MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 9

Let

$\displaystyle I_1 = \int_1^2 \frac{1}{\sqrt{1 + x^2}} dx$ and

$I_2 \displaystyle = \int_1^2 \frac{1}{x} dx$ . Then

0%
$I_1 > I_2$
0%
$I_2 > I_1$
0%
$I_1 = I_2$
0%
$I_1 > 2 I_2$

Explanation

$\displaystyle { I }_{ 1 }=\int _{ 1 }^{ 2 }{ \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } dx$

$={ \left( \log { \left( x+\sqrt { 1+{ x }^{ 2 } } \right) } \right) }_{ 1 }^{ 2 }$

$\displaystyle =\log { \left( 2+\sqrt { 5 } \right) } -\log { \left( 1+\sqrt { 2 } \right) } =\log { \left( \frac { \left( 2+\sqrt { 5 } \right) }{ \left( 1+\sqrt { 2 } \right) } \right) }$

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

$\therefore { I }_{ 2 }>{ I }_{ 1 }$

The value (s) of

$\displaystyle \int_{0}^{1}\frac{x^{4}\left ( 1-x \right )^{4}}{1+x^{2}} dx$ is (are)

0%
$\displaystyle \frac{22}{7}-\pi$
0%
$\displaystyle \frac{2}{105}$
0%
$0$
0%
$\displaystyle \frac{71}{15}-\frac{3\pi}{2}$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 }{ \left( 1-x \right) }^{ 4 } }{ 1+{ x }^{ 2 } } } dx$

$\displaystyle =\int _{ 0 }^{ 1 }{ \frac { \left( { x }^{ 4 }-1 \right) { \left( 1-x \right) }^{ 4 }+{ \left( 1-x \right) }^{ 4 } }{ \left( 1+{ x }^{ 2 } \right) } dx }$

$\displaystyle =\int _{ 0 }^{ 1 }{ \left( { x }^{ 2 }-1 \right) } { \left( 1-x \right) }^{ 4 }dx+\int _{ 0 }^{ 1 }{ \frac { { \left( 1+{ x }^{ 2 }-2x \right) }^{ 2 } }{ \left( 1+{ x }^{ 2 } \right) } dx } \\$

$\displaystyle =\int _{ 0 }^{ 1 }{ \left\{ \left( { x }^{ 2 }-1 \right) { \left( 1-x \right) }^{ 4 }+\left( 1+{ x }^{ 2 } \right) -4x+\frac { { 4x }^{ 2 } }{ \left( 1+{ x }^{ 2 } \right) } \right\} dx }$

$\displaystyle =\int _{ 0 }^{ 1 }{ \left( \left( { x }^{ 2 }-1 \right) { \left( 1-x \right) }^{ 4 }+\left( 1+{ x }^{ 2 } \right) -4x+4\frac { 4 }{ 1-{ x }^{ 2 } } \right) dx }$

$\displaystyle =\int _{ 0 }^{ 1 }{ \left( { x }^{ 6 }-{ 4x }^{ 5 }+{ 5x }^{ 4 }-{ 4x }^{ 2 }+4-\frac { 4 }{ 1+{ x }^{ 2 } } \right) dx }$

$\displaystyle =\left[ \frac { { x }^{ 7 } }{ 7 } -\frac { { 4x }^{ 6 } }{ 6 } +\frac { { 5x }^{ 5 } }{ 5 } -\frac { { 4x }^{ 3 } }{ 3 } +4x-4\tan ^{ -1 }{ x } \right] _{ 0 }^{ 1 }$

$\displaystyle =\frac { 1 }{ 7 } -\frac { 4 }{ 6 } +\frac { 5 }{ 5 } -\frac { 4 }{ 3 } +4-4\left( \frac { \pi }{ 4 } -0 \right) =\frac { 22 }{ 7 } -\pi$

$\displaystyle I= \int_{0}^{1}\frac{x dx}{8+x^{3}}$ then the smallest interval in which

$I$ lies is

0%
$\displaystyle \left ( 0,\frac{1}{8} \right )$
0%
$\displaystyle \left ( 0,\frac{1}{9} \right )$
0%
$\displaystyle \left ( 0,\frac{1}{10} \right )$
0%
$\displaystyle \left ( 0,\frac{1}{7} \right )$

Explanation

$\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { x }{ 8+{ x }^{ 3 } } } dx=\int _{ 0 }^{ 1 }{ \frac { x }{ \left( x+2 \right) \left( { x }^{ 2 }-2x+4 \right) } } dx$

$\displaystyle =\int _{ 0 }^{ 1 }{ \left( \frac { x+2 }{ 6\left( { x }^{ 2 }-2x+4 \right) } -\frac { 1 }{ 6\left( x+2 \right) } \right) } dx$

$\displaystyle =\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \left( \frac { 2x-2 }{ 2\left( { x }^{ 2 }-2x+4 \right) } +\frac { 3 }{ 6{ x }^{ 2 }-2x+4 } \right) dx } -\frac { 1 }{ 6 } \int { \frac { 1 }{ x+2 } } dx$

$\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { 2x-2 }{ { x }^{ 2 }-2x+4 } } dx+\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ 2 }-2x+4 } } -\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \frac { 1 }{ x+2 } } dx$

$\displaystyle =\left[ \frac { 1 }{ 12 } \log { \left( { x }^{ 2 }-2x+4 \right) } -\frac { 1 }{ 6 } \log { \left( x+2 \right) } \right] +\frac { \tan ^{ -1 }{ \left( \frac { x-1 }{ \sqrt { 3 } } \right) } }{ 2\sqrt { 3 } }$

$\displaystyle =\frac { 1 }{ 12 } \log { \left( 3 \right) } -\frac { 1 }{ 6 } \log { \left( 3 \right) } +\frac { \tan ^{ -1 }{ \left( \frac { 2 }{ \sqrt { 3 } } \right) } }{ 2\sqrt { 3 } }$

$\displaystyle -\frac { 1 }{ 12 } \log { \left( 4 \right) -\frac { 1 }{ 6 } \log { \left( 2 \right) } } +\frac { \tan ^{ -1 }{ \left( \frac { -1 }{ \sqrt { 3 } } \right) } }{ 2\sqrt { 3 } }$

And this lies between

$\displaystyle \left( 0,\frac { 1 }{ 9 } \right)$

$\displaystyle I_{n} =\int_{0}^{\frac{\pi }{4}}\tan ^{n}xdx$

then

$\displaystyle \frac{1}{I_{2}+I_{4}},\frac{1}{I_{3}+I_{5}},\frac{1}{I_{4}+I_{6}}$ are in?

