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CBSE Questions for Class 12 Commerce Maths Integrals Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Integrals
Quiz 8
$$\displaystyle \int_{\frac{5}{2}}^{5}\frac{\sqrt{(25-x^2)^3}}{x^4}\:dx$$ is equal to
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$$\displaystyle \frac{\pi }{6}$$
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$$\displaystyle \frac{2\pi }{3}$$
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$$\displaystyle \frac{5\pi }{6}$$
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$$\displaystyle \frac{\pi }{3}$$
Explanation
$$\displaystyle I=\int_{\frac{5}{2}}^{5}\frac{\sqrt{(25-x^2)^3}}{x^4}dx$$
Let $$x= 5\sin\theta$$
$$\displaystyle \therefore dx= 5 \cos\theta d\theta$$
$$\displaystyle \therefore I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{5^3\cos ^3\theta.5\cos\theta }{5^4 \sin^4\theta}d\theta$$
$$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta(\ cosec^2\theta -1)d\theta$$
$$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2
\theta\ cosec^2\theta
d\theta-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta d \theta$$
$$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2
\theta\ cosec^2\theta d\theta- \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\
cosec^2 \theta-1)$$
$$\displaystyle =\left [ -\frac{\cot^3\theta}{3} +\cot\theta+\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$$
$$\displaystyle =-0+0+\frac{\pi}{2}-\left ( -\frac{3\sqrt{3}}{3}+\sqrt{3}+\frac{\pi}{6} \right )$$ $$\displaystyle =\frac{\pi}{3}$$
The minimum value of the function f(x) = $$\int^x_0 \frac{d \theta}{cos \theta} + \int^{\pi/2}_x \frac{d \theta}{sin \theta} $$ where $$x \in [0, \frac{\pi}{2}], $$ is
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$$2ln(\sqrt{2} + 1)$$
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$$ln(2\sqrt{2} + 2)$$
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$$ln(\sqrt{3} + 2)$$
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$$ln(\sqrt{2} + 3)$$
Explanation
f(x) = $$\int^x_0 \frac{d \theta}{cos \theta} + \int^{\pi/2}_x \frac{d \theta}{sin \theta} $$
$$ f^{'}(x) = \frac{1}{cos x} - \frac{1}{sin x}$$
= $$\frac{sin x - cos x}{sin xcos x}$$
$$ f^{'}(x)$$ change sign $$-t_0 + at x = \frac{\pi}{4}$$
so f(x) takes minimum value at x = $$\frac{\pi}{4}$$
$$f(x) = \int^{\pi /4}_0 sec \theta d\theta+ \int^{\pi /2}_{\pi /4} cosec\theta d\theta$$
$$ = (ln(sec \theta + tan \theta))^{\pi/4}_0 + (ln(cosec \theta - cot \theta))^{\pi/2}_{\pi/4} $$
$$ = ln(\sqrt{2} + 1) - ln(1 + 0) + ln 1 - ln(\sqrt{2} - 1)$$
$$ = ln[\frac{\sqrt{2} + 1}{\sqrt{2} - 1}] = ln(\sqrt{2} + 1)^2 = 2 ln(\sqrt{2} + 1)$$
If $$ \displaystyle \int_{0}^{1} \frac{\tan^{-1} x}{x}dx$$ is equal to
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$$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}dx$$
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$$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$$
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$$\displaystyle \frac{1}{2} \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}dx$$
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$$\displaystyle \frac{1}{2} \int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$$
Explanation
$$ \displaystyle I= \int_{0}^{1} \frac{\tan^{-1} x}{x}dx$$
Putting $$x=\tan \theta$$ or $$dx= \sec^2 \theta d\theta$$, we get
$$I\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{\theta}{\tan \theta}\sec^2 \theta d \theta$$
$$\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{2\theta}{\sin 2\theta}d\theta$$
Putting $$2\theta=t$$,i.e.,$$2d\theta=dt$$, we get
$$I\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{t}{\sin t}dt$$
$$\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$$
The value of the integral $$\displaystyle \int_{0 }^{\log5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\:dx$$ is
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$$3+2\pi$$
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$$4-\pi$$
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$$2+\pi$$
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none of these
Explanation
Putting $$e^x-1=t^2$$ in the given integral, we have
$$ \displaystyle \int_{0
}^{\log
5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=2\int_{0}^{2}\frac{t^2}{t^2+4}dt=2\left
( \int_{0}^{2} 1 dt -4\int_{0}^{2}\frac{dt}{t^2+4}\right )$$
$$ \displaystyle =2\left [ \left ( t-2\tan^{-1}\left ( \frac{t}{2} \right ) \right )_0^2 \right ]$$
$$\displaystyle =2[(2-2\times \frac{\pi}{4})]=4-\pi$$
The value of the integral $$\displaystyle \int_{0}^{\frac{1}{\sqrt{3}}}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$$
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$$\displaystyle \frac{\pi }{2\sqrt{2}}$$
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$$\displaystyle \frac{\pi }{4\sqrt{2}}$$
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$$\displaystyle \frac{\pi }{8\sqrt{2}}$$
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none of these
Explanation
Orl putting $$x= \sin \theta$$ ,we get $$dx =\cos \theta d \theta$$
Integral (without limits)$$\displaystyle =\int \dfrac{\cos\theta d\theta}{(1+ \sin^2\theta )(\cos \theta)}$$
$$\displaystyle =\int \dfrac{d\theta}{1+ \sin^2\theta}=\int\dfrac{\ cosec{^2}\theta d \theta }{2+\cot^2}$$
$$\displaystyle=\int \dfrac{-dt}{2+t^2}$$ where $$t=\cot \theta$$
$$=-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{t}{\sqrt{2}}$$
$$=-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{\cot\theta}{\sqrt{2}}$$
$$=-\dfrac{1}{\sqrt{2}} \tan^{-1}\dfrac{1}{\sqrt{2}}\left ( \dfrac{\sqrt{1-x^2}}{x} \right )$$
$$\therefore$$ Definite integral $$= -\dfrac{1}{\sqrt{2}}\tan^{-1}1+\dfrac{1}{\sqrt{2}}\tan^{-1}{\infty }$$
$$=-\dfrac{\pi}{4\sqrt{2}}+\dfrac{\pi}{2\sqrt{2}}=\dfrac{\pi}{4\sqrt{2}}$$
Let $$ \displaystyle I_1=\int_{0}^{1}\frac{e^x dx}{1+x}$$ and $$\displaystyle I_2=\int_{0}^{1}\dfrac{x^2 dx}{e^{x^3}(2-x^3)}$$.
