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CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Definite Integrals
Quiz 4
If $$\displaystyle f\left ( \frac{1}{x} \right )+x^{2}f\left ( x \right )=0$$ for $$x> 0,$$
and $$\displaystyle I=\int_{1/x}^{x}f\left ( z \right )dz, \frac{1}{2}\leq x\leq 2$$
then $$\displaystyle I$$ is?
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$$\displaystyle f\left ( 2 \right )-f\left ( 1/2 \right )$$
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$$\displaystyle f\left ( 1/2 \right )-f\left ( 2 \right )$$
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$$0$$
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None of these
Explanation
Substitute $$\displaystyle z=\dfrac { 1 }{ t } $$
$$\displaystyle I=\int _{ x }^{ \dfrac { 1 }{ t } }{ f\left( \dfrac { 1 }{ t } \right) } .\left( \dfrac { -1 }{ { t }^{ 2 } } \right) dt=\int _{ x }^{ \dfrac { 1 }{ x } }{ \left( -{ t }^{ 2 }.f\left( t \right) \right) } \left( \dfrac { -1 }{ { t }^{ 2 } } \right) $$
$$\displaystyle =\int _{ x }^{ \dfrac { 1 }{ x } }{ f\left( t \right) } dt=-\int _{ \dfrac { 1 }{ x } }^{ x }{ f\left( t \right) } dt=-I\\ \therefore 2I=0$$
If $$\displaystyle I_{1}= \int_{-4}^{-5}e^{\left ( x +5 \right )^{2}}dx$$ and $$\displaystyle I_{2}= 3\int_{{1}/{3}}^{{2}/{3}}e^{\left ( 3x -2 \right )^{2}}dx$$
then $$I_{1}+I_{2}$$ equals?
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$$\displaystyle \frac{1}{3}$$
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$$\displaystyle -\frac{1}{3}$$
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$$0$$
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None of these
Explanation
$$I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx$$
$$\displaystyle x=\frac { -y-3 }{ 3 } $$
$$\displaystyle dx=-\frac { dy }{ 3 } $$
$$\displaystyle \Rightarrow I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx=-\int _{ -4 }^{ -5 } e^{ \left( y+5 \right) ^{ 2 } }dx=-I_{ 1 }$$
$$\displaystyle \Rightarrow I_{ 1 }+I_{ 2 }=0$$
If $$f\left ( a+x \right )= f\left ( x \right )$$ , then $$\forall$$ $$ a> 0, n\epsilon N$$ the value of $$\displaystyle \int_{0}^{n a}f\left ( x \right )dx$$ equals ?
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$$\displaystyle \left ( n-1 \right )\int_{0}^{a}f\left ( x \right )dx$$
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$$\displaystyle \left ( 1-n \right )\int_{0}^{a}f\left ( x \right )dx$$
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$$\displaystyle n\int_{0}^{a}f\left ( x \right )dx$$
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None of the above
Explanation
$$I=\displaystyle \int _{ 0}^{ na }{ f\left( x \right) dx } $$
Substitute $$x=a+t\Rightarrow dx=dt$$
$$I=\displaystyle \int _{ 0 }^{ \left( n-1 \right) a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+x \right) dx } =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( x \right) dx } $$
If $$\displaystyle f\left ( x \right )=\int_{-1}^{1}\frac{\sin x}{1+t^{2}}dt$$ then $$\displaystyle {f}'\left ( \frac{\pi }{3} \right )$$ is
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nonexistent
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$$\displaystyle \pi /4$$
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$$\displaystyle \pi \sqrt{3/4}$$
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none of these
Explanation
$$\displaystyle f\left( x \right) =\int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ t }^{ 2 } } } dt=\sin { x } \int _{ -1 }^{ 1 }{ \frac { 1 }{ 1+{ t }^{ 2 } } } dt$$
$$\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t } \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\sin { x } \left[ \frac { \pi }{ 2 } \right] $$
$$\displaystyle \therefore f'\left( x \right) =\frac { \pi }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi }{ 3 } \right) =\frac { \pi }{ 2 } \frac { 1 }{ 2 } =\frac { \pi }{ 4 } $$
$$\displaystyle \int_{0}^{1}\frac{2^{x+1}-3^{x-1}}{6^{x}}dx$$
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$$\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{6}\log _{2}e.$$
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$$\displaystyle -\frac{4}{3}\log _{3}e+\frac{1}{6}\log _{2}e.$$
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$$\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{3}\log _{2}e.$$
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$$\displaystyle -\frac{2}{3}\log _{3}e+\frac{1}{6}\log _{2}e.$$
Explanation
$$ \displaystyle \int _{ 0 }^{ 1 } \frac { 2^{ x+1 }-3^{ x-1 } }{ 6^{ x } } dx=\int _{ 0 }^{ 1 } \left( 2.3^{ -x }-\frac { 1 }{ 3 } 2^{ -x } \right) dx$$
$$ \displaystyle =\left[ -2\frac { 3^{ -x } }{ \log 3 } +\frac { 1 }{ 3 } \frac { 2^{ -x } }{ \log 2 } \right] ^{ 1 }$$
$$ \displaystyle =-\frac { 2 }{ \log 3 } \left( \frac { 1 }{ 3 } -1 \right) +\frac { 1 }{ 3\log 2 } \left( \frac { 1 }{ 2 } -1 \right) $$
$$\displaystyle =\frac { 4 }{ 3 } \log _{ 3 } e-\frac { 1 }{ 6 } \log _{ 2 } e$$
Hence, option 'A' is correct.