0%
$A.P$
0%
$H.P$
0%
$G.P$
0%
None of these

Explanation

$\displaystyle I_{n}= \int_{0}^{\pi /4}\tan ^{n-2}x\tan ^{2}xdx$

$\displaystyle= \int_{0}^{\pi /4}\tan ^{n-2}x\left ( \sec ^{2}x-1 \right )dx=\int_{0}^{\pi /4}\tan ^{n-2}x \sec ^{2}x dx -\int_{0}^{\pi /4}\tan ^{n-2}xdx$

$\displaystyle I_{n}= \left [ \frac{\tan ^{n-1}x}{n-1} \right ]_{0}^{\pi /4}-I_{n-2}$

$\displaystyle \therefore I_{n}+I_{n-2}= \frac{1}{n-1}$
Substitute

$n=4,5,6, ....$ we get

$\displaystyle I_{4}+I_{2}, I_{5}+I_{3}, I_{6}+I_{4},....$ are respectively

$\displaystyle \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\cdots$ which are in H.P. and hence their reciprocals are in A.P.

$\displaystyle I_{t}=\int_{0}^{\dfrac{\pi }{2}}\frac{\sin^{2}tx}{\sin^{2}x}dx$ then ,

$I_{1},I_{2},I_{3}$ are in

0%
A.P.
0%
H.P.
0%
G.P.
0%
None of these

Explanation

Consider

$\displaystyle I_{t+2}+I_{t}-2I_{t+1}$

$\displaystyle \int_{0}^{\pi /2}\frac{\left ( \sin ^{2}tx+\sin^{2}\left ( t+2 \right )x-2\sin ^{2}\left ( t+1 \right )x \right )dx}{\sin ^{2}x}$

$\displaystyle \int_{0}^{\pi /2}\frac{\left ( \sin ^{2}\left ( t+2 \right )x-\sin^{2}\left ( t+1 \right )x-\sin ^{2}\left ( t+1 \right )x+\sin^{2}tx \right )dx}{\sin ^{2}x}$

$\displaystyle \int_{0}^{\pi /2}\frac{\left [ \sin \left ( 2t+3 \right )x-\sin \left ( 2t+1 \right )x \right ]dx}{\sin x}$

$\displaystyle =2\int_{0}^{\pi /2}\cos \left ( 2t+2 \right )x dx=2\left ( \frac{\sin \left ( 2t+2 \right )x}{2\left ( t+1 \right )} \right )^{\pi /2}_{0}$

$\displaystyle = \frac{1}{\left ( t+1 \right )}\left ( 0 \right )=0$

$\displaystyle \therefore I_{t},I_{t+1},I_{t+2}\epsilon A.P.$
Short Cut Method :
Substitute

$t=1, 2, 3$ in given integral we get

$\displaystyle I_{1}= \frac{\pi}{2},I_{2}=\pi , I_{3}=\frac{3\pi }{2}$

$\displaystyle I_{1},I_{2}, I_{3}\epsilon A.P.$

The value of

$\displaystyle \int_{0}^{\pi /2}\sin \theta \log \left ( \sin \theta \right )\:d\theta$ equals

0%
$\displaystyle \log_{e}\left ( \frac{1}{e} \right )$
0%
$\displaystyle \log _{2}e$
0%
$\displaystyle \log_{e}{2}-1$
0%
$\displaystyle \log_{e}\left ( \frac{e}{2} \right )$

Explanation

Let

$I=\displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}\sin \theta \log(\sin \theta )d\theta =\displaystyle \frac{1}{2}\int_{0}^{\displaystyle \frac{\pi}{2}}\sin \theta \log(\sin^{2} \theta )d\theta \left ( \because \log_{e}a^{m}=m\log_{e}a \right )$

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

Take

$t=\cos\theta$

$\Rightarrow \sin \theta d\theta=-dt$

$\theta=0,t=1$ and

$\theta =\displaystyle\frac{\pi}{2},t=0$
Then,

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

$=-\displaystyle \frac{1}{2}\left [ \int_{0}^{1}\left ( t^{2}+\frac{t^{4}}{2}+\frac{t^{6}}{3}+...\infty \right )dt \right ]$

$=-\displaystyle\frac{1}{2}\left [\displaystyle\frac{1}{3}+\frac{1}{2\cdot5}+\frac{1}{7\cdot3}+...\infty \right ]$

$=-\left [\displaystyle\frac{1}{2.3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+....\infty \right ]$

$=-\left [\left (\displaystyle\frac{1}{2}-\frac{1}{3}\right )+\left (\displaystyle\frac{1}{4}-\displaystyle\frac{1}{5} \right )+\left (\displaystyle\frac{1}{6}-\displaystyle \frac{1}{7} \right )+....\infty\right ]$

$=-\displaystyle\frac{1}{2}+\frac{1}{3}-\displaystyle\frac{1}{4}+\frac{1}{5}-\displaystyle \frac{1}{6}+\frac{1}{7}+....\infty$

$=\left(1-\displaystyle\frac{1}{2}+\frac{1}{3}-\displaystyle\frac{1}{4}+\frac{1}{5}-\displaystyle \frac{1}{6}+\frac{1}{7}+....\infty\right)-1$

$=\log_{e}2-1$

$=\log_{e}2-\log e=\log_{e}\left (\displaystyle\frac{2}{e} \right )$

Ans: C

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

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

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

The value(s) of

$\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 4 }{ \left( 1-x \right) }^{ 4 } }{ 1+{ x }^{ 2 } } } dx$ is (are)

0%
$\cfrac{22}{7}-\pi$
0%
$\cfrac{2}{105}$
0%
$0$
0%
$\cfrac{71}{15}-\cfrac{3\pi}{2}$

$\displaystyle 2f(x)+f(-x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$ , then the value of

$\displaystyle\int_{\frac{1}{e}}^{e}{f(x)dx}$ , is

0%
$1$
0%
$0$
0%
$e$
0%
$-1$

Explanation

Since,

$\displaystyle 2f(x)+f(-x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$ ...(i)

$\displaystyle\therefore 2f(-x)+f(x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$ (replace

$x$ by

$-x$ ) ...(ii)

$\implies f(x)=f(-x)$ [subtracting equations (i) and (ii)]

$\displaystyle\therefore 3f(x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$
Hence,

$\displaystyle I=\int_{\frac{1}{e}}^{e}{f(x)dx}=\frac{1}{3}\int_{\frac{1}{e}}^{e}{\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}dx}$
Now, put

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

$\displaystyle\therefore I=\frac{1}{3}\int_{e}^{\frac{1}{e}}{t\sin{\left(\frac{1}{t}-t\right)}.\left(-\frac{1}{t^2}\right)dt}$

$\displaystyle=\frac{1}{3}\int_{e}^{\frac{1}{e}}{\frac{1}{t}\sin{\left(t-\frac{1}{t}\right)}dt}$

$\displaystyle=-\frac{1}{3}\int_{\frac{1}{e}}^{e}{\frac{1}{t}\sin{\left(t-\frac{1}{t}\right)}dt}$

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

$\displaystyle\therefore\int_{\frac{1}{e}}^{e}{f(x)dx}=0$ .