Then $$\dfrac{I_1}{I_2}$$ is equal to?
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$$\displaystyle \dfrac{3}{e}$$
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$$\displaystyle \frac{e}{3}$$
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$$\displaystyle {3}{e}$$
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$$\displaystyle \dfrac{1}{3e}$$
Explanation
In $$\displaystyle{ I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ { x }^{ 3 } }\left( 2-{ x }^{ 3 } \right) } } $$
Substitute $$1-{ x }^{ 3 }=t\Rightarrow -3{ x }^{ 2 }dx=dt$$
We get $$\displaystyle{ I }_{ 2 }=-\cfrac { 1 }{ 3 } \int _{ 1 }^{ 0 }{ \cfrac { dt }{ { e }^{ 1-t }\left( 1+t \right) } } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t }dt }{ 1+t } } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ 1+x } } $$
As $$\displaystyle{ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ 1+x } } $$
Therefore $$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =3e$$
The value of $$\displaystyle \int_{0}^{\infty} \frac {dx}{1 + x^4}$$ is
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same as that of $$\displaystyle \int_{0}^{\infty} \frac {x^2 + 1dx}{1 + x^4}$$
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$$\displaystyle \frac {\pi}{2\sqrt{2}}$$
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same as that of $$\displaystyle \int_{0}^{\infty} \frac {x^2 \: dx}{1 + x^4}$$
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$$\displaystyle \frac {\pi}{\sqrt{2}}$$
Explanation
$$\displaystyle I = \int_{0}^{\infty} \frac {dx}{1 + x^4}$$ ....(1)
$$\displaystyle \int_{0}^{\infty} \frac {x^2 + 1 - x^2}{1 + x^4} dx$$
$$\displaystyle = \int_{0}^{\infty} \frac {x^2}{1 + x^4} dx + \int_{0}^{\infty} \frac {1 - x^2}{1 + x^4} dx $$ ....(2)
Let $$ I_1 =\int_{0}^{\infty} \frac {x^2}{1 + x^4} dx$$
and $$ I_2= \int_{0}^{\infty} \frac {1 - x^2}{1 + x^4} dx$$
$$\displaystyle I_2 = \int_{0}^{\infty} \frac {\frac {1}{x^2}-1}{\frac {1}{x^2}+x^2} dx$$
$$I_{ 2 }=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } -1 }{ \frac { 1 }{ x^{ 2 } } +x^{ 2 }+2-2 } dx$$
$$I_{ 2 }=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } -1 }{ (x+\frac { 1 }{ x } )^{ 2 }-2 } dx$$
Put $$\displaystyle x + \frac {1}{x} = y$$
$$\Rightarrow dy=(1-\frac { 1 }{ x^{ 2 } } )dx$$
Also, when $$x=0\Rightarrow y=\infty $$
And when $$x=\infty \Rightarrow y=\infty $$
$$\displaystyle \therefore I_2 = \int_{\infty}^{\infty} \displaystyle \frac {-1}{y^2 - 2} dy = 0$$
$$\therefore I=\int _{ 0 }^{ \infty } \displaystyle \frac { x^{ 2 }dx }{ 1+x^{ 4 } } $$ .....(3)
Adding equations (1) and (3), we get
$$2I=\int _{ 0 }^{ \infty } \displaystyle \frac { 1+x^{ 2 } }{ 1+x^{ 4 } } dx$$
$$= \int_{0}^{\infty} \displaystyle \frac {\frac {1}{x^2}+1}{\frac {1}{x^2}+x^2}dx$$ $$\displaystyle \:$$
$$=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } +1 }{ \frac { 1 }{ x^{ 2 } } +x^{ 2 }+2-2 } dx$$
$$=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } +1 }{ (x-\frac { 1 }{ x } )^{ 2 }+2 } dx$$
Put $$\displaystyle x - \frac {1}{x} = t$$
$$\Rightarrow dt=(1+\frac { 1 }{ x^{ 2 } } )dx$$
$$\displaystyle 2I = \int_{-\infty}^{\infty} \frac {dt}{t^2 + 2}$$
$$\displaystyle = [\displaystyle\frac {1}{\sqrt {2}} tan^{-1} \frac {t}{\sqrt{2}}]_{- \infty}^{\infty} = \frac {\pi}{\sqrt {2}}$$
$$\therefore \displaystyle I = \frac {\pi}{2 \sqrt {2}}$$
Evaluate $$\displaystyle\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{9+16\{1-{(\sin{x}-\cos{x})}^2\}}dx}$$
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$$\displaystyle\frac{1}{20}(\log{\sqrt{3}})$$
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$$\displaystyle\frac{1}{10}(\log{3})$$
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$$\displaystyle\frac{1}{20}(\log{3})$$
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$$\displaystyle\frac{1}{20}(\log{3}-\log{2})$$
Explanation
Let $$\displaystyle I=\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{9+16\{1-{(\sin{x}-\cos{x})}^2\}}dx}$$
$$\displaystyle\therefore I=\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{25-16{(\sin{x}-\cos{x})}^2}dx}$$,
Substitute $$4(\sin{x}-\cos{x})=t\implies4(\cos{x}+\sin{x})dx=dt$$
$$\displaystyle\frac{1}{4}\int_{\displaystyle-4}^0{\frac{dt}{25-t^2}}$$