$$\displaystyle \int_{1}^{2}\left ( x+\frac{1}{x} \right )^{3/2}\frac{x^{2}-1}{x^{2}}dx$$
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$$\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}+\frac{8}{5}\sqrt{2}$$
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$$\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}$$
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$$\displaystyle \sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}$$
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$$\displaystyle \frac{3}{2}\sqrt{\left ( \frac{3}{2} \right )}-\frac{8}{5}\sqrt{2}$$
Explanation
Let $$\displaystyle I=\int _{ 1 }^{ 2 } \left( x+\frac { 1 }{ x } \right) ^{ 3/2 }\frac { x^{ 2 }-1 }{ x^{ 2 } } dx$$
Put $$\displaystyle x+\frac { 1 }{ x } =t\Rightarrow \left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx\Rightarrow dt$$
Therefore
$$\displaystyle I=\int _{ 2 }^{ 5/2 } t^{ 3/2 }dt=\frac { 2 }{ 5 } t^{ 5/2 }=\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 5/2 }-2^{ 5/2 } \right] $$
$$\displaystyle =\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 2 }\sqrt { \left( \frac { 5 }{ 2 } \right) } -2^{ 2 }\sqrt { 2 } \right] =\frac { 5 }{ 2 } \sqrt { \left( \frac { 5 }{ 2 } \right) } -\frac { 8 }{ 5 } \sqrt { 2 } $$
Hence, option 'B' is correct.
Evaluate the integrals $$\displaystyle \int_{a}^{b}\frac{\log x}{x}dx$$
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$$\displaystyle \frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}$$
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$$\displaystyle \frac{1}{4} \log\left ( ab\right )\cdot \log \frac{b}{a}$$
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$$\displaystyle \log\left ( ab\right )\cdot \log \frac{b}{a}$$
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$$\displaystyle -\frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}$$
Explanation
Substitute $$\displaystyle \log x=t \therefore \left ( 1/x \right )dx=dt$$ and adjust the limits
$$\displaystyle \therefore I= \int t\:dt=\left [ \frac{1}{2}t^{2} \right ]_{\log a}^{\log b}$$
$$\displaystyle \therefore I= \frac{1}{2}\left [ \left ( \log b \right )^{2}-\left ( \log a \right )^{2} \right ]$$
$$\displaystyle = \frac{1}{2}\left [ \left ( \log b+\log a \right )\left ( \log b-\log a \right ) \right ]$$
$$\displaystyle \therefore I= \frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}.$$
Given $$\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,$$ the value of $$\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx } $$ is?
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$${ e }^{ 4 }-e$$
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$${ e }^{ 4 }-a$$
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$$2{ e }^{ 4 }-a$$
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$$2{ e }^{ 4 }-e-a$$
Explanation
Given $$\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,$$
Let $$I=\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx } $$
Put $$\displaystyle \ln { \left( x \right) } ={ t }^{ 2 }\Rightarrow \frac { 1 }{ x } dx=2tdt$$
$$\therefore I=\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } .{ 2t }^{ 2 }dt={ \left( t{ e }^{ { t }^{ 2 } } \right) }_{ 1 }^{ 2 }-\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } dt={ 2e }^{ 4 }-e-a.$$
$$\displaystyle \int_{0}^{\pi /2}\frac{dx}{\sin x}$$equals
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$$0$$
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$$\displaystyle \frac{1}{2}$$
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$$1$$
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$$3/2$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { 1 }{ sin{ x } } dx } =\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \csc { x } } dx$$
Multiply numerator and denominator by $$cot{ x }+\csc { x } $$, we get
$$\displaystyle \therefore I=-\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { -cot{ x }\csc { x } -\csc ^{ 2 }{ x } dx }{ cot{ x }+\csc { x } } dx } $$
Substitute $$t=cot{ x }+\csc { x } \Rightarrow dt=\left( -\csc ^{ 2 }{ x } -cot{ x }\csc { x } \right) dx$$
$$\displaystyle \therefore I=-\int _{ 0 }^{ 1 }{ \dfrac { 1 }{ t } } dt=\left[ -\log { t } \right] _{ 0 }^{ 1 }=0$$
The value of $$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}+2x}}$$ is
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$$\pi /6$$
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$$\pi /3$$
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$$\pi /2$$
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$$\pi $$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { x }^{ 2 }+2x } } dx } =\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { \left( x+1 \right) }^{ 2 }-1 } } dx } $$
Substitute $$u=x+1\Rightarrow du=dx$$
$$\displaystyle\therefore I=\int _{ 1 }^{ 2 }{ \frac { 1 }{ u\sqrt { { u }^{ 2 }-1 } } du } $$
Substitute
$$\displaystyle s=\sqrt { { u }^{ 2 }-1 } \Rightarrow ds=\frac { u }{ \sqrt { { u }^{ 2 }-1 } } du$$
$$\displaystyle\therefore I=\int _{ 0 }^{ \sqrt { 3 } }{ \frac { 1 }{ { s }^{ 2 }+1 } ds } ={ \left[ \tan ^{ -1 }{ s } \right] }_{ 0 }^{ \sqrt { 3 } }=\frac { \pi }{ 3 } $$