$\displaystyle I= \int_{1/\pi }^{\pi }\frac{1}{x}\cdot \sin \left ( x-\frac{1}{x} \right )dx$ then I is equal to

0%
$0$
0%
$\displaystyle \pi$
0%
$\displaystyle \pi -\frac{1}{\pi }$
0%
$\displaystyle \pi +\frac{1}{\pi }$

Explanation

Let

$\displaystyle I=\int _{ \dfrac { 1 }{ \pi } }^{ \pi } \frac { 1 }{ x } \cdot \sin \left( x-\dfrac { 1 }{ x } \right) dx$

Substitute

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

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

$\displaystyle =-\int _{ \dfrac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ t } \sin { \left( t-\dfrac { 1 }{ t } \right) dt } } -I\\ \Rightarrow 2I=0\Rightarrow I=0$

$x$ satisfies the equation

$\displaystyle\left(\int_0^1{\frac{dt}{t^2+2t\cos{\alpha}+1}}\right)x^2-\left(\int_{-3}^3{\frac{t^2\sin{2t}}{t^2+1}dt}\right)x-2=0$

for

$(0<\alpha<\pi)$

then the value of

$x$ is?

0%
$\displaystyle\pm\sqrt{\frac{\alpha}{2\sin{\alpha}}}$
0%
$\displaystyle\pm\sqrt{\frac{2\sin{\alpha}}{\alpha}}$
0%
$\displaystyle\pm\sqrt{\frac{\alpha}{\sin{\alpha}}}$
0%
$\displaystyle\pm2\sqrt{\frac{\sin{\alpha}}{\alpha}}$

Explanation

$\displaystyle \left( \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha } +1 } } \right) x^{ 2 }-\left( \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t } }{ t^{ 2 }+1 } dt } \right) x-2=0$

Here,

$\displaystyle \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t } }{ t^{ 2 }+1 } dt } =0 \left(\because f(t)=\frac { t^{ 2 }\sin { 2t } }{ t^{ 2 }+1 } \text{ is odd}\right)$

So, the equation becomes

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

Now,

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha } +1 } } =\int _{ 0 }^{ 1 }{ \frac { dt }{ (t+\cos { \alpha } )^{ 2 }+\sin ^{ 2 }{ \alpha } } }$

$=\displaystyle \frac { 1 }{ \sin { \alpha } } \left[{ \tan ^{ -1 }{ \left(\frac { t+\cos { \alpha} }{ \sin { \alpha } } \right) } }\right]_{ 0 }^{ 1 }$

$\displaystyle =\frac { 1 }{ \sin { \alpha } } \left[\tan ^{ -1 }{\cot {\dfrac{\alpha}2 } -\tan ^{ -1 }{ \cot { \alpha } } }\right]$

$\displaystyle =\frac { 1 }{ \sin { \alpha } } \left[\frac { \pi }{ 2 }-\frac { \alpha }{ 2 } -\frac { \pi }{ 2 } +\alpha\right]$

$\displaystyle =\frac { \alpha }{ 2\sin { \alpha } }$

So, equation (1) becomes

$\displaystyle\frac { \alpha }{ 2\sin { \alpha } } x^{ 2 }-2=0$

$\Rightarrow \displaystyle x=\pm 2 \sqrt {\frac{\sin \alpha}{\alpha}}$

Hence, option D.

The tangent to the graph of the function

$\displaystyle y = f(x)$ at the point with abscissa x = a forms with the x-axis an angle of

$\displaystyle \pi/3$ and at the point with abscissa x = b at an angle of

$\displaystyle \pi/4$ , then the value of the integral,

$\displaystyle \int_{a}^{b} f'(x).f''(x)dx$ is equal to

0%
1
0%
0
0%
$\displaystyle -\sqrt 3$
0%
-1

Explanation

Given , at

$x=a , \frac{dy}{dx}=\tan {\frac{\pi}{3}}=\sqrt{3}$

$\Rightarrow f'(a)=\sqrt{3}$
Also, at

$x=b, \frac{dy}{dx}=\tan {\frac{\pi}{4}}=1$

$\Rightarrow f'(b)=1$
Now,

$\displaystyle \int_{a}^{b} f'(x).f''(x)dx$
Put

$f'(x)=t \Rightarrow f''(x)dx=dt$

$=\displaystyle \int_{\sqrt{3}}^{1} t dt$

$=-\displaystyle \int_{1}^{\sqrt{3}} t dt$

$\displaystyle =-[\frac{t^{2}}{2}]_{1}^{\sqrt{3}}$

$=-1$

Let

$\displaystyle F\left ( x \right )=f\left ( x \right )+f\left ( \frac{1}{x} \right )$ where

$\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\log t}{1+t}dt$

Then

$F(e)$ is equal to?

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

Explanation

$\displaystyle F\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt+\int_{1}^{1/x}\frac{\ln t}{1+t}dt$

$\displaystyle F\left ( x \right )=\int_{1}^{x}\left ( \frac{\ln t}{1+t}+\frac{\ln t}{\left ( 1+t \right )t} \right )dt=\int_{1}^{t}\frac{\ln t}{t}dt=\frac{1}{2}\left ( \ln x \right )^{2}$

$F\left ( e \right )=1/2$

The value of

$\displaystyle \int _{ 0 }^{ \pi /2 }{ \frac { d\theta }{ 5+3\cos { \theta } } }$ is?

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

Explanation

Let

$I=\displaystyle \int _{ 0 }^{ \pi /2 }{ \frac { d\theta }{ 5+3\cos { \theta } } } =\int _{ 0 }^{ \pi /2 }{ \frac { d\theta }{ 5+3.\frac { 1-\tan ^{ 2 }{ \frac { \theta }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { \theta }{ 2 } } } } }$

$\displaystyle =\int _{ 0 }^{ \pi /2 }{ \frac { \sec ^{ 2 }{ \frac { \theta }{ 2d\theta } } }{ 8+2\tan ^{ 2 }{ \frac { \theta }{ 2 } } } }$

$\displaystyle =\int _{ 0 }^{ \pi /2 }{ \frac { \frac { 1 }{ 2 } \sec ^{ 2 }{ \left( \frac { \theta }{ 2 } \right) } d\theta }{ 4+\tan ^{ 2 }{ \left( \frac { \theta }{ 2 } \right) } } } =\int _{ 0 }^{ 1 }{ \frac { dt }{ 4+{ t }^{ 2 } } }$

$[$ Put

$\displaystyle \tan { \frac { \theta }{ 2 } } =t$

So,

$\dfrac { 1 }{ 2 } \sec ^{ 2 }{ \dfrac { \theta }{ 2 } } d\theta =dt]$

$\therefore I=\displaystyle =\frac { 1 }{ 2 } \left[ \tan ^{ -1 }{ \frac { t }{ 2 } } \right] _{ 0 }^{ 1 }$