$$\displaystyle I=\frac{1}{4}.\frac{1}{2(5)}{\left(\log{\left|\frac{5-t}{5+t}\right|}\right)}_{\displaystyle-4}^0$$
$$\displaystyle=\frac{1}{40}\left[\log{\left|\frac{5-0}{5+0}\right|}-\log{\left|\frac{5+4}{5-4}\right|}\right]$$
$$\displaystyle=\frac{1}{40}\left[\log{1}-\log{\left(\frac{1}{9}\right)}\right]$$, (where $$\log{1}=0$$)
$$\displaystyle=\frac{1}{40}[\log{9}]=\frac{2}{40}\log{3}=\frac{1}{20}(\log{3})$$
If $$\displaystyle 2f(x)+f(-x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$$, then the value of $$\displaystyle\int_{\frac{1}{e}}^{e}{f(x)dx}$$, is
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$$1$$
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$$0$$
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$$e$$
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$$-1$$
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 $$\displaystyle x=\frac{1}{t}\Rightarrow\displaystyle dx=-\frac{1}{t^2}dt$$
$$\displaystyle\therefore I=\frac{1}{3}\int_{e}^{\frac{1}{e}}{t\sin{\left(\frac{1}{t}-t\right)}.\left(-\frac{1}{t^2}\right)dt}$$
$$\displaystyle=\frac{1}{3}\int_{e}^{\frac{1}{e}}{\frac{1}{t}\sin{\left(t-\frac{1}{t}\right)}dt}$$
$$\displaystyle=-\frac{1}{3}\int_{\frac{1}{e}}^{e}{\frac{1}{t}\sin{\left(t-\frac{1}{t}\right)}dt}$$
$$\Rightarrow I=-I\Rightarrow 2I=0\Rightarrow I=0$$
$$\displaystyle\therefore\int_{\frac{1}{e}}^{e}{f(x)dx}=0$$.
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
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$$I_1 > I_2$$
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$$I_2 > I_1$$
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$$I_1 = I_2$$
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$$I_1 > 2 I_2$$
Explanation
$$\displaystyle { I }_{ 1 }=\int _{ 1 }^{ 2 }{ \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } dx$$
$$={ \left( \log { \left( x+\sqrt { 1+{ x }^{ 2 } } \right) } \right) }_{ 1 }^{ 2 }$$
$$\displaystyle =\log { \left( 2+\sqrt { 5 } \right) } -\log { \left( 1+\sqrt { 2 } \right) } =\log { \left( \frac { \left( 2+\sqrt { 5 } \right) }{ \left( 1+\sqrt { 2 } \right) } \right) } $$
$$\displaystyle { I }_{ 2 }=\int _{ 1 }^{ 2 }{ \frac { 1 }{ x } } dx={ \left( \log { x } \right) }_{ 1 }^{ 2 }=\log { 2 } -\log { 1 } =\log { 2 } $$
$$\therefore { I }_{ 2 }>{ I }_{ 1 }$$
Let $$\displaystyle\frac{d}{dx}(F(x))=\frac{e^{\displaystyle\sin{x}}}{x}$$, $$x>0$$. If $$\displaystyle\int_1^4{\frac{2e^{\displaystyle\sin{x^2}}}{x}dx}=F(k)-F(1)$$, then the possible value of $$k$$ is
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10
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14
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16
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18
Explanation
Substitute $${ x }^{ 2 }=z$$
Then $$\displaystyle \int _{ 1 }^{ 4 }{ \frac { 2{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =\int _{ 1 }^{ 16 }{ \frac { { e }^{ \sin { z } } }{ z } dz } =\int _{ 1 }^{ 16 }{ \frac { d }{ dz } \left\{ F\left( z \right) \right\} } dz$$
$$={ \left( F\left( z \right) \right) }_{ 1 }^{ 16 }=F\left( 16 \right) -F\left( 1 \right) $$
$$\therefore F\left( 16 \right) -F\left( k \right) =F\left( 16 \right) -F\left( 1 \right) $$
Gives $$k=16$$
The evaluation of $$\displaystyle \int \frac{pX^{p+2q-1}-qX^{q-1}}{X^{2p+2q}+2X^{p+q}+1}dx$$ is
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$$-\displaystyle \frac{X^p}{X^{p+q}+1}+C$$
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$$\displaystyle \frac{X^q}{X^{p+q}+1}+C$$
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$$-\displaystyle \frac{X^q}{X^{p+q}+1}+C$$
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$$\displaystyle \frac{X^p}{X^{p+q}+1}+C$$
Explanation
Let us take $$\displaystyle \frac{x^{m}}{x^{(p+q)}+1}=t$$ where m is either p or q as seen from the options,
$$\displaystyle \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{(x^{(p+q)}+1)mx^{m-1}-x^{m}(p+q)x^{p+q-1}}{x^{2p+2q}+1+2x^{p+q}}$$
$$\displaystyle =\frac{(mx^{(p+q+m-1)}+mx^{m-1}-px^{p+q+m-1}-qx^{p+q+m-1})}{x^{2p+2q}+1+2x^{p+q}}$$
we want powers of $$p+2q-1$$ and $$q-1$$, so $$m=q,$$
$$\displaystyle =\frac{(qx^{(p+2q-1)}+qx^{q-1}-px^{p+2q-1}-qx^{p+2q-1})}{x^{2p+2q}+1+2x^{p+q}}$$