Value of $$\displaystyle \int_{0}^{\pi /4}\left ( \sqrt{\tan x}-\sqrt{\cot x} \right )\: dx$$ is
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$$\sqrt{2}\log \left ( \sqrt{2}-1 \right )$$
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$$\sqrt{2}\log \left ( \sqrt{2}+1 \right )$$
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$$\log \left ( \sqrt{2}+1 \right )$$
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$$\log \left ( \sqrt{2}-1 \right )$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \left( \sqrt { \tan { x } } -\sqrt { cot{ x } } \right) } dx=-\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \frac { cos{ x }-sin{ x } }{ \sqrt { cos{ x }sin{ x } } } dx } $$
Substitute $$\left( sin{ x+cos{ x } } \right) =t\Rightarrow 2sin{ x }cos{ x }={ t }^{ 2 }-1$$
$$\displaystyle \therefore I=\sqrt { 2 } \int _{ 1 }^{ \sqrt { 2 } }{ \frac { dt }{ \sqrt { { t }^{ 2 }-1 } } } \\ =\left[ \sqrt { 2 } \log { \left( t+\sqrt { { t }^{ 2 } } -1 \right) } \right] _{ 0 }^{ \sqrt { 2 } }\\ =\sqrt { 2 } \log { \left( \sqrt { 2 } -1 \right) } $$
The value of $$\displaystyle \int_{a}^{b}\displaystyle \frac{\log x}{x}\: dx$$ is
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$$\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )$$
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$$\displaystyle \frac{1}{2}\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )$$
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$$\log \left ( a^{2}-b^{2} \right )$$
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$$\left ( a+b \right )\log \left ( a+b \right )$$
Explanation
$$\displaystyle I=\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx={ \left[ \log { x } .\log { x } \right] }_{ a }^{ b }-\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx\\ \Rightarrow 2I={ \left[ { \left( \log { x } \right) }^{ 2 } \right] }_{ a }^{ b }={ \left( \log { b } \right) }^{ 2 }-{ \left( \log { a } \right) }^{ 2 }$$
$$\displaystyle \Rightarrow I=\frac { 1 }{ 2 } \left( \log { b } +\log { a } \right) \left( \log { b } -\log { a } \right) =\frac { 1 }{ 2 } \log { ab } \log { \frac { b }{ a } } $$
If $$f\left ( x \right )= A\sin \left ( \dfrac {\pi x}{2} \right )\: +\: B,f{}'\left ( \dfrac 12 \right )= \sqrt{2}$$ and $$\displaystyle \int_{0}^{1}f\left ( x \right )dx= \displaystyle \frac{2A}{\pi }$$,
then the constants $$A$$ and $$B$$ are
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$$A=\dfrac {\pi}{2}, B=\dfrac {\pi}{2}$$
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$$A=\dfrac {2}{\pi}, B=3\pi $$
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$$A=0, B=-4\pi $$
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$$A=\dfrac {4}{\pi}, B=0$$
Explanation
$$f(x)=A\sin{(\pi x/2)}+B$$
$$\implies f^{'}(x)=\dfrac{A\pi}{2}\cos{(\pi x/2)}$$
$$\implies f^{'}(1/2)=\dfrac{A\pi}{2}\cos{(\pi/4)=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}}$$
$$\implies\sqrt{2}=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}$$
Hence, $$ A=\dfrac{4}{\pi}$$
$$\displaystyle\int_{0}^{1} f(x)dx=\left[-\dfrac{2A}{\pi}\cos{(\pi x/2)}+Bx\right]_{0}^{1}$$
$$\implies\dfrac{2A}{\pi}=\dfrac{2A}{\pi}+B$$
Hence, $$B=0$$
Hence, answer is option-(D).
The value of the integral $$\displaystyle \int_{0}^{\pi /4}\displaystyle \frac{\sin x+\cos x}{3+\sin 2x}dx$$ is
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$$\log 2$$
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$$\log 3$$
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$$\left ( 1/4 \right )\log 3$$
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$$\left ( 1/8 \right )\log 3$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ \pi /4 } \frac { \sin x+\cos x }{ 3+\sin 2x } dx=\int _{ 0 }^{ \pi /4 } \frac { \sin { x } +\cos { x } }{ 4-{ \left( \sin { x } -\cos { x } \right) }^{ 2 } } dx$$
Substitute $$\sin { x } -\cos { x } =t$$
$$\displaystyle I=\int _{ -1 }^{ 0 }{ \frac { dt }{ 4-{ t }^{ 2 } } } =\frac { 1 }{ 4 } \log { 3 } $$
The value of $$\displaystyle \int_{0}^{\pi }\displaystyle \frac{dx}{1-2\alpha \cos x+\alpha ^{2}}$$ is
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$$\displaystyle \frac{\pi }{1+\alpha ^{2}}$$ if $$\alpha > 1$$
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$$\displaystyle \frac{\pi }{\alpha ^{2}-1}$$ if $$\alpha > 1$$
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$$\displaystyle \frac{\pi }{1+\alpha ^{2}}$$ if $$\alpha < 1$$
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$$\displaystyle \frac{\pi }{\alpha ^{2}-1}$$ if $$\alpha < 1$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } } =\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } } $$
Substitute $$\displaystyle \tan { \frac { x }{ 2 } } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt$$
$$\displaystyle I=\int _{ 0 }^{ \infty } \dfrac { 1 }{ 1-2\alpha \left( \dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } \right) +\alpha ^{ 2 } } .\dfrac { 2t }{ 1+{ t }^{ 2 } } dt$$
$$\displaystyle =\int _{ 0 }^{ \infty } \dfrac { 2t }{ 1+{ t }^{ 2 }-2\alpha \left( 1-{ t }^{ 2 } \right) +\alpha ^{ 2 }\left( 1+{ t }^{ 2 } \right) } dt$$
$$\displaystyle =\frac { \pi }{ { \alpha }^{ 2 }-1 } $$ if $$\alpha >1$$