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

$= \displaystyle \frac { 1 }{ 2 } \tan ^{ -1 }{ \dfrac { 1 }{ 2 } }$

Evaluate :

$\displaystyle \underset{0}{\overset{\infty}{\int}} \dfrac{dx}{(1 + x^2)^4}$

0%
$\dfrac{\pi}{32}$
0%
$\dfrac{3 \pi}{32}$
0%
$\dfrac{5 \pi}{32}$
0%
$\dfrac{7 \pi}{32}$

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

The value of

$\int _{ 0 }^{ \infty }{ \cfrac { \log { x } }{ { a }^{ 2 }+{ x }^{ 2 } } }$

0%
$\cfrac{2\pi\log{a}}{a}$
0%
$\cfrac{\pi\log{a}}{2a}$
0%
$\pi \log{a}$
0%
$0$

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

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

$\displaystyle \int_{0}^{\pi /2}\sin x\log \left ( \sin x \right )dx=$

0%
$\displaystyle \log _{e}e$
0%
$\displaystyle \log _{e}2$
0%
$\displaystyle \log _{e}\left ( e/2 \right )$
0%
$\displaystyle \log _{e}\left ( 2/e \right )$

Explanation

$\displaystyle I= \int_{0}^{\pi /2}\sin x\log \left ( \sin x \right )dx$

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

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

$\displaystyle = \frac{1}{2}\int_{0}^{1}\log \left ( 1-t^{2} \right )dt$ where

$\displaystyle t= \cos x$

$\displaystyle = \frac{1}{2}\int_{0}^{1}\left [ -t^{2}-\frac{\left ( t^{2} \right )^{2}}{2}-\frac{\left ( t^{2} \right )^{3}}{3}-\cdots \right ]dt$

$\displaystyle = -\left [ \frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\cdots \right ]$

$\displaystyle = -\left [ \left ( \frac{1}{2}-\frac{1}{3} \right )+\left ( \frac{1}{4}-\frac{1}{5} \right )+\left ( \frac{1}{6}-\frac{1}{7} \right )+\cdots \right ]$

$\displaystyle = -1+\left ( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots \right )$ .

$\displaystyle = -\log _{e}e+\log _{e}\left ( 1+1 \right )= \log _{e}\left ( 2/e \right )$

$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{n}}dx,\:\forall\:n\:> 1$ is equal to?

0%
$\displaystyle2 \int_{0}^{\infty}\frac{1}{1+x}dx$
0%
$\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^{n}}dx$
0%
$\displaystyle \int_{1}^{\infty}\frac{dx}{(x^{n}-1)^{1/n}}$
0%
$\displaystyle \int_{0}^{1}\frac{1}{(1-x^{n})^{1/n}}$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \infty } \frac { 1 }{ 1+x^{ n } } dx=\int _{ \infty }^{ 0 } \frac { -dx }{ 1+x^{ n } }$

Substitute

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

$\displaystyle I=\int _{ 0 }^{ \infty } \frac { 1 }{ y^{ 2 } } \frac { dy }{ 1+\left( \frac { 1 }{ y } \right) ^{ n } } =\int _{ 0 }^{ \infty } \frac { y^{ n-1 } }{ y(1+y^{ n }) } dy$

Substitute

$t^{ n }-1+y^{ n }\Rightarrow n{ t }^{ n-1 }dt=n{ y }^{ n-1 }dy$

$\displaystyle I=\int _{ 1 }^{ \infty } \frac { dt }{ t(t^{ n }-t)^{ 1/n } } =\int _{ \infty }^{ 1 } -\frac { 1 }{ t^{ 2 } } .\frac { dt }{ \left( 1-\frac { 1 }{ t^{ n } } \right) ^{ 1/n } }$

Substitute

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

$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dz }{ (1-z^{ n })^{ 1/n } } =\int _{ 0 }^{ 1 } \frac { dx }{ (1-x^{ n })^{ 1/n } }$

$\displaystyle I_1=\int_{0}^{\pi /2}\frac{x}{\sin x}dx$ and

$\displaystyle I_2=\int_{0}^{\pi /2}\frac{\tan ^{-1}x}{x}dx,$ then

$\displaystyle \frac{I_{1}}{I_{2}}=$

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

Explanation

Substitute

$\displaystyle x=\tan \theta \therefore I_2=\int_{0}^{\pi /4}\frac{\theta }{\tan \theta }.\sec ^{2}\theta d\theta$ or

$\displaystyle I_{2}=\int_{0}^{\pi /4}\frac{2\theta }{2\sin \theta \cos \theta }d\theta =\int_{0}^{\pi /4} \frac{2\theta }{\sin 2\theta }d\theta$
Now substitute

$\displaystyle 2\theta =t \therefore I_{2}=\frac{1}{2}\int_{0}^{\pi /2}\frac{t}{\sin t}dt=\frac{1}{2}I_{1}$

$\displaystyle \therefore \frac{I_{1}}{I_{2}}=2\Rightarrow \left ( C \right )$

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

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

0%
$\displaystyle\frac{\pi }{4}$
0%
does not exist
0%
$\displaystyle \frac{\pi \sqrt{3}}{4}$
0%
None of these

Explanation

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

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

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

The value of

$\displaystyle \int_{1}^{1/e}f(x)dx+\int_{1}^{e}f(x)dx$ where

$f(x)$ is given as

$\displaystyle \frac{log\:x}{1+x}$ equals

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

Explanation

Let

$\displaystyle l_{ 1 }=\int _{ 1 }^{ 1/e } \frac { logx }{ 1+x } dx$ and

$\displaystyle l_{ 2 }=\int _{ 1 }^{ 0 } \frac { log\: x }{ 1+x } dx$

Now for

$\displaystyle l_{ 1 }=\int _{ 1 }^{ 1/e } \frac { log\: t }{ 1+x }$

Substitute

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

$l_{ 1 }=\int _{ 1 }^{ e } \dfrac { log\: t }{ t(1+t) } dt=\int _{ 1 }^{ e } \dfrac { log\: x }{ x(1+x) } dx$ ( Replacing

$t\rightarrow \: x$ )

$\displaystyle \therefore l_{ 1 }+l_{ 2 }=\int _{ 1 }^{ e } \frac { log\: x }{ x(1+x) } dx+\int _{ 1 }^{ e } \frac { log\: x }{ 1+x } dx$

$\displaystyle =\int _{ 1 }^{ e } \frac { (log(x))(1+x) }{ x(1+x) } dx=\int _{ 1 }^{ e } \frac { log\: x }{ x } dx$

$\displaystyle =\frac { 1 }{ 2 } \left[ (log\: x) \right] _{ 1 }^{ e }=\frac { 1 }{ 2 } [log_{ e }e-log_{ e }1]=\frac { 1 }{ 2 }$

The value of

$\displaystyle \int_{1/e}^{\tan x}\displaystyle \frac{t}{1+t^{2}}\, dt\, +\, \displaystyle \int_{1/e}^{\cot x}\displaystyle \frac{dt}{t\left ( 1+t^{2} \right )}$ is