$$\displaystyle =\frac{qx^{q-1}-px^{p+2q-1}}{x^{2p+2q}+1+2x^{p+q}}$$
$$\displaystyle =\frac{\mathrm{d} t}{\mathrm{d} x}$$
so the integral will be $$-\dfrac{x^{q}}{x^{(p+q)}+1}$$
Let $$\displaystyle \frac{df\left ( x \right )}{dx}=\frac{e^{\sin x}}{x}, x> 0$$. If $$\displaystyle \int_{1}^{4}\displaystyle \frac{3e^{\sin x^{3}}}{x}dx=f\left ( k \right )-f\left ( 1 \right )$$ then one of the possible values of $$k$$ is
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$$16$$
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$$63$$
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$$64$$
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$$15$$
Explanation
Substitute $${ x }^{ 3 }=z$$
Then $$\displaystyle \int _{ 1 }^{ 4 }{ \frac { 3{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =\int _{ 1 }^{ 16 }{ \frac { { e }^{ \sin { z } } }{ z } dz } =\int _{ 1 }^{ 64 }{ \frac { d }{ dz } \left\{ F\left( z \right) \right\} } dz$$
$$={ \left( F\left( z \right) \right) }_{ 1 }^{ 64 }=F\left( 64 \right) -F\left( 1 \right) $$
$$\therefore F\left( 64 \right) -F\left( k \right) =F\left( 64 \right) -F\left( 1 \right) $$
Gives $$k=64$$
$$\displaystyle \int_{1}^{e^{37}}\frac{\pi \sin \left ( \pi \log _{e}x \right )}{x}dx$$ is equal to
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$$2$$
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$$\displaystyle -2$$
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$$\displaystyle 2/\pi $$
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$$\displaystyle 2\pi $$
Explanation
Let $$\displaystyle I=\int _{ 1 }^{ e^{ 37 } } \dfrac { \pi \sin \left( \pi \log _{ e } x \right) }{ x } dx$$
Substitute $$\displaystyle \pi \log { x } =t\Rightarrow \dfrac { \pi }{ x } dx=dt$$
$$\displaystyle\therefore I=\int _{ 0 }^{ 37\pi }{ \sin { t } dt=-{ \left[ \cos { t } \right] }_{ 0 }^{ 37\pi} } \\ =-\left[ \cos { 37\pi } -\cos { 0 } \right] =2$$
$$\displaystyle \int_{0}^{2}\sqrt{\frac{2+x}{2-x}}dx$$ is equal to
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$$\displaystyle \pi +1$$
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$$\displaystyle 1+\pi /2$$
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$$\displaystyle \pi +3/2$$
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none of these
Explanation
Let $$\displaystyle I=\sqrt { \frac { 2+x }{ 2-x } } dt$$
Put $$\displaystyle t=\frac { 2+x }{ 2-x } \Rightarrow dt=\left( \frac { 1 }{ 2-x } -\frac { -x-2 }{ { \left( 2-x \right) }^{ 2 } } \right) dx$$
$$\displaystyle \therefore I=4\int { \frac { \sqrt { t } }{ { \left( t+1 \right) }^{ 2 } } dt } $$
Put $$\displaystyle u=\sqrt { t } \Rightarrow du=\frac { 1 }{ 2\sqrt { t } } dt$$
$$\displaystyle \therefore I=8\int { \frac { { u }^{ 2 } }{ { \left( { u }^{ 2 }+1 \right) }^{ 2 } } } du=8\int { \left( \frac { 2 }{ { u }^{ 2 }+1 } -\frac { 1 }{ { \left( { u }^{ 2 }+1 \right) }^{ 2 } } \right) du } $$
$$\displaystyle =\frac { 4\left( \left( { u }^{ 2 }+1 \right) \tan ^{ -1 }{ u-u } \right) }{ { u }^{ 2 }+1 } +C$$
$$\displaystyle =\frac { 4\left( \left( t+1 \right) \tan ^{ -1 }{ \left( \sqrt { t } \right) -\sqrt { t } } \right) }{ t+1 } +C$$
$$\displaystyle =\sqrt { \frac { 2+x }{ 2-x } } \left( x-2 \right) +4\tan ^{ -1 }{ \left( \sqrt { \frac { 2+x }{ 2-x } } \right) } +C$$
Therefore
$$\displaystyle \int _{ 0 }^{ 2 }{ \sqrt { \frac { 2+x }{ 2-x } } dx } =\left[ \sqrt { \frac { 2+x }{ 2-x } } \left( x-2 \right) +4\tan ^{ -1 }{ \left( \sqrt { \frac { 2+x }{ 2-x } } \right) } \right] _{ 0 }^{ 2 }$$
$$\displaystyle =\left[ -2+4\frac { \pi }{ 2 } - \right] -0=2\pi -2$$
$$\displaystyle\int_0^\infty{f\left(x+\frac{1}{x}\right).\frac{\ln{x}}{x}dx}$$ is equal to
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0
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1
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$$\displaystyle\frac{1}{2}$$
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cannot be evaluated
Explanation
$$I=\int _{ 0 }^{ \infty }{ f\left( x+\dfrac { 1 }{ x } \right) .\dfrac { \ln { x } }{ x } dx } $$
Let $$x={ e }^{ t }$$
$$\Rightarrow dx={ e }^{ t }dt$$
$$I=\int _{ -\infty }^{ \infty }{ f\left( { e }^{ t }+{ e }^{ -t } \right) tdt } =0$$ since, $$f\left( { e }^{ t }+{ e }^{ -t } \right) t$$ is an odd function.