If $$I= \displaystyle \int_{1}^{2}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}-1}}$$ then $$I$$ is equal to
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$$1$$
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$$1/2$$
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$$1/\sqrt{2}$$
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$$1/\sqrt{3}$$
Explanation
As form of integrand is $$\displaystyle \dfrac { 1 }{ L\sqrt { Q } } $$
Substitute $$\displaystyle L=\dfrac { 1 }{ t } \Rightarrow x+1=\dfrac { 1 }{ t } $$
$$\displaystyle I=\int _{ \dfrac { 1 }{ 2 } }^{ \dfrac { 1 }{ 3 } }{ \dfrac { \left( \dfrac { -1 }{ { t }^{ 2 } } \right) dt }{ \dfrac { 1 }{ t } \sqrt { { \left( \dfrac { 1 }{ t } -1 \right) }^{ 2 }-1 } } } =\int _{ \dfrac { 1 }{ 3 } }^{ \dfrac { 1 }{ 2 } }{ \dfrac { dt }{ \sqrt { 1-2t } } } $$
$$\displaystyle ={ \left[ \dfrac { { \left( 1-2t \right) }^{ \dfrac { 1 }{ 2 } } }{ \left( -2 \right) \left( \dfrac { 1 }{ 2 } \right) } \right] }_{ \dfrac { 1 }{ 3 } }^{ \dfrac { 1 }{ 2 } }{ = }\dfrac { 1 }{ \sqrt { 3 } } $$
The value of $$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{e^{x}+e^{-x}}$$ is
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$$\tan ^{-1}e$$
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$$\tan ^{-1}\left ( e \right )-\pi /4$$
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$$\tan ^{-1}\left ( e \right )-\tan ^{-1}\left ( 1/e \right )$$
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$$\tan ^{-1}\left ( 1/e \right )+\pi /4$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dx }{ e^{ x }+e^{ -x } } =\int _{ 0 }^{ 1 } \frac { { e }^{ x }dx }{ e^{ 2x }+1 } $$
Substitute $${ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt$$
$$\displaystyle I=\int _{ 1 }^{ e }{ \frac { t }{ { t }^{ 2 }+1 } dt } ={ \left[ \tan ^{ -1 }{ t } \right] }_{ 1 }^{ e }=\tan ^{ -1 }{ e } -\frac { \pi }{ 4 } $$
The value of $$\displaystyle \int_{8}^{15}\displaystyle \frac{dx}{\left ( x-3 \right )\sqrt{x+1}}$$ is
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$$\log \displaystyle \frac{5}{3}$$
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$$\displaystyle \frac{1}{2}\log \displaystyle \frac{5}{3}$$
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$$2\log \displaystyle \frac{5}{3}$$
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$$\sqrt{2}\log \left ( \sqrt{2}-1 \right )$$
Explanation
$$\displaystyle I=\int _{ 8 }^{ 15 } \frac { dx }{ \left( x-3 \right) \sqrt { x+1 } } $$
Substitute $$\sqrt { x+1 } =t\Rightarrow x+1={ t }^{ 2 }$$
$$\displaystyle I=\int _{ 3 }^{ 4 }{ \frac { 2t }{ \left( { t }^{ 2 }-4 \right) t } dt } ={ \left[ \frac { 2 }{ 2.2 } \log { \left( \frac { t-2 }{ t+2 } \right) } \right] }_{ 3 }^{ 4 }$$
$$\displaystyle =\frac { 1 }{ 2 } \left[ \log { \frac { 1 }{ 3 } } -\log { \frac { 1 }{ 5 } } \right] =\frac { 1 }{ 2 } \log { \frac { 5 }{ 3 } } $$
If $$b > a,$$ and $$\displaystyle I = \int _{a}^{b} \sqrt{\frac{ x-a}{b-x} }dx,$$ then $$I$$ equals
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$$ \displaystyle \frac{\pi}{2} (b -a)$$
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$$\pi (b -a)$$
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$$\pi/2 $$
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$$ 2\pi (b-a)$$
Explanation
Substitute
$$ b -x = t^{2} $$ so that
$$\displaystyle I = \int _{\sqrt{b-a}}^{0}\sqrt{\frac{b-t^{2}-a}{t^{2}}} (-2t) dt$$
$$ \displaystyle= 2 \int_{0}^{c} \sqrt{c^{2}-t^{2}} dt $$, where $$ c= \sqrt{b-a}$$
$$=\displaystyle
2 \left [ \frac{1}{2} t \sqrt{c^{2}-t^{2}}+ \frac{c^{2}}{2} \sin ^{-1}
\left( \frac{t}{c}\right) \right] _{0}^{c}$$
$$ =0 + c^{2} \sin^{-1} (1) -0$$ $$\displaystyle =\frac{\pi}{2}(b-a) $$
Value of $$\displaystyle \int_{0}^{2a}\dfrac{x^{3/2}}{\sqrt{2a-x}} dx$$ is
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$$\displaystyle \frac{3\pi a^{2}}{2}$$
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$$\pi a^{3}$$
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$$\sqrt{2}\pi a^{3}$$
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$$2\pi a^{3}$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 2a }{ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \sqrt { 2a-x } } dx } $$
Substitute
$$\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx$$
$$\displaystyle I=2\int _{ 0 }^{ \sqrt { 2a } }{ \frac { { u }^{ 4 } }{ \sqrt { 2a-{ u }^{ 2 } } } du } $$
Substitute
$$u=\sqrt { 2a } \sin { t } \Rightarrow du=\sqrt { 2a } \cos { t } dt$$
$$\displaystyle I=2\sqrt { 2a } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2\sqrt { 2 } { a }^{ \frac { 3 }{ 2 } }\sin ^{ 4 }{ t } \right) dt } =8{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \sin ^{ 4 }{ t } dt } $$
Using reduction formulae
$$\displaystyle \int { \sin ^{ m }{ x } dx } =\frac { -\cos { x } \sin ^{ m-1 }{ x } }{ m } +\frac { m-1 }{ m } \int { \sin ^{ m-2 }{ x } dx } $$
$$\displaystyle I={ \left[ -2{ a }^{ 2 }\sin ^{ 3 }{ t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2 }{ t } dt } $$
$$\displaystyle =0+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \frac { 1 }{ 2 } -\frac { 1 }{ 2 } \cos { 2t } \right) dt } $$