0%
$1/2$
0%
$1$
0%
$\pi /4$
0%
none of these

Explanation

$\displaystyle \int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \frac { dt }{ t\left( 1+t^{ 2 } \right) }$

$\displaystyle =\int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \left( \frac { 1 }{ t } -\frac { 1 }{ 1+{ t }^{ 2 } } \right) dt$

$=\dfrac { 1 }{ 2 } { \left[ \log { \left( 1+{ t }^{ 2 } \right) } \right] }_{ 1/e }^{ \tan { x } }+{ \left[ \log { t } \right] }_{ 1/e }^{ \cot { x } }-{ \left[ \tan ^{ -1 }{ t } \right] }_{ 1/e }^{ \cot { x } } =1$

Evaluate :

$\displaystyle \int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\times \left [ \sin ^{-1}\left ( 1-2x^{2} \right ) +\cos ^{-1}\left ( 2x\sqrt{1-x^{2}} \right )\right ]dx$

0%
$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$
0%
$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$
0%
$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$
0%
$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$

Explanation

$\displaystyle N^{r}$ put

$\displaystyle x=\sin \theta ,$ then

$\displaystyle \sin ^{-1}\left ( 1-2\sin ^{2}\theta \right )+\cos ^{-1}\left ( 2\sin \theta \cos \theta \right )=\sin ^{-1}\left ( \cos 2\theta \right )+\cos ^{-1}\left ( \sin 2\theta \right )$

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

$\displaystyle =\pi -2\theta -2\theta =\pi -4\sin ^{-1}x$

$\displaystyle \therefore I=\int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\left [ \pi -4\sin -\mid x \right ]dx$

$\displaystyle =2 \int_{0}^{\frac{1}{\sqrt{2}}}\pi \frac{x^{8}}{1-x^{4}}dx+0$ by Prop.V

$\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}\frac{x^{8}-1+1}{1-x^{4}}dx$

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

$\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}-\left ( x^{4}+1 \right )+\frac{1}{2}\left \{ \frac{1}{1-x^{2}}+\frac{1}{1+x^{2}} \right \}dx$

$\displaystyle =2x\left [ -\left ( \frac{x^{5}}{5}+x \right )+\frac{1}{4}\log \frac{1+x}{1-x} +\frac{1}{2}\tan -1x\right ]^{1/\sqrt{2}}_{0}$

$\displaystyle =\pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$

Ans: A

Solve :-

$\int_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}} = }$

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

Solve:-

$\int\limits_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}}}$

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

$I= \displaystyle \int_{0}^{1/\sqrt{3}}\displaystyle \frac{dx}{\left ( 1+x^{2} \right )\sqrt{1-x^{2}}}$ then

$I$ is equal to

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

Explanation

Let

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

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

$\displaystyle I=\int { \frac { 1 }{ \sin ^{ 2 }{ t } +1 } dt }$
Multiply numerator and denominator by

$\csc ^{ 2 }{ t }$

$\displaystyle I=\int { \frac { \csc ^{ 2 }{ t } }{ \csc ^{ 2 }{ t } +1 } dt } =\int { \frac { \csc ^{ 2 }{ t } }{ \cot ^{ 2 }{ t } +2 } dt }$
Substitute

$u=\cot { t } \Rightarrow du=-\csc ^{ 2 }{ t } dt$

$\displaystyle I=-\int { \frac { 1 }{ { u }^{ 2 }+2 } du } =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \frac { u }{ \sqrt { 2 } } }$

$\displaystyle =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \frac { \cot { t } }{ \sqrt { 2 } } } =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \frac { \cot { \sin ^{ -1 }{ x } } }{ \sqrt { 2 } } }$
Therefore

$\displaystyle \int _{ 0 }^{ 1/\sqrt { 3 } }{ \frac { 1 }{ \left( 1+{ x }^{ 2 } \right) \sqrt { 1-{ x }^{ 2 } } } dx } =\frac { \pi }{ 4\sqrt { 2 } }$

$\displaystyle \int_{0}^{\pi /3}\frac{\cos \theta }{3+4\sin \theta }d\theta =\lambda \log \frac{3+2\sqrt{3}}{3}$ then

$\displaystyle \lambda$ equals

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{8}$

Explanation

Given

$\displaystyle \int_{0}^{\pi /3}\frac{\cos \theta }{3+4\sin \theta }d\theta =\lambda \log \frac{3+2\sqrt{3}}{3}$ ....(1)

Consider,

$I= \int _{ 0 }^{ \pi /3 } \dfrac { \cos \theta }{ 3+4\sin \theta } d\theta$
Substitute

$\sin \theta =t$

$\Rightarrow \cos \theta d\theta =dt$

$I=\displaystyle \int _{ 0 }^{ \sqrt { 3 } /2 } \frac { dt }{ 3+4t }$
Substitute

$3+4t=u$

$4dt=du$

$\therefore I=\displaystyle \frac {1}{4}\int _{ 3 }^{ 3+2\sqrt { 3 } } \frac { du }{ u }$

$\displaystyle =\frac{1}{4}[\log { u } ]_{ 3 }^{3+ 2\sqrt { 3 } }$

$\displaystyle I=\frac{1}{4} [\log { (3+2\sqrt { 3 } )-\log { 3 } }]$

$\Rightarrow I=\displaystyle \frac{1}{4}\log \frac { 3+2\sqrt { 3 } }{ 3 }$
So, on comparing with (1), we get

$\lambda =\displaystyle \frac{1}{4}$

The value of

$\displaystyle \int ^{\tan x}_{1/e}\displaystyle \frac{t\, dt}{1+t^{2}}+\displaystyle \int ^{\cot x}_{1/e}\displaystyle \frac{dt}{t\left ( 1+t^{2} \right )}$ is

0%
$\displaystyle \frac{1}{2+tan^{2}x}$
0%
$1$
0%
$\pi /4$
0%
$\displaystyle \frac{2}{\pi }\displaystyle \int ^{1}_{-1}\displaystyle \frac{dt}{1+t^{2}}$

Explanation

$\displaystyle \int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \frac { dt }{ t\left( 1+t^{ 2 } \right) }$

$\displaystyle =\int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \left( \frac { 1 }{ t } -\frac { 1 }{ 1+{ t }^{ 2 } } \right) dt$

$=\frac { 1 }{ 2 } { \left[ \log { \left( 1+{ t }^{ 2 } \right) } \right] }_{ 1/e }^{ \tan { x } }+{ \left[ \log { t } \right] }_{ 1/e }^{ \cot { x } }-{ \left[ \tan ^{ -1 }{ t } \right] }_{ 1/e }^{ \cot { x } }\\ =1$