Ans: 0
$$ \displaystyle \int_{1}^{\infty }\frac{\log \left ( t-1 \right )}{t^2\log t+\log \left ( \frac{t}{t-1} \right )}\:dt $$ equals
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$$ \displaystyle \frac{1}{2} $$
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$$ \displaystyle \frac{1}{3} $$
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$$ \displaystyle \frac{2}{3} $$
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None of these
Explanation
$$\displaystyle\int_{1}^{\infty}\frac{\log(t-1)}{t^{2}\log
t\log\displaystyle \frac{t}{(t-1)}}dt=\int_{0}^{1}\displaystyle
\frac{\log\left (\displaystyle \frac{1}{x}-1 \right )}{\log
x\log(1-x)}dx$$
(By substituting $$t=\displaystyle\frac{1}{x} dt=\frac{1}{x^{2}dx} \mbox{and} t=1\Leftrightarrow x=1 \mbox{and} t=\infty \Leftrightarrow x=0$$)
$$\therefore I = \displaystyle\int_{0}^{1}\frac{\log(1-x)-\log x}{\log x\log(1-x)}dx$$
$$=\displaystyle\int_{0}^{1}\left (\displaystyle\frac{1}{\log x}-\frac{1}{\log(1-x)}\right )dx$$
$$=\displaystyle\int_{0}^{1}\left (\displaystyle\frac{1}{\log(1-x)}-\frac{1}{\log x}\right )dx = -I\therefore I = 0$$
$$\left [ \displaystyle\because \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx \right ]$$
If $$\displaystyle I_{t}=\int_{0}^{\dfrac{\pi }{2}}\frac{\sin^{2}tx}{\sin^{2}x}dx$$ then ,$$I_{1},I_{2},I_{3}$$ are in
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A.P.
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H.P.
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G.P.
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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
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$$ \displaystyle \log_{e}\left ( \frac{1}{e} \right ) $$
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$$ \displaystyle \log _{2}e $$
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$$ \displaystyle \log_{e}{2}-1 $$
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$$ \displaystyle \log_{e}\left ( \frac{e}{2} \right ) $$
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
If $$\displaystyle I= \int_{1/\pi }^{\pi }\frac{1}{x}\cdot \sin \left ( x-\frac{1}{x} \right )dx$$ then I is equal to
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$$0$$
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$$\displaystyle \pi $$
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$$\displaystyle \pi -\frac{1}{\pi }$$
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$$\displaystyle \pi +\frac{1}{\pi }$$
Explanation
Let $$\displaystyle I=\int _{ \dfrac { 1 }{ \pi } }^{ \pi } \frac { 1 }{ x } \cdot \sin \left( x-\dfrac { 1 }{ x } \right) dx$$
Substitute $$\displaystyle x=\dfrac { 1 }{ t } $$
$$\displaystyle \therefore I=\int _{ \pi }^{ \dfrac { 1 }{ \pi } }{ t\sin { \left( \dfrac { 1 }{ t } -t \right) } \left( -\dfrac { 1 }{ { t }^{ 2 } } \right) dt } $$
$$\displaystyle =-\int _{ \dfrac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ t } \sin { \left( t-\dfrac { 1 }{ t } \right) dt } } -I\\ \Rightarrow 2I=0\Rightarrow I=0$$
If $$x$$ 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?
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$$\displaystyle\pm\sqrt{\frac{\alpha}{2\sin{\alpha}}}$$
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$$\displaystyle\pm\sqrt{\frac{2\sin{\alpha}}{\alpha}}$$
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$$\displaystyle\pm\sqrt{\frac{\alpha}{\sin{\alpha}}}$$
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$$\displaystyle\pm2\sqrt{\frac{\sin{\alpha}}{\alpha}}$$
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.
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?
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$$1$$
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?
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$$\displaystyle \tan ^{ -1 }{ \frac { 1 }{ 2 } } $$
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$$\displaystyle \tan ^{ -1 }{ \frac { 1 }{ 3 } } $$
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$$\displaystyle \frac{1}{2}\tan ^{ -1 }{ \frac { 1 }{ 2 } } $$
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$$\displaystyle \frac{1}{3}\tan ^{ -1 }{ \frac { 1 }{ 3 } } $$
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, $$\dfrac { 1 }{ 2 } \sec ^{ 2 }{ \dfrac { \theta }{ 2 } } d\theta =dt]$$
$$\therefore I=\displaystyle =\frac { 1 }{ 2 } \left[ \tan ^{ -1 }{ \frac { t }{ 2 } } \right] _{ 0 }^{ 1 }$$
$$\displaystyle =\frac { 1 }{ 2 } \left( \tan ^{ -1 }{ \frac { 1 }{ 2 } } -\tan ^{ -1 }{ 0 } \right)$$
$$ = \displaystyle \frac { 1 }{ 2 } \tan ^{ -1 }{ \dfrac { 1 }{ 2 } } $$
$$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{n}}dx,\:\forall\:n\:> 1$$ is equal to?