$$\displaystyle ={ \left[ 3{ a }^{ 2 }t-3{ a }^{ 2 }\sin { t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { 3\pi { a }^{ 2 } }{ 2 } $$
Value of $$\displaystyle \int_{0}^{25}\displaystyle \frac{1}{\sqrt{4+\sqrt{x}}}\: dx$$ is
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$$2\left ( \sqrt{29}-1 \right )$$
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$$2\left ( \sqrt{29}-5 \right )$$
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$$3 \sqrt{29}-1 $$
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none of these
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 25 }{ \frac { 1 }{ \sqrt { \sqrt { x } +4 } } dx } $$
Substitute
$$\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx$$
$$\displaystyle I=2\int _{ 0 }^{ 5 }{ \frac { u }{ \sqrt { u+4 } } du } $$
Substitute
$$ s=u+4\Rightarrow ds=du$$
$$\displaystyle I=2\int _{ 4 }^{ 9 }{ \frac { s-4 }{ \sqrt { s } } ds } =2\int _{ 4 }^{ 9 }{ \left( \sqrt { s } -\frac { 4 }{ \sqrt { s } } \right) ds } $$
$$\displaystyle={ \left[ \frac { 4{ s }^{ 3/2 } }{ 3 } \right] }_{ 4 }^{ 9 }-{ \left[ 16\sqrt { s } \right] }_{ 4 }^{ 9 }=36-\frac { 32 }{ 3 } -48-32=-\frac { 164 }{ 3 } $$
$$\displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx$$
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is equal to zero
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is equal to one
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is equal to $$\displaystyle \frac{1}{2}$$
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can not be evaluated
Explanation
$$I = \displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx $$
Substitute
$$\displaystyle x = \frac{1}{t}\Rightarrow dx =-\frac{1}{t^2}dt$$
$$I = \displaystyle \int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln (1/t)}{1/t} .\frac{-dt}{t^2} $$
$$\quad \displaystyle =\int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln t}{t} dt$$
$$\quad \displaystyle =\int_{\infty }^0 f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x} dx=-I$$
$$\Rightarrow 2I = 0\Rightarrow I = 0$$
If$$\displaystyle \int_{0}^{\pi /3}\frac{\cos }{3+4\sin x}dx=K\log \frac{\left ( 3+2\sqrt{3} \right )}{3}$$ then K is
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{3}$$
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{1}{8}$$
Explanation
$$\displaystyle I=\int _{ 0 }^{ \pi /3 }{ \cfrac { \cos { x } }{ 3+4\sin { x } } dx } $$
Substituting $$t=3+4\sin { x } \Rightarrow dt=4\cos { x } $$
$$\displaystyle I=\frac { 1 }{ 4 } \int _{ 3 }^{ 3+2\sqrt { 3 } }{ \frac { 1 }{ t } dt } ={ \left[ \frac { \log { t } }{ 4 } \right] }_{ 3 }^{ 3+2\sqrt { 3 } }$$
$$\displaystyle=\frac { \log { \left( 3+2\sqrt { 3 } \right) } -\log { 3 } }{ 4 } =\frac { 1 }{ 4 } \log { \left( \frac { 3+2\sqrt { 3 } }{ 3 } \right) } \Rightarrow K=\frac { 1 }{ 4 } $$
Suppose that F(x) is an antiderivative of f(x)$$\displaystyle =\frac{\sin x}{x},x> 0$$ then $$\displaystyle \int_{1}^{3}\frac{\sin 2x}{x}$$ can be expressed as
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$$F(6) - F(2)$$
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$$\displaystyle \frac{1}{2}\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )$$
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$$\displaystyle \frac{1}{2}\left ( F\left ( 3 \right )-F\left ( 1 \right ) \right )$$
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$$\displaystyle 2\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )$$
Explanation
$$\displaystyle F\left ( x \right )=\int \frac{\sin x}{x}dx$$ Now $$\displaystyle I = \int_{1}^{3}\frac{\sin 2x}{x}dx\left [ put2x=t \right ]=\int_{2}^{6}\frac{2}{2}\frac{\sin }{t}dt=\left [ F\left ( x \right ) \right ]_{2}^{6}=F(6)-F(2)$$
If $$0 < \alpha < 1$$ and $$\displaystyle I=\int _{-1}^{1} \frac{dx}{\sqrt{1-2\alpha x+\alpha^{2}}} $$ then $$I$$ equals
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$$1/\alpha $$
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$$ 2/\alpha $$
0%
$$ 3/\alpha$$
0%
none of these
Explanation
$$\displaystyle I =\int_{-1}^{1} (1 -2\alpha x + \alpha^{2} )^{-1/2} dx$$
$$\displaystyle =\left. \frac{2(1-2\alpha x+\alpha^{2})^{1/2}}{-2\alpha}\right ]_{-1}^{1}$$
$$\displaystyle =-\frac{1}{\alpha}[([(1 -2\alpha + \alpha^{2})^{1/2} -(1 + 2 \alpha + \alpha ^{2})^{1/2} ]$$
$$=-\dfrac{1}{\alpha}[\sqrt{(\alpha-1)^2}-\sqrt{(\alpha+1)^2}]$$
$$=-\dfrac1\alpha\left[(1-\alpha)-(1+\alpha)\right]\ldots(\because0<\alpha<1)$$
$$=2$$
Ans: D
$$\displaystyle \int_{1/2}^{2}\frac{1}{x}\sin \left ( x-\frac{1}{x} \right )dx$$ has the value equal to
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$$0$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{5}{4}$$
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$$2$$
Explanation
$$\displaystyle I=\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ x } } \sin { \left( x-\frac { 1 }{ x } \right) dx } $$
Put $$\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=\frac { 1 }{ { t }^{ 2 } } dt$$
$$\displaystyle I=\int _{ 2 }^{ \frac { 1 }{ 2 } }{ t\sin { \left( \frac { 1 }{ t } -t \right) \left( \frac { -1 }{ { t }^{ 2 } } \right) dt } } =\int _{ 2 }^{ \frac { 1 }{ 2 } }{ \frac { 1 }{ t } \sin { \left( t-\frac { 1 }{ t } \right) dt } } $$