$\displaystyle I_{n}= \int_{0}^{\pi /4}\tan ^{n}x\times \sec ^{2}x dx,$ then

$\displaystyle I_{1}, I_{2}, I_{3},$ .......are in

0%
A.P.
0%
G.P.
0%
H.P.
0%
none

Explanation

$I_{ n }=\int _{ 0 }^{ \pi /4 } \tan ^{ n }{ x } \sec ^{ 2 } xdx$
Let,

$t=\tan { x }$

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

Therefore,

$I_{ n }=\int _{ 0 }^{ 1 }{ { t }^{ n } } dt={ \left( \dfrac { t }{ n+1 } \right) }_{ 0 }^{ 1 }=\frac { 1 }{ n }$
Therefore,

$I_{ 1 },I_{ 2 },I_{ 3 }\cdots$ are

$\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 3 } ,\dfrac { 1 }{ 4 } ,\cdots$ which are in H.P

Ans: C

The value of the integral

$\displaystyle \int_{\alpha }^{\beta }\displaystyle \dfrac{dx}{\sqrt{\left ( x-\alpha \right )\left ( \beta -x \right )}}$ for

$\beta > \alpha$ , is

0%
$\sin ^{-1}\: \alpha /\beta$
0%
$\pi /2$
0%
$\sin ^{-1}\beta /2\alpha$
0%
$\pi$

Explanation

Given :

$\displaystyle \int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { (x-\alpha )(\beta -x) } } }$

$\displaystyle I=\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { (x-\alpha )(\beta -x) } } } =\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { -\alpha \beta +x(\beta +\alpha )-{ x }^{ 2 } } } }$

$\displaystyle I=\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { \cfrac { { (\alpha +\beta ) }^{ 2 } }{ 4 } -\alpha \beta -\left[ x-\cfrac { \beta +\alpha }{ 2 } \right] ^{ 2 } } } }$

$\displaystyle I=\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { \cfrac { (\beta -\alpha )^{ 2 } }{ 4 } -\left[ x-\cfrac { (\beta +\alpha ) }{ 2 } \right] ^{ 2 } } } }$

we know that,

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

$I=\sin ^{ -1 }{ \left[ \cfrac { x-\left( \cfrac { \beta +\alpha }{ 2 } \right) }{ \left( \cfrac { \beta -\alpha }{ 2 } \right) } \right] } _{ \alpha }^{ \beta }=\sin ^{ -1 }{ \left[ \cfrac { \left( \cfrac { \beta -\alpha }{ 2 } \right) }{ \left( \cfrac { \beta +\alpha }{ 2 } \right) } \right] } -\sin ^{ -1 }{ \left[ \cfrac { \left( \cfrac { \alpha -\beta }{ 2 } \right) }{ \left( \cfrac { \beta -\alpha }{ 2 } \right) } \right] }$

$I=\sin ^{ -1 }{ (1)-\sin ^{ -1 }{ (-1) } }$

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

Hence the correct answer is

$\pi$

$\displaystyle \int_{\pi /6}^{\pi /4}\frac{dx}{\sin 2x}$ is equal to

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

Explanation

Let

$\displaystyle I=\int { \frac { 1 }{ \sin { 2x } } dx } =\int { \csc { 2x } dx }$
Put

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

$\displaystyle I=\frac { 1 }{ 2 } \int { \csc { t } dt }$
Multiply numerator and denominator by

$\cot { t } +\csc { t }$

$\displaystyle I=\frac { 1 }{ 2 } \int { -\frac { -\csc ^{ 2 }{ t } -\cot { t } \csc { t } }{ \cot { t } +\csc { t } } dt }$
Put

$\cot { t } +\csc { t } =u\Rightarrow \left( -\csc ^{ 2 }{ t } -\cot { t } \csc { t } \right) dt=du$

$\displaystyle I=-\frac { 1 }{ 2 } \int { \frac { 1 }{ u } du } =-\frac { 1 }{ 2 } \log { u } =-\frac { 1 }{ 2 } \log { \left( \cot { t } +\csc { t } \right) }$

$\displaystyle =-\frac { 1 }{ 2 } \log { \left( \cot { 2x } +\csc { 2x } \right) } =-\frac { 1 }{ 2 } \log { \left( \cot { x } \right) }$
Hence

$\displaystyle \int _{ \pi /6 }^{ \pi /4 }{ \frac { 1 }{ \sin { 2x } } dx } =-\frac { 1 }{ 2 } { \left[ \log { \left( \cot { x } \right) } \right] }_{ \pi /6 }^{ \pi /4 }=\frac { 1 }{ 2 } \log { \sqrt { 3 } }$

Value of

$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\left ( 1+x^{2} \right )\sqrt{1-x^{2}}}$ is?

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

Explanation

Let

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

Substitute

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

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

Substitute

${ t }^{ 2 }-1={ u }^{ 2 }\Rightarrow 2tdt=2udu$

$\displaystyle I=-\int { \frac { udu }{ \left( { u }^{ 2 }+1 \right) u } } =-\int { \frac { 1 }{ { u }^{ 2 }+{ \left( \sqrt { 2 } \right) }^{ 2 } } du }$

$\displaystyle =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \frac { u }{ \sqrt { 2 } } \right) } =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \frac { \sqrt { 2 } x }{ \sqrt { 1-{ x }^{ 2 } } } \right) }$

Therefore

$\displaystyle \int _{ 0 }^{ 1 }{ Idx } =\frac { \pi }{ 2\sqrt { 2 } }$

$3+2\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ e }^{ -x^{ 2 } } } dx=\displaystyle\int _{ 0 }^{ 1 }{ { e }^{ -x^{ 2 } } } dx$ then the value of

$\beta$ is

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

$I_n=\displaystyle\int^1_0\dfrac{dx}{(1+x^2)^n}; n\in N$ , then which of the following statements hold good?

0%
$2n I_{n+1}=2^{-n}+(2n-1)I_n$
0%
$I_2=\dfrac{\pi}{8}+\dfrac{1}{4}$
0%
$I_2=\dfrac{\pi}{8}-\dfrac{1}{4}$
0%
$I_3=\dfrac{\pi}{16}-\dfrac{5}{48}$

$\int { \frac { { x }^{ 1/2 } }{ { x }^{ 1/2 }-{ x }^{ 1/3 } } } dx=Ax\frac { 6 }{ 5 } { x }^{ 5/6 }+\frac { 3 }{ 2 } { x }^{ 2/3 }+B\sqrt { x } +{ Cx }^{ 1/3 }+{ Dx }^{ 1/6 }+E$ In

$\left( { x }^{ 1/6 }-1 \right) =k,$ then A+B+C+D+E=

0%
6
0%
12
0%
18
0%
17

The value of

$\displaystyle \int_{1/2}^{1}\displaystyle \frac{dx}{x\sqrt{3x^{2}+2x-1}}$ is?