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$$\displaystyle2 \int_{0}^{\infty}\frac{1}{1+x}dx$$
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$$\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^{n}}dx$$
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$$\displaystyle \int_{1}^{\infty}\frac{dx}{(x^{n}-1)^{1/n}}$$
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$$\displaystyle \int_{0}^{1}\frac{1}{(1-x^{n})^{1/n}}$$
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
$$\displaystyle z=\frac { 1 }{ t } \Rightarrow dx=-\frac { 1 }{ { t }^{ 2 } } dt$$
$$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dz }{ (1-z^{ n })^{ 1/n } } =\int _{ 0 }^{ 1 } \frac { dx }{ (1-x^{ n })^{ 1/n } } $$
If $$\displaystyle I_1=\int_{0}^{\pi /2}\frac{x}{\sin x}dx $$ and $$\displaystyle I_2=\int_{0}^{\pi /2}\frac{\tan ^{-1}x}{x}dx, $$ then $$\displaystyle \frac{I_{1}}{I_{2}}= $$
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$$\displaystyle \frac{1}{2} $$
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1
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2
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$$\displaystyle \frac{\pi }{2} $$
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
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$$\displaystyle\frac{\pi }{4}$$
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does not exist
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$$ \displaystyle \frac{\pi \sqrt{3}}{4} $$
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None of these
Explanation
$$\displaystyle f\left( x \right) =\int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ t }^{ 2 } } } dt=\sin { x } \int _{ -1 }^{ 1 }{ \frac { 1 }{ 1+{ t }^{ 2 } } } dt$$
$$\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t } \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\sin { x } \left[ \frac { \pi }{ 2 } \right] $$
$$\displaystyle \therefore f'\left( x \right) =\frac { \pi }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi }{ 3 } \right) =\frac { \pi }{ 2 } \frac { 1 }{ 2 } =\frac { \pi }{ 4 } $$
If $$\displaystyle I_{n} =\int_{0}^{\frac{\pi }{4}}\tan ^{n}xdx$$
then $$\displaystyle \frac{1}{I_{2}+I_{4}},\frac{1}{I_{3}+I_{5}},\frac{1}{I_{4}+I_{6}}$$ are in?
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$$A.P$$
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$$H.P$$
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$$G.P$$
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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.
$$\displaystyle\int_{0}^{a}x^{4}\left ( a^{2}-x^{2} \right )^{1/2} dx$$ equals
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$$\displaystyle\frac{\pi a^{5}}{32}$$
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$$\displaystyle\frac{\pi a^{6}}{32}$$
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$$\displaystyle \frac{\pi a^{2}}{32}$$
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None of these
Explanation
$$\displaystyle I=\int _{ 0 }^{ a }{ { x }^{ 4 }\sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } =\int _{ 0 }^{ a }{ { x }^{ 5 }\sqrt { { \left( \frac { a }{ x } \right) }^{ 2 }-1 } dx } $$
Put $$\displaystyle { \left( \frac { a }{ x } \right) }^{ 2 }=t\Rightarrow \left( -\frac { 2{ a }^{ 2 } }{ { x }^{ 3 } } \right) dx=dt$$
$$\displaystyle I=-\int _{ \infty }^{ 1 }{ \frac { { a }^{ 8 } }{ { t }^{ 4 } } \sqrt { { t }^{ 2 }-1 } dt } =\int _{ 1 }^{ \infty }{ \frac { { a }^{ 8 }\sqrt { { t }^{ 2 }-1 } }{ \sqrt { { t }^{ 2 } } } dt } $$
$$\displaystyle ={ a }^{ 8 }\int _{ 1 }^{ \infty }{ \sqrt { 1-\frac { 1 }{ { t }^{ 2 } } } dt } =\frac { \pi { a }^{ 6 } }{ 32 } $$
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
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$$1/2$$
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$$1$$
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$$\pi /4$$
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none of these
Explanation
$$\displaystyle \int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \frac { dt }{ t\left( 1+t^{ 2 } \right) } $$
$$\displaystyle =\int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \left( \frac { 1 }{ t } -\frac { 1 }{ 1+{ t }^{ 2 } } \right) dt$$
$$=\dfrac { 1 }{ 2 } { \left[ \log { \left( 1+{ t }^{ 2 } \right) } \right] }_{ 1/e }^{ \tan { x } }+{ \left[ \log { t } \right] }_{ 1/e }^{ \cot { x } }-{ \left[ \tan ^{ -1 }{ t } \right] }_{ 1/e }^{ \cot { x } } =1$$
$$\displaystyle \int_{0}^{\pi /2}\sin x\log \left ( \sin x \right )dx= $$
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$$\displaystyle \log _{e}e$$
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$$\displaystyle \log _{e}2$$
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$$\displaystyle \log _{e}\left ( e/2 \right )$$
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$$\displaystyle \log _{e}\left ( 2/e \right )$$
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 $$\displaystyle t= \cos x$$
$$\displaystyle = \frac{1}{2}\int_{0}^{1}\left [ -t^{2}-\frac{\left ( t^{2} \right )^{2}}{2}-\frac{\left ( t^{2} \right )^{3}}{3}-\cdots \right ]dt$$
$$\displaystyle = -\left [ \frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\cdots \right ]$$
$$\displaystyle = -\left [ \left ( \frac{1}{2}-\frac{1}{3} \right )+\left ( \frac{1}{4}-\frac{1}{5} \right )+\left ( \frac{1}{6}-\frac{1}{7} \right )+\cdots \right ]$$
$$\displaystyle = -1+\left ( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots \right )$$.