$$\displaystyle =-\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ t } } \sin { \left( t-\frac { 1 }{ t } \right) dt } =-I.$$
$$\Rightarrow 2I=0\quad \Rightarrow I=0$$
The value of the definite integral $$\displaystyle \int_{1}^{\infty}\left ( e^{x+1}+e^{3-x} \right )^{-1}dx$$ is
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$$\displaystyle \frac{\pi }{4e^{2}}$$
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$$\displaystyle \frac{\pi }{4e}$$
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$$\displaystyle \frac{1}{e^{2}}\left ( \frac{\pi }{2}-\tan ^{-1}\frac{1}{e} \right )$$
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$$\displaystyle \frac{\pi }{2e^{2}}$$
Explanation
$$I=\int _{ 1 }^{ \infty }{ \cfrac { 1 }{ { e }^{ x+1 }+{ e }^{ 3-x } } dx } =\int _{ 1 }^{ \infty }{ \cfrac { { e }^{ x-1 } }{ \left( { e }^{ x }-ie \right) \left( { e }^{ x }+ie \right) } dx } $$
Substituting $${ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt$$
$$I=\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \cfrac { 1 }{ \left( t-ie \right) \left( t+ie \right) } dt } =\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \cfrac { 1 }{ \left( e-it \right) \left( e+it \right) } dt } \\ =\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \left( \cfrac { i }{ 2\left( et+i{ e }^{ 2 } \right) } -\cfrac { i }{ 2\left( et-i{ e }^{ 2 } \right) } \right) } dt\\ ={ \left[ \cfrac { i\log { \left( et+i{ e }^{ 2 } \right) } }{ 2{ e }^{ 2 } } -\cfrac { i\log { \left( et-i{ e }^{ 2 } \right) } }{ 2{ e }^{ 2 } } \right] }_{ e }^{ \infty }\\ ={ \left[ \cfrac { \tan ^{ -1 }{ \left( { e }^{ x-1 } \right) } }{ { e }^{ 2 } } \right] }_{ e }^{ \infty }=\cfrac { \pi }{ 4{ e }^{ 2 } } $$
$$\displaystyle \int ^{\displaystyle \frac{3 \pi}{10}}_{\displaystyle \frac{\pi}{5}} \frac{sin x}{sin x + cos x}dx $$ is equal to
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$$\pi$$
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$$\displaystyle \frac{\pi}{2}$$
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$$\displaystyle \frac{\pi}{4}$$
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$$\displaystyle \frac{\pi}{20}$$
Explanation
Let $$\displaystyle I=\int { \frac { \sin { x } }{ \sin { x } +\cos { x } } dx } $$
Multiply numerator and denominator by $$I=\int { \dfrac { \sin { x } }{ \sin { x } +\cos { x } } dx } \\ \csc ^{ 3 }{ x } $$
$$\displaystyle I=\int { \frac { \csc ^{ 2 }{ x } }{ \csc ^{ 2 }{ x } +\cot { x } \csc ^{ 2 }{ x } } dx } =\int { \frac { \csc ^{ 2 }{ x } }{ \cot ^{ 3 }{ x } +\cot ^{ 2 }{ x } +\cot { x } +1 } dx } $$
Put $$t=\cot { x } \Rightarrow dt=-\csc ^{ 2 }{ x } dx$$
$$\displaystyle I=-\int { \frac { 1 }{ { t }^{ 3 }+{ t }^{ 2 }+t+1 } dt } =-\int { \frac { 1-t }{ 2\left( { t }^{ 2 }+1 \right) } dt } -\int { \frac { 1 }{ 2\left( t+1 \right) } dt } $$
$$\displaystyle =\frac { 1 }{ 2 } \int { \frac { t }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ t+1 } dt } $$
$$\displaystyle =\frac { 1 }{ 4 } \log { \left( { t }^{ 2 }+1 \right) } -\frac { 1 }{ 2 } \tan ^{ -1 }{ t } -\frac { 1 }{ 2 } \log { \left( t+1 \right) } $$
$$\displaystyle =\frac { 1 }{ 2 } \left( x-\log { \left( \sin { x } +\cos { x } \right) } \right) $$
Hence $$\displaystyle \int _{ \frac { \pi }{ 5 } }^{ \frac { 3\pi }{ 10 } } \frac { sinx }{ sinx+cosx } dx=\frac { \pi }{ 20 } $$
If $$\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt$$ where $$x > 0$$, then the value(s) of $$x$$ satisfying the equation, $$f(x) +f(1/x)=2$$ is
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$$2$$
0%
$$e$$
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$$\displaystyle e^{-2}$$
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$$\displaystyle e^{2}$$
Explanation
$$\displaystyle f\left( x \right)=\int _{ 1 }^{ x }{ \frac { \ln { t } }{ 1+t } dt } $$
$$\displaystyle \Rightarrow f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ 1/x }{ \cfrac { \ln { t } }{ 1+t } dt } $$
Substituting $$t=\dfrac { 1 }{ u } \Rightarrow dt=\left( -\dfrac { 1 }{ { u }^{ 2 } } \right) du$$
Therefore $$\displaystyle f\left( \dfrac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \dfrac { \ln { \left( \dfrac { 1 }{ u } \right) \left( -1 \right) } }{ \left( 1+\dfrac { 1 }{ u } \right) { u }^{ 2 } } du }$$
$$\displaystyle=\int _{ 1 }^{ x }{ \frac { \ln { u } }{ u\left( u+1 \right) } du } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt } $$
Now, $$\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \cfrac { \ln { t } }{ \left( 1+t \right) } dt } +\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt }$$
$$\displaystyle=\int _{ 1 }^{ x }{ \frac { \left( 1+t \right) \ln { t } }{ t\left( 1+t \right) } dt } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t } dt } =\frac { 1 }{ 2 } { \left( \ln { x } \right) }^{ 2 }$$
Hence $$\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =2\Rightarrow x={ e }^{ \pm 2 }$$
Choose a function $$f(x)$$ such that it is integrable over every interval on the real line
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$$f(x) = [x]$$
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$$f(x)=x|x|$$
