\displaystyle \frac{\sin 2x}{\sqrt{1+\sin 2\alpha \sin x}}$ is

0%
$-\displaystyle \frac{4}{3}\tan \alpha \sec \alpha$
0%
$-\displaystyle \frac{4}{3}\cot \alpha cosec\alpha$
0%
$-\displaystyle \frac{4}{3}\tan \alpha cosec\alpha$
0%
$-\displaystyle \frac{4}{3}\cot \alpha \sec \alpha$

Explanation

Let

$\displaystyle I=\int _{ -\pi /2 }^{ \pi /2 } \frac { \sin 2x }{ \sqrt { 1+\sin 2\alpha \sin x } } =2\int _{ -\pi /2 }^{ \pi /2 } \frac { \sin { x } \cos { x } }{ \sqrt { 1+\sin 2\alpha \sin x } } dx$

Substitute

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

$\displaystyle I=2\int _{ -1 }^{ 1 }{ \frac { t }{ \sqrt { t\sin { 2\alpha } +1 } } dt }$

Substitute

$u=t\sin { 2\alpha } \Rightarrow du=\sin { 2\alpha } dt$

$\displaystyle I=\int _{ 1-\sin { 2\alpha } }^{ 1+\sin { 2\alpha } }{ 2\csc ^{ 2 }{ 2\alpha } \frac { u-1 }{ \sqrt { u } } }$

$\displaystyle { \left[ \frac { 4 }{ 3 } { u }^{ \frac { 3 }{ 2 } }\csc ^{ 2 }{ 2\alpha } -4\sqrt { 5 } \csc ^{ 2 }{ 2\alpha } \right] }_{ 1-\sin { 2\alpha } }^{ 1+\sin { 2\alpha } }$

$\displaystyle =-\frac { 4 }{ 3 } \cot { \alpha } \csc { \alpha }$

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

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

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

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

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

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

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

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

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

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

$\displaystyle 2u=\pi /3$

$\displaystyle\int { \cfrac { { x }^{ 2 }-1 }{ { x }^{ 4 }+{ x }^{ 2 }+1 } } dx$ is equal to

0%
$\log { \left( { x }^{ 4 }+{ x }^{ 2 }+1 \right) } +c$
0%
$\log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c }$
0%
$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c }$
0%
$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 } +c }$

Explanation

Let

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

$\Rightarrow I=\int { \cfrac { 1-\cfrac { 1 }{ { x }^{ 2 } } }{ { x }^{ 2 }+1+\cfrac { 1 }{ { x }^{ 2 } } +1-1 } } dx$

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

$x+\cfrac { 1 }{ x } =t\quad \Rightarrow 1-\cfrac { 1 }{ { x }^{ 2 } } dx=dt$

$\therefore I=\int { \cfrac { 1 }{ { t }^{ 2 }-1 } } dt$

$=\cfrac { 1 }{ 2\times 1 } \log { \cfrac { t-1 }{ t+1 } +c }$

$=\cfrac { 1 }{ 2 } \log { \cfrac { x+\cfrac { 1 }{ x } -1 }{ x+\cfrac { 1 }{ x } +1 } } +c$

$=\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } } +c$

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

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

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

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

Let

$f(x)$ be a positive function. Let

$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$ ,

$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$ ,
where

$2k - 1 > 0$ , then

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

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

Explanation

$I_{1} = \displaystyle \int_{1 - k}^{k} xf \left \{x (1 - x)\right \}dx$

$= \displaystyle \int_{1 - k}^{k} (1 - x)f\left \{(1 - x)(1 -(1- x))\right \}dx$

$(\because \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx)$

$=\displaystyle \int_{1 - k}^{k} (1 - x)f\left \{(1 - x)x\right \}dx$

$= \displaystyle \int_{1 - k}^{k} f\left \{x(1 - x)\right \} dx$

$- \displaystyle \int_{1 - k}^{k} xf \left \{x(1 - x)\right \}dx$
Therefore,

$I_{1} = I_{2} - I_{1}\Rightarrow 2I_{1} = I_{2}$

$\Rightarrow \dfrac {I_{1}}{I_{2}} = \dfrac {1}{2}$

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.

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

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

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

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

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

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

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

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

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

Consider

$I=\displaystyle \int^{\pi}_{0}\displaystyle\frac{xdx}{1+\sin x}$ . What is I equal to?