$$\displaystyle = -\log _{e}e+\log _{e}\left ( 1+1 \right )= \log _{e}\left ( 2/e \right )$$
Evaluate : $$\displaystyle \int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\times \left [ \sin ^{-1}\left ( 1-2x^{2} \right ) +\cos ^{-1}\left ( 2x\sqrt{1-x^{2}} \right )\right ]dx$$
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$$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$$
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$$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$$
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$$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$$
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$$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$$
Explanation
In $$\displaystyle N^{r}$$ 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
If $$\displaystyle \int_{\log 2}^{x}\displaystyle \frac{dx}{\sqrt{e^{x}-1}}= \displaystyle \frac{\pi }{6}$$, the value of $$x$$ is
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$$4$$
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$$\log 8$$
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$$\log 4$$
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none of these
Explanation
Let $$\displaystyle I=\int _{ \log 2 }^{ x } \frac { dx }{ \sqrt { e^{ x }-1 } } =\frac { \pi }{ 6 } $$
Put $${ e }^{ x }-1={ t }^{ 2 }$$
$$\displaystyle \Rightarrow 2\tan ^{ -1 }{ \sqrt { { e }^{ 2 }-1 } } -\frac { \pi }{ 2 } =\frac { \pi }{ 6 } \Rightarrow \sqrt { { e }^{ 2 }-1 } =\tan { \frac { \pi }{ 3 } } =\sqrt { 3 } $$
$$\Rightarrow { e }^{ x }=4\Rightarrow x=\log { 4 } $$
$$\displaystyle \int_{0}^{\pi /3}\frac{\cos \theta }{3+4\sin \theta }d\theta =\lambda \log \frac{3+2\sqrt{3}}{3}$$ then $$\displaystyle \lambda $$ equals
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$$\dfrac{1}{2}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{8}$$
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
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$$\displaystyle \frac{1}{2+tan^{2}x}$$
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$$1$$
0%
$$\pi /4$$
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$$\displaystyle \frac{2}{\pi }\displaystyle \int ^{1}_{-1}\displaystyle \frac{dt}{1+t^{2}}$$
Explanation
$$\displaystyle \int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \frac { dt }{ t\left( 1+t^{ 2 } \right) } $$
$$\displaystyle =\int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \left( \frac { 1 }{ t } -\frac { 1 }{ 1+{ t }^{ 2 } } \right) dt$$
$$=\frac { 1 }{ 2 } { \left[ \log { \left( 1+{ t }^{ 2 } \right) } \right] }_{ 1/e }^{ \tan { x } }+{ \left[ \log { t } \right] }_{ 1/e }^{ \cot { x } }-{ \left[ \tan ^{ -1 }{ t } \right] }_{ 1/e }^{ \cot { x } }\\ =1$$
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
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$$\sin ^{-1}\: \alpha /\beta $$
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$$\pi /2$$
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$$\sin ^{-1}\beta /2\alpha $$
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$$\pi $$
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
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$$\displaystyle \frac{1}{2}\log \left ( -1 \right )$$
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$$\displaystyle \log \left ( -1 \right )$$
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$$\displaystyle \log 3$$
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$$\displaystyle \frac {1}{2} \log \sqrt{3}$$
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?
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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 $$\displaystyle \int _{ 0 }^{ 1 }{ Idx } =\frac { \pi }{ 2\sqrt { 2 } } $$
The value of $$\displaystyle \int_{1/2}^{1}\displaystyle \frac{dx}{x\sqrt{3x^{2}+2x-1}}$$ is?
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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 $$\displaystyle t=\sqrt { 3 } x+\frac { 1 }{ \sqrt { 3 } } \Rightarrow dt=\sqrt { 3 } dt$$
$$\displaystyle I=\frac { 1 }{ \sqrt { 3 } } \int { -\frac { 1 }{ \left( \sqrt { 3 } -3t \right) \sqrt { { t }^{ 2 }-\frac { 4 }{ 3 } } } } dt$$
$$\displaystyle =-3\int { \frac { 1 }{ \left( \sqrt { 3 } -3t \right) \sqrt { { t }^{ 2 }-\frac { 4 }{ 3 } } } } dt$$
$$\displaystyle I=3\int { \frac { 2 }{ 3\sqrt { 3 } { s }^{ 2 }+\sqrt { 3 } } ds } =2\sqrt { 3 } \int { \frac { 1 }{ 3{ s }^{ 2 }+1 } ds } $$
$$\displaystyle =2\tan ^{ -1 }{ \left( \sqrt { 3 } s \right) } =\tan ^{ -1 }{ \left( \frac { x-1 }{ \sqrt { 3{ x }^{ 2 }+2x-1 } } \right) } $$
$$\displaystyle \therefore \int _{ 1/2 }^{ 1 } { \frac { 1 }{ x\sqrt { { 3x }^{ 2 }+2x-1 } } } { dx } =\frac { \pi }{ 6 } $$
If $$I= \displaystyle \int_{1}^{\infty }\displaystyle \frac{x^{2}-2}{x^{3}\sqrt{x^{2}-1}}\: dx$$, then $$I$$ equals
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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 $$x=\sec { u } \Rightarrow dx=\tan { u } \sec { u } du$$
$$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 2 }{ u\left( \sec ^{ 2 }{ u } -2 \right) } du } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 1-\sin ^{ 2 }{ u } \right) \left( \sec ^{ 2 }{ u } -2 \right) du } $$