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$$f(x)=[sinx]$$
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$$f(x)=\dfrac{|x-1|}{x-1}$$
$$\displaystyle \int_{0}^{\infty }\frac{x}{\left ( 1+x \right )\left ( 1+x^{2} \right )}dx$$
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$$\displaystyle \frac{\pi }{4}$$
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$$\displaystyle \frac{\pi }{2}$$
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is sme as $$\displaystyle \int_{0}^{\infty }\frac{dx}{\left ( 1+x \right )\left ( 1+x^{2} \right )}$$
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cannot be evaluated
Explanation
Let $$I=\int _{ 0 }^{ \infty }{ \cfrac { xdx }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } } $$
Using partial fraction
$$=\int _{ 0 }^{ \infty }{ \left( \cfrac { x+1 }{ 2\left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2\left( 1+x \right) } \right) dx } \\ ={ \left( \lim _{ b\rightarrow \infty }{ \cfrac { 1 }{ 2 } \log { \left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+x \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }x } \right) }_{ 0 }^{ b }\\ =\lim _{ b\rightarrow \infty }{ \left( \cfrac { 1 }{ 2 } \log { \left( 1+b \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+b \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }b \right) } \\ =\cfrac { \pi }{ 4 } $$
State true or false:
The average value of the function $$f(x) = sin^2xcos^3x$$ on the interval $$[ -\pi ,\pi ]$$ is 0.
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True
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False
Explanation
$$\int _{ -\pi }^{ \pi }{ f\left( x \right) dx } =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos ^{ 3 }{ x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \left( 1-\sin ^{ 2 }{ x } \right) \cos { x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos { x } dx } -\int _{ -\pi }^{ \pi }{ \sin ^{ 4 }{ x } \cos { x } dx } $$
$$\displaystyle ={ \left[ \frac { \sin ^{ 3 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }-{ \left[ \frac { \sin ^{ 5 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }=0$$
$$I_{1}$$ is equal to
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$$\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta$$
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$$\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta$$
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$$\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta$$
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$$\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta$$
Explanation
$$\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3}dx=(1-x^{2})^{3}_{0}+\frac {2}{3} \int_{0}^{1}\frac {x^{2}}{(1-x^{2})^{2/3}}dx$$
$$x=sin\theta, dx=cos\theta d\theta$$
$$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}\frac {sin^{2}\theta cos\theta}{(cos\theta)^{4/3}}d\theta$$
$$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}sin^{2}\theta (cos\theta)^{-1/3}d\theta$$
Ans: $$A$$
Evaluate $$\displaystyle \int_{0}^{\pi /2} \frac{dx}{2+\sin 2x}$$
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$$\displaystyle \frac{2\pi }{{3}}$$
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$$\displaystyle \frac{\pi }{{3}}$$
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$$\displaystyle \frac{2\pi }{{5}}$$
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None of these
Explanation
$$\displaystyle \int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\sin { 2x } } } =\int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\frac { 2\tan { x } }{ 1+\tan ^{ 2 }{ x } } } } dx$$
$$\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ 2x }{ \frac { \sec ^{ 2 }{ x } }{ \tan ^{ 2 }{ x } +\tan { x+1 } } dx } $$
Put $$\tan { x } =t\Rightarrow \sec ^{ 2 }{ x } dx=dt$$
$$\displaystyle I=\frac { 1 }{ 2 } \int { \frac { 1 }{ { { t }^{ 2 }+t+1 } } } dt.=\frac { 1 }{ 2 } \int { \frac { 1 }{ { \left( t+\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 3 }{ 2 } } } =\frac { 2\pi }{ 3 } $$
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0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect
Explanation
Reason is true
Assertion
Put $$\displaystyle x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^{2}}dt$$
$$\displaystyle l=-\int_{3}^{1/3}tcosec ^{99}\left ( \frac{1}{t}-t \right )\frac{1}{t^{2}}dt$$
$$\displaystyle l=-\int_{1/3}^{3}\frac{1}{t}cosec^{99}\left ( t-\frac{1}{t} \right )dt$$
$$\displaystyle l=-1$$
$$\Rightarrow 2l=0$$
$$\Rightarrow l=0$$
$$\displaystyle \int_{\tfrac{1}{\sqrt{3}}}^{0}\dfrac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$$
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$$\displaystyle - \tan^{-1} \dfrac{1}{2}$$
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$$\displaystyle \tan^{-1} 1$$
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$$\displaystyle - \tan^{-1} \dfrac{1}{3}$$
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$$\displaystyle \tan^{-1} \dfrac{1}{\sqrt 2}$$
Explanation
$$I=\displaystyle \int_{\frac{1}{\sqrt{3}}}^{0}\frac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$$