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

Explanation

$I = \displaystyle \int_{0}^{\pi}{\dfrac{xdx}{1+\sin x}}$

by property,

$I = \displaystyle \int_{0}^{\pi}{\dfrac{(\pi-x)dx}{1+\sin x}}$

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

$2I =\displaystyle \int_{0}^{\pi}{\dfrac{\pi }{1+\sin x}} \times \dfrac{1-\sin x}{1-\sin x}dx$

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

$I = \dfrac{\pi}{2}\displaystyle \int _{0}^{\pi} [\sec^{2} x \tan x \sec x]dx$

$I = \dfrac{\pi}{2} [\tan{x} - \sec{x}]_{0}^{\pi}$

$I = \dfrac{\pi}{2} [0+2]$

$I = \pi$

$\int { { x }^{ 4 }{ e }^{ 2x } } dx=$

0%
$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$
0%
$\cfrac { { e }^{ 2x } }{ 2 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$
0%
$\cfrac { { e }^{ 2x } }{ 8 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$
0%
$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$

Explanation

$\int x^4.e^{2x}dx$

Applying by parts rule of integration,

$\dfrac{x^4}{2} e^{2x}-\dfrac{4}{2}\int x^3.e^{2x}dx$

Applying by-parts again, multiple times till the time powers in

$x$ in the integral vanishes,

$\dfrac{x^4}{2} e^{2x}-2[\dfrac{x^3}{2} e^{2x}-\dfrac{3}{2}\int x^2.e^{2x}dx]$

$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3[\dfrac{x^2}{2}.e^{2x}-\int x.e^{2x} dx]$

$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3\dfrac{x^2}{2}.e^{2x}-3[\dfrac{x}{2}.e^{2x}-\int\dfrac{e^{2x}}{2}dx]$

$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3\dfrac{x^2}{2}.e^{2x}-3\dfrac{x}{2}.e^{2x}+3\dfrac{e^{2x}}{4}+c$

$\rightarrow \cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$

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%
$\dfrac{4}{3}$

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.

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

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

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

$\displaystyle I=\int _{ { -\pi }/{ 6 } }^{ { \pi }/{ 6 } }{ \dfrac { \pi +4{ x }^{ 5 } }{ 1-\sin { \left( \left| x \right| +\dfrac { \pi }{ 6 } \right) } } } dx$ , 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 $\left( 0,\pi /2 \right)$

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

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

Let

$y = y(x), y(1)=1\ and\ y(e) ={e^2}$ . Consider

$J = \int {{{x + y} \over {xy}}} dy$ ,

$I = \int {{{x + y} \over {{x^2}}}} dx$ ,

$J - I = g\left( x \right)$ and g(1) = 1, then the value of g(e) is

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

Explanation

$I=\int { \dfrac { x+y }{ xy } } dy\quad ;\quad I=\int { \dfrac { x+y }{ { x }^{ 2 } } } dx$

$I=\int { \left( \dfrac { 1 }{ y } +\dfrac { 1 }{ x } \right) } dy\quad ;\quad I=\int { \left( \dfrac { 1 }{ x } +\dfrac { y }{ { x }^{ 2 } } \right) } dx$

$I\Rightarrow Iny+\dfrac { y }{ x } \quad ;\quad I=Inx-\dfrac { y }{ x }$

$y\left( x \right) =J-I$

$\Rightarrow Iny-Inx+\dfrac { 2y }{ x }$

$g\left( x \right) \Rightarrow In\left( y/x \right) +\dfrac { 2y }{ x } =In\left( \dfrac { y\left( x \right) }{ 2 } \right) +\dfrac { 2y\left( x \right) }{ x }$

$g\left( 1 \right) =Iny+2y+1$

$g\left( 2 \right) =In\left| \dfrac { { e }^{ 2 } }{ e } \right| +\dfrac { 2{ e }^{ 2 } }{ e }$

$\Rightarrow 2e+1$

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

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

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

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

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%
$1$

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

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

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

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

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

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

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

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

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

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

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%
$None\ of\ these$

One of the roots of the equation

$2000x^6+100x^5+10x^3+x-2=0$ is of the form

$\dfrac{m+\sqrt{n}}{r}$ . When 'm' is non zero integer and n and r relatively prime natural numbers. Then