0%
$\pi /2$
0%
$\pi /3$
0%
$\pi /6$
0%
$\pi /\sqrt{2}$

Explanation

Let

$\displaystyle I=\int { \frac { 1 }{ x\sqrt { { 3x }^{ 2 }+2x-1 } } } dx=\int { \frac { 1 }{ x\sqrt { { \left( \sqrt { 3 } x+\frac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }-\frac { 4 }{ 3 } } } } dx$

Put

$\displaystyle t=\sqrt { 3 } x+\frac { 1 }{ \sqrt { 3 } } \Rightarrow dt=\sqrt { 3 } dt$

$\displaystyle I=\frac { 1 }{ \sqrt { 3 } } \int { -\frac { 1 }{ \left( \sqrt { 3 } -3t \right) \sqrt { { t }^{ 2 }-\frac { 4 }{ 3 } } } } dt$

$\displaystyle =-3\int { \frac { 1 }{ \left( \sqrt { 3 } -3t \right) \sqrt { { t }^{ 2 }-\frac { 4 }{ 3 } } } } dt$

$\displaystyle I=3\int { \frac { 2 }{ 3\sqrt { 3 } { s }^{ 2 }+\sqrt { 3 } } ds } =2\sqrt { 3 } \int { \frac { 1 }{ 3{ s }^{ 2 }+1 } ds }$

$\displaystyle =2\tan ^{ -1 }{ \left( \sqrt { 3 } s \right) } =\tan ^{ -1 }{ \left( \frac { x-1 }{ \sqrt { 3{ x }^{ 2 }+2x-1 } } \right) }$

$\displaystyle \therefore \int _{ 1/2 }^{ 1 } { \frac { 1 }{ x\sqrt { { 3x }^{ 2 }+2x-1 } } } { dx } =\frac { \pi }{ 6 }$

$I= \displaystyle \int_{1}^{\infty }\displaystyle \frac{x^{2}-2}{x^{3}\sqrt{x^{2}-1}}\: dx$ , then

$I$ equals

0%
$-1$
0%
$0$
0%
$\pi /2$
0%
$\pi -\sqrt{3}$

Explanation

$\displaystyle I=\int _{ 1 }^{ \infty }{ \frac { { x }^{ 2 }-2 }{ { x }^{ 3 }\sqrt { { x }^{ 2 }-1 } } dx }$
Put Put

$x=\sec { u } \Rightarrow dx=\tan { u } \sec { u } du$

$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 2 }{ u\left( \sec ^{ 2 }{ u } -2 \right) } du } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 1-\sin ^{ 2 }{ u } \right) \left( \sec ^{ 2 }{ u } -2 \right) du }$

$\displaystyle =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2\sin ^{ 2 }{ u } -\tan ^{ 2 }{ u } +\sec ^{ 2 }{ u } -2 \right) } du=-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } +2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2 }{ u } du }$

$\displaystyle =-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } -\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { 2udu } } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } =-{ \left[ \frac { \sin { 2u } }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=0$

$0< \alpha < \pi /2$ then the value of

$\displaystyle \int_{0}^{\alpha }\displaystyle \frac{dx}{1-\cos x\cos \alpha }$ is

0%
$\pi /\alpha$
0%
$\pi /2\sin \alpha$
0%
$\pi /2\cos \alpha$
0%
$\pi /2\alpha$

Explanation

Let

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

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

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

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

Substitute

$\displaystyle \tan { \dfrac { x }{ 2 } } =t$

$\displaystyle I=\int _{ 0 }^{ \tan { \dfrac { \alpha }{ 2 } } }{ \dfrac { \sec ^{ 2 }{ \dfrac { \alpha }{ 2 } } }{ \tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } +{ t }^{ 2 } } dt } =\sec ^{ 2 }{ \dfrac { \alpha }{ 2 } } \cot { \dfrac { \alpha }{ 2 } } { \left[ \tan ^{ -1 }{ \dfrac { 1 }{ \tan { \dfrac { \alpha }{ 2 } } } } \right] }_{ 0 }^{ \tan { \dfrac { \alpha }{ 2 } } }$

$\displaystyle =\dfrac { \pi }{ 2\sin { \alpha } }$

Value of

$\displaystyle \int_{a}^{\infty }\displaystyle \frac{dx}{x^{4}\sqrt{a^{2}+x^{2}}}$ is

0%
$\displaystyle \frac{2+\sqrt{2}}{3a^{4}}$
0%
$\displaystyle \frac{2-\sqrt{2}}{3a^{2}}$
0%
$\displaystyle \frac{2-\sqrt{2}}{3a^{4}}$
0%
$\displaystyle \frac{\sqrt{2}+1}{3a^{2}}$

Explanation

Let

$\displaystyle I=\int _{ a }^{ \infty }{ \frac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+x^{ 2 } } } } dx$

Substitute

$x=a\tan { t\Rightarrow } dx=a\sec ^{ 2 }{ t } dt$

$\displaystyle I=a\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \frac { \cot ^{ 3 }{ t } \csc { t } }{ { a }^{ 5 } } } dt$

$\displaystyle =\frac { 1 }{ { a }^{ 4 } } \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \cot { t } \csc { t } \left( \csc ^{ 2 }{ t-1 } \right) dt }$

Substitute

$\upsilon =\csc { t } \Rightarrow d\upsilon =-\cot { t } \csc { t } dt$

$\displaystyle I=\frac { 1 }{ { a }^{ 4 } } \int _{ \sqrt { 2 } }^{ 1 }{ \left( { \upsilon }^{ 2 }-1 \right) } d\upsilon$

$\displaystyle =\frac { 1 }{ { a }^{ 2 } } \left[ \frac { { \upsilon }^{ 2 } }{ 3 } -\upsilon \right] _{ \sqrt { 2 } }^{ 1 }{ =\frac { 2-\sqrt { 2 } }{ { 3a }^{ 4 } } }$

The value of

$\displaystyle \int_{-4}^{-5}e^{\left ( x+5 \right )^{2}}dx+3\displaystyle \int_{1/3}^{2/3}e^{9\left ( x-2/3 \right )^{2}}dx$ is

0%
$2/5$
0%
$1/5$
0%
$1/2$
0%
none of these

Explanation

Let

$I=\int _{ -4 }^{ -5 } e^{ \left( x+5 \right) ^{ 2 } }dx+3\int _{ 1/3 }^{ 2/3 } e^{ 9\left( x-2/3 \right) ^{ 2 } }dx={ I }_{ 1 }+{ I }_{ 2 }$
Where