Substitute
$$x=\displaystyle \frac{1}{z}$$
$$\Rightarrow\displaystyle dx=-\frac{1}{z^2}dz$$
$$I=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -\dfrac { 1 }{ { z }^{ 2 } } dz }{ \left( \dfrac { 2 }{ { z }^{ 2 } } +1 \right) \sqrt { \dfrac { 1 }{ { z }^{ 2 } } +1 } } $$
$$=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -zdz }{ \left( 2+{ z }^{ 2 } \right) \sqrt { 1+{ z }^{ 2 } } } $$
Substitute
$$\sqrt { 1+{ z }^{ 2 } } =t$$
$$\Rightarrow z^2+1=t^2$$
$$\Rightarrow zdz=tdt$$
$$I=\displaystyle \int _{ 2 }^{ \infty } \frac { -tdt }{ \left( 1+{ t }^{ 2 } \right) t } $$
$$I=[\cot^{-1}t]_{2}^{\infty}$$
$$\Rightarrow I=-\cot^{-1}2$$
$$\Rightarrow I=-\tan^{-1}{\dfrac{1}{2}}$$
The value of definite integral $$\displaystyle \int_{\infty }^{0}\frac{Ze^{-z}}{\sqrt{1-e^{-2z}}}dz$$
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$$\displaystyle -\frac{\pi }{2}ln2$$
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$$\displaystyle \frac{\pi }{2}ln2$$
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$$\displaystyle -\pi ln2$$
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$$\displaystyle \pi ln\frac{1}{\sqrt{2}}$$
Explanation
$$\displaystyle l=\int_{\infty }^{0}\frac{ze^{-z}}{\sqrt{1-e^{-2z}}}dz$$ put $$\displaystyle e^{-z}=\sin \theta $$
$$\displaystyle l=-\int_{0}^{\pi /2}\frac{ln\left ( \sin \theta \right )\left ( -\cos \theta \right )d\theta }{\sqrt{1-\sin ^{2}\theta }}=\int_{0}^{\pi /2}ln\sin \theta d\theta $$
$$\displaystyle \frac{-\pi }{2}ln2$$
The derivative of $$F(x)=\displaystyle\int^{x^3}_{x^2}\frac{dt}{log t}(x >0)$$ is
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$$\displaystyle\frac{1}{3log x}$$
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$$\displaystyle\frac{1}{3log x}-\frac{1}{2log x}$$
0%
$$(log x)^{-1}x(x-1)$$
0%
$$\displaystyle\frac{3x^2}{log x}$$
Explanation
Given, $$F(x)=\displaystyle\int^{x^3}_{x^2}\frac{dt}{log t}$$
$$\Rightarrow F'(x)=\displaystyle\frac{d}{dx}F(x)$$
$$=\displaystyle\frac{d}{dx}(x^3)\cdot\frac{1}{log x^3}-\frac{d}{dx}(x^2)\cdot\frac{1}{log x^2}$$
$$=\displaystyle\frac{3x^2}{3log x}-\frac{2x}{2log x}=\frac{x^2-x}{log x}$$
$$=x(x-1)(log x)^{-1}$$
If $$\displaystyle \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0$$, then
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$$1<\alpha <2$$
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$$\alpha <2$$
0%
$$0<\alpha<1$$
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$$\alpha=0$$
Explanation
$$\displaystyle\because \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0,$$
$$\therefore { e }^{ { x }^{ 2 } }\left( x-\alpha \right) $$ must be $$+ve$$ and $$-ve$$ both for $$x\in (0,1)$$
i.e. $${ e }^{ x }\left( x-\alpha \right) =0$$ for one $$x\in (0,1)$$
$$\therefore \alpha \in(0,1)$$
Trapezoidal rule for evaluation of $$\displaystyle\int _{ a }^{ b }{ f(x)dx } $$ requires the interval $$(a,b)$$ to be divided into
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$$2n$$ sub-intervals of equal width
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any number of sub intervals of not equal width
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any number of sub intervals of equal width
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$$3n$$ sub-intervals of equal width
Explanation
Trapezoidal rule for evaluation of $$\int _{ a }^{ b }{ f(x)dx } $$ requires the interval $$(a,b)$$ to be divided into any number of sub-intervals of equal width.
$$\displaystyle \int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } $$ is equal to
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$$\displaystyle \frac { \pi }{ 2 } +1$$
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$$\displaystyle \frac { \pi }{ 2 } -1$$
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$$\displaystyle 1-\frac { \pi }{ 2 } $$
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none of these
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } =\int _{ 0 }^{ \pi /2 }{ \frac { a\cos { \theta } d\theta }{ a+a\cos { \theta } } } $$
By putting $$x=a\sin { \theta } \Rightarrow dx=a\cos { \theta } d\theta $$
$$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \frac { \cos { \theta } }{ 1+\cos { \theta } } d\theta } =\int _{ 0 }^{ \pi /2 }{ \left( 1-\frac { 1 }{ 1+\cos { \theta } } \right) d\theta } $$
$$\displaystyle =\int _{ 0 }^{ \pi /2 }{ d\theta } -\frac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \sec ^{ 2 }{ \frac { \theta }{ 2 } } d\theta } $$
$$\displaystyle ={ \left[ \theta -\tan { \frac { \theta }{ 2 } } \right] }_{ 0 }^{ \pi /2 }=\frac { \pi }{ 2 } -\tan { \frac { \pi }{ 4 } } =\left( \frac { \pi }{ 2 } -1 \right) $$
The value of the integral $$\int_0^{\dfrac{\pi}{2}}\,sin^5x\,dx$$ is
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$$\dfrac{4}{15}$$
0%
$$\dfrac{8}{5}$$
0%
$$\dfrac{8}{15}$$
0%
$$\dfrac{4}{5}$$
Explanation
$$I=\int_0^{\dfrac{\pi}{2}}sin^4x.sin\,x\,dx$$
$$=\int_0^{\dfrac{\pi}{2}}(1-cos^2\,x)^2\,sin\,x\,dx$$
Put $$cos\,x=t$$
$$\Rightarrow\;-sin\,x\,dx=dt$$
$$=-\int_1^0(1-t^2)^2dt=\int_0^(t^4-2t^2+1)dt$$
$$[\because\,-\int_a^bf(x)dx=\int_a^bf(x)dx]$$
$$=\dfrac{1}{5}(t^5)_0^1-\dfrac{