$\dfrac{m+n+r}{100}=?$

0%
$100$
0%
$2$
0%
$3$
0%
$0$

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

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

The value of the integer

$I=\int_{1}^{\infty} \dfrac{(x^{2}-x)}{x^{3}\sqrt{(x^{2}-1)}}dx$ is

0%
$0$
0%
$2/3$
0%
$4/3$
0%
$none\ of\ these$

Explanation

Let,

$x = \sec t \implies dx = \sec t \tan t dt$

$x : 1 \rightarrow \infty, \sec t : 0 \rightarrow \pi/2$

$I = \int_{1}^{\infty} \dfrac{(x^2 - x)}{x^3 \sqrt{x^2 - 1}} dx$

$I = \int_{1}^{\infty} \dfrac{(x - 1)}{x^2 \sqrt{x^2 - 1}} dx$

$I = \int_{0}^{\pi/2} \dfrac{(\sec t - 1)}{\sec^2 t \sqrt{\sec^2 t - 1}} \sec t \tan t dt$

$I = \int_{0}^{\pi/2} \dfrac{(\sec t - 1)}{\sec^2 t . \tan t} \sec t \tan t dt$

$I = \int_{0}^{\pi/2} \dfrac{(\sec t - 1)}{\sec t } dt$

$I = \int_{0}^{\pi/2} (1 - \cos t ) dt$

$I = [t - \sin t]_{0}^{\pi/2}$

$I = \dfrac{\pi }{2} - 1$ (Ans)

Option D is correct.

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

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

The integral

$\int _{ \tfrac { \pi }{ 12 } }^{ \tfrac { \pi }{ 4 } }{ \dfrac { 8\cos { 2x } }{ { \left( \tan { x } +\cot { x } \right) }^{ 3 } } dx }$ equals:

0%
$\dfrac {15}{128}$
0%
$\dfrac {5}{64}$
0%
$\dfrac {13}{32}$
0%
$\dfrac {13}{256}$

Explanation

$\displaystyle\int^{{\tfrac{\pi}{4}}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x}{{(tan x + cot x)^3}}dx$

$\displaystyle \int^{\tfrac{\pi}{4}}_{{\tfrac{\pi}{12}}} \frac{8\ cos2x}{(\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx})^3}dx$

$\displaystyle \int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x\ sin^3x \ cos^3x}{(sin^2x +cos^2x )^3}dx$

$\displaystyle \int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x \ (8\ sin^3x\ cos^3x)}{(sin^2x +cos^2x )^3}\dfrac{dx}{8}$

$\displaystyle \int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x (sin2x)^3 }{(sin^2x +cos^2x )^3}\frac{dx}{8}$

$\int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} {cos2x (sin2x)^3 }{}dx$

put

$\displaystyle sin 2x=t\rightarrow2\ cos2x.dx=dt$

when

$x=\frac{\pi}{12}\rightarrow t= \frac12$

and

$x=\frac{\pi}{4}\rightarrow t= 1$

$\displaystyle\int^{1}_{\frac12} {cos2x.t^3 }{}\frac{dx}{2cos2x}$

$\displaystyle \int^{1}_{\frac12} {\frac{t^3 }{2}}{}dx$

$\displaystyle \frac12[\frac{t^4}{4}]^{1}_{\frac12}$

$\displaystyle \frac18[1-\frac{1}{16}]$

$\displaystyle \frac{15}{128}$

Hence option A is correct

Solve

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

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

Explanation

Let,

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

$x\rightarrow 0 \implies t \rightarrow 0$

$x\rightarrow \infty \implies t \rightarrow \pi/2$

Now,

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

$= \int_{0}^{\pi/2} \dfrac{\tan t . (t)}{1+\tan^2 t} dt$

$= \int_{0}^{\pi/2} \dfrac{\tan t . (t)}{\sec^2 t} dt$

$= \int_{0}^{\pi/2} \sin t \cos t .(t) dt$

$= \dfrac{1}{2}\int_{0}^{\pi/2} 2\sin t \cos t .(t) dt$

$= \dfrac{1}{2}\int_{0}^{\pi/2} \sin 2t .(t) dt$

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

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

$= \dfrac{\pi}{8}$ (Ans)

$\displaystyle \int_{0}^{100 \pi}\sqrt {1-\cos 2x}dx$ is

0%
$200\sqrt {2}$
0%
$100\sqrt {2}$
0%
$0$
0%
$none\ of\ these$

Explanation

Let $I = \int\limits_0^{100\pi } {\sqrt {1 - \cos 2x} dx}$

$I = \int\limits_0^{100\pi } {\sqrt {2{{\sin }^2}x} dx}$

$= \int\limits_0^{100\pi } {\sqrt 2 \sin xdx}$

$= - \sqrt 2 \left[ {\cos x} \right]_0^{100\pi }$

$= - \sqrt 2 \left( {\cos 100\pi - \cos 0} \right)$

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

$= 0$

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

The value of

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

0%
$0$
0%
$\dfrac {\pi}{\sqrt {3}}$
0%
$\dfrac {2\pi}{\sqrt {3}}$
0%
$cannot\ be\ evaluated$