$$\displaystyle =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2\sin ^{ 2 }{ u } -\tan ^{ 2 }{ u } +\sec ^{ 2 }{ u } -2 \right) } du=-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } +2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2 }{ u } du } $$
$$\displaystyle =-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } -\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { 2udu } } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } =-{ \left[ \frac { \sin { 2u } }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=0$$
If $$0< \alpha < \pi /2$$ then the value of $$\displaystyle \int_{0}^{\alpha }\displaystyle \frac{dx}{1-\cos x\cos \alpha }$$ is
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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
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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
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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 $$-3x+2=u$$
$${ I }_{ 2 }=\int _{ 0 }^{ 1 }{ { e }^{ { u }^{ 2 } } } du$$
$$\therefore I=0$$
If $$\displaystyle \int _{ 0 }^{ 1 }{ \frac { \sin { t } }{ 1+t } dt } =\alpha $$, then the value of the integral $$\displaystyle \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 4\pi +2-t } dt } $$ in terms of $$\alpha$$ is given by
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0%
$$2\alpha$$
0%
$$-2\alpha$$
0%
$$\alpha$$
0%
$$-\alpha$$
Explanation
$$\displaystyle \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 4\pi +2-t } dt } =\frac { 1 }{ 2 } \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 1+\left( 2\pi -\dfrac { t }{ 2 } \right) } dt } $$
$$\displaystyle =2.\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { \sin { \left( 2\pi -u \right) } }{ 1+u } du } $$
$$($$ Substitute $$\displaystyle 2\pi -\frac { t }{ 2 } =u\Rightarrow dt=-2du)$$
$$\displaystyle =\int _{ 0 }^{ 1 }{ \frac { \sin { u } }{ 1+u } du } =-\int _{ 0 }^{ 1 }{ \frac { \sin { t } }{ 1+t } dt } =-\alpha $$
Value of $$\displaystyle \int_{0}^{16}\displaystyle \frac{x^{1/4}}{1+x^{1/2}}\: dx$$ is
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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 $$\displaystyle u=\sqrt [ 4 ]{ x } \Rightarrow du=\frac { 1 }{ 4{ x }^{ \frac { 3 }{ 4 } } } dx$$
$$\displaystyle I=4\int _{ 0 }^{ 2 }{ \left( { u }^{ 2 }+\frac { 1 }{ { u }^{ 2 }+1 } -1 \right) } du$$
$$\displaystyle =4\int _{ 0 }^{ 2 }{ { u }^{ 2 } } du+4\int _{ 0 }^{ 2 }{ \frac { 1 }{ { u }^{ 2 }+1 } du } -4\int _{ 0 }^{ 2 }{ du } $$
$$\displaystyle =4\left[ \frac { { u }^{ 3 } }{ 3 } \right] _{ 0 }^{ 2 }{ + }4\left[ \tan ^{ -1 }{ u } \right] _{ 0 }^{ 2 }-4\left[ u \right] _{ 0 }^{ 2 }$$
$$\displaystyle =4\left( \frac { 2 }{ 3 } +\tan ^{ -1 }{ 2 } \right) $$
If $$I_{1}= \displaystyle \int_{x}^{1}\displaystyle \frac{dt}{1+t^{2}}$$ and $$I_{2}= \displaystyle \int_{1}^{1/x}\displaystyle \frac{dt}{1+t^{2}}$$ for $$x> 0$$, then
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0%
$$I_{1}= I_{2}$$
0%
$$I_{1}> I_{2}$$
0%
$$I_{2}> I_{1}$$
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$$I_{2}= \left ( \pi /2 \right )-\tan ^{-1}x$$
If $$I_{1}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{dx}{1+x^{4}}$$ and $$I_{2}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{x^{2}}{1+x^{4}}\: dx$$, then
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0%
$$I_{1}= I_{2}$$
0%
$$I_{1}=2 I_{2}$$
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$$2I_{1}= I_{2}$$
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 $$t=\dfrac{1}{x}\implies d{t}=-\dfrac{d{x}}{x^2}$$
$$I_{2}=\displaystyle\int _{\infty}^{0} \dfrac{-d{t}}{1+t^4}=\int_{0}^{\infty} \dfrac{d{x}}{1+x^4}=I_{1}$$
If $$\displaystyle \int_{0}^{\infty }\displaystyle \frac{\log \left ( 1+x^{2} \right )}{1+x^{2}}\: dx= \lambda \displaystyle \int_{0}^{1}\displaystyle \frac{\log \left ( 1+x \right )}{1+x^{2}}\: dx$$ then $$\lambda $$ equals
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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 $$\lambda = \cfrac{I_1}{I_2}=8$$
If $$\displaystyle I=\int _{8}^{15} \frac{dx}{(x-3)\sqrt{x+1} }$$ then$$ I$$ equals
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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
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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
As $$x $$ 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
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0%
$$\cfrac 13$$
0%
$$\cfrac 14$$
0%
$$\cfrac 23$$
0%
$$\cfrac 12$$
Explanation
$$\displaystyle \int_{0}^{1}f(x) dx =\sum_{r=1}^{\infty} \int_{2^{-r}}^{2^{-(r-1)}} \frac{1}{2^{r-1}}dx $$
$$\displaystyle =\sum_{r=1}^{\infty}\frac{1}{2^{r-1}}[2^{-(r-1)}-2^{-r}]$$
$$\displaystyle =\sum_{1}^{\infty}2^{-2(r-1)}- \sum _{1}^{\infty} 2^{-2r+1}$$
$$\displaystyle =(2^{2}-2)\sum_{1}^{\infty}2^{-2r}=2\cdot \frac{1}{4}\cdot \frac{1}{1-1/4}=\frac{2}{3}\cdot $$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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