${ I }_{ 1 }=\int _{ -4 }^{ -5 } e^{ \left( x+5 \right) ^{ 2 } }dx$
Put

$x+5=t$

${ I }_{ 1 }=-\int _{ 0 }^{ 1 }{ { e }^{ { t }^{ 2 } } } dt$
And

${ I }_{ 2 }=3\int _{ 1/3 }^{ 2/3 } e^{ 9\left( x-2/3 \right) ^{ 2 } }dx$
Put

$-3x+2=u$

${ I }_{ 2 }=\int _{ 0 }^{ 1 }{ { e }^{ { u }^{ 2 } } } du$

$\therefore I=0$

Value of

$\displaystyle \int_{0}^{16}\displaystyle \frac{x^{1/4}}{1+x^{1/2}}\: dx$ is

0%
$\displaystyle \frac{8}{3}$
0%
$\displaystyle \frac{4}{3}\tan ^{-1}\: 2$
0%
$4\displaystyle \left ( \displaystyle \frac{2}{3}+\tan ^{-1}\: 2 \right )$
0%
$4\displaystyle \left ( \displaystyle \frac{2}{3}-\tan ^{-1}\: 2 \right )$

Explanation

$\displaystyle I=\int _{ 0 }^{ 16 }{ \frac { \sqrt [ 4 ]{ x } }{ 1+\sqrt { x } } dx }$

Put

$\displaystyle u=\sqrt [ 4 ]{ x } \Rightarrow du=\frac { 1 }{ 4{ x }^{ \frac { 3 }{ 4 } } } dx$

$\displaystyle I=4\int _{ 0 }^{ 2 }{ \left( { u }^{ 2 }+\frac { 1 }{ { u }^{ 2 }+1 } -1 \right) } du$

$\displaystyle =4\int _{ 0 }^{ 2 }{ { u }^{ 2 } } du+4\int _{ 0 }^{ 2 }{ \frac { 1 }{ { u }^{ 2 }+1 } du } -4\int _{ 0 }^{ 2 }{ du }$

$\displaystyle =4\left[ \frac { { u }^{ 3 } }{ 3 } \right] _{ 0 }^{ 2 }{ + }4\left[ \tan ^{ -1 }{ u } \right] _{ 0 }^{ 2 }-4\left[ u \right] _{ 0 }^{ 2 }$

$\displaystyle =4\left( \frac { 2 }{ 3 } +\tan ^{ -1 }{ 2 } \right)$

The solution of

$x$ of the equation

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

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

$\int_{0}^{1}\frac{1}{ax+b}dx is$

0%
zero
0%
$log_{e}\frac{a+b}{b}$
0%
$log_{e}(ax+b)$
0%
$\frac{1}{a}log_{e}\left ( \frac{a+b}{b} \right )$

$\int _{ 0 }^{ 8 }{ \left[ \sqrt { t } \right] } dt$ at equals to (where [.] greatest integer function.)

0%
28
0%
11
0%
2
0%
8

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

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

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

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

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

$I_{1}= \displaystyle \int_{x}^{1}\displaystyle \frac{dt}{1+t^{2}}$ and

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

$x> 0$ , then

0%
$I_{1}= I_{2}$
0%
$I_{1}> I_{2}$
0%
$I_{2}> I_{1}$
0%
$I_{2}= \left ( \pi /2 \right )-\tan ^{-1}x$

$I_{1}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{dx}{1+x^{4}}$ and

$I_{2}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{x^{2}}{1+x^{4}}\: dx$ , then

0%
$I_{1}= I_{2}$
0%
$I_{1}=2 I_{2}$
0%
$2I_{1}= I_{2}$
0%
none of these

Explanation

$I_{2}=\displaystyle \int _{0}^{\infty} \dfrac{x^2}{1+x^4}d{x}=\int^{\infty}_{0}\dfrac{\dfrac{1}{x^2}}{1+\dfrac{1}{x^4}}d{x}$

Put

$t=\dfrac{1}{x}\implies d{t}=-\dfrac{d{x}}{x^2}$

$I_{2}=\displaystyle\int _{\infty}^{0} \dfrac{-d{t}}{1+t^4}=\int_{0}^{\infty} \dfrac{d{x}}{1+x^4}=I_{1}$

$\displaystyle \int_{0}^{\infty }\displaystyle \frac{\log \left ( 1+x^{2} \right )}{1+x^{2}}\: dx= \lambda \displaystyle \int_{0}^{1}\displaystyle \frac{\log \left ( 1+x \right )}{1+x^{2}}\: dx$ then

$\lambda$ equals

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

Explanation

Let

$I_1=\displaystyle \int_{0}^{\infty }\displaystyle \frac{\log \left ( 1+x^{2} \right )}{1+x^{2}}\: dx$ and

$I_2= \displaystyle \int_{0}^{1}\displaystyle \frac{\log \left ( 1+x \right )}{1+x^{2}}\: dx$
Put

$x=\tan\theta$ in both integral

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

$I_1=\displaystyle \int_{0}^{\frac{\pi}{2} }\displaystyle \frac{\log \left ( \sec^2\theta \right )}{\sec^2\theta}\cdot \sec^2\theta d\theta=-\int_0^{\frac{\pi}{2}}\log\cos^2\theta d\theta ......(1)$
Also using

$\left( \int_0^a f(x)dx=\int_0^af(a-x)dx \right)$

$I_1 =\displaystyle -\int_0^{\frac{\pi}{2}}\log \sin^2\theta d\theta ........(2)$
Adding (1) and (2) we get

$2I_1 = \displaystyle -2 \int_0^{\frac{\pi}{2}}\log (\sin\theta\cos\theta)d\theta$

$\Rightarrow\displaystyle I_1=-\int_0^{\frac{\pi}{2}}\log\frac{\sin2\theta}{2} d\theta =\log 2\int_0^{\frac{\pi}{2}}d\theta- \int_0^{\frac{\pi}{2}} \log(\sin2\theta)d\theta$
Put

$2\theta =x\Rightarrow 2d\theta=dx$

$\Rightarrow \displaystyle I_1= \frac{\pi}{2}\log 2 -\frac{1}{2}\int_0^{\pi}\log(\sin\theta)d\theta=\frac{\pi}{2}\log 2 -\int_0^{\frac{\pi}{2}}\log(\sin\theta)d\theta=\frac{\pi}{2}\log 2+\frac{I_1}{2}$ , using (2)

$\therefore \displaystyle I_1 =\pi\log 2$
Now

$I_2=\displaystyle \int_0^{\frac{\pi}{4}}\log(1+\tan\theta)d\theta$

$\Rightarrow \displaystyle I_2 =\int_0^{\frac{\pi}{4}}\log(1+\tan(\frac{\pi}{4}-\theta))d\theta=\int_0^{\frac{\pi}{4}}\log\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)d\theta$

$\displaystyle \Rightarrow I_2=\int_0^{\frac{\pi}{4}}\log\left(\frac{2}{1+\tan\theta}\right)d\theta=\log2\int_0^{\frac{\pi}{4}}d\theta-\displaystyle \int_0^{\frac{\pi}{4}}\log(1+\tan\theta)d\theta$

$\Rightarrow \displaystyle I_2=\frac{\pi}{4}\log 2-I_2 \Rightarrow I_2=\frac{\pi}{8}\log 2$
Hence

$\lambda = \cfrac{I_1}{I_2}=8$

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 9 - 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 9 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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