$\displaystyle \int^{\pi/2}_{-\pi/2}\sqrt {\cos x-\cos^{3}x}dx=$

0%
$1$
0%
$4/3$
0%
$-1/3$
0%
$0$

$f(x) =\displaystyle \underset{1}{\overset{x}{\int}} \dfrac{\tan^{-1} t}{t} dt \, \, \forall \in R$ , then the value of

$f(e^2) - f \left (\dfrac{1}{e^2}\right)$ is

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

Let

$f:R\rightarrow R^{+}$ and

$I_{I}=\int^{k}_{1-k}\,xf(x(1-x))\,dx,I_2=\int^{k}_{1-k}f(x(1-x))\,dx$ where

$2k-1>0$ . Then

$\dfrac{I_I}{I_2}$ is

0%
2
0%
k
0%
1/2
0%
1

$\int {x\left( {\frac{{\ln {a^{{a^{\frac{x}{2}}}}}}}{{3{a^{\frac{{5x}}{2}}}{b^{3x}}}} + \frac{{\ln {b^{{b^{\frac{x}{2}}}}}}}{{2{a^{2x}}{b^{4x}}}}} \right)} dx\left( {where\,a,b \in R} \right)$ is equal to

0%
$\frac{1}{{6\ln {a^2}{b^3}}}{a^{2x}}{b^{3x}}\ln \frac{{{a^{2x}}{b^{3x}}}}{e} + k$
0%
$\frac{1}{{6\ln {a^2}{b^3}}}{a^{2x}}{b^{3x}}\ln \frac{1}{{e{a^{2x}}{b^{3x}}}} + k$
0%
$\frac{1}{{6\ln {a^2}{b^3}}}\frac{1}{{{a^{2x}}{b^{3x}}}}\ln \left( {e{a^{2x}}{b^{3x}}} \right) + k$
0%
$- \frac{1}{{6\ln {a^2}{b^3}}}\frac{1}{{{a^{2x}}{b^{3x}}}}\ln \left( {e{a^{2x}}{b^{3x}}} \right) + k$

$\displaystyle \int_{-1}^{1} \cot^{-1} \left(\dfrac{x+x^{3}}{1+x^{4}}\right)dx$ is equal to

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

$I=\int { \left\{ \log _{ e }( \log _{ e }{ x }) +\cfrac { 1 }{ { \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} } dx$ is equal to

0%
$x\log _{ e }( \log _{ e }{ x } )+c$
0%
$x\log _{ e }( \log _{ e }{ x })-\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$
0%
$x\log _{ e }{ x } \log _{ e }{ x } +\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$
0%
none of these

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

0%
$(x+1)e^{x+x^{-1}}+c$
0%
$(x-1)e^{x+x^{-1}}+c$
0%
$-xe^{x+x^{-1}}+c$
0%
$xe^{x+x^{-1}}+c$

Solve:

$\int\dfrac{sin^32x}{cos^52x}dx$

0%
$\dfrac{{{{\tan }^4}2x}}{8} + C$
0%
$\dfrac{{{{\cos}^4}2x}}{8} + C$
0%
$\dfrac{{{{\sin}^4}2x}}{8} + C$
0%
$\dfrac{{{{\sec}^4}2x}}{8} + C$

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

0%
$1/3$
0%
$1/2$
0%
$1/6$
0%
$1/4$

$\int _{ 1/2 }^{ 2 }{ \dfrac { 1 }{ x } \csc { ^{ 101 }\left( x-\dfrac { 1 }{ x } \right) dx } \\ }$ is equal to

0%
$1/4$
0%
$1$
0%
$0$
0%
$\dfrac{101}{2}$

Explanation

Let

$I=\displaystyle \int _{1/2}^2 \dfrac {1}{x} \csc ^{101} \left (x-\dfrac {1}{x} \right)dx$

Let

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

$\therefore \ I=-\displaystyle \int _2^{1/2} \dfrac {t}{t^2} \csc ^{101} \left (\dfrac {1}{t} -t \right) dt$

$=-\displaystyle \int _2^{1/2} \csc^{101} \left (\dfrac {1}{t} - t \right) dt$

$=-I$

or,

$2I=0\ \Rightarrow \ I=0$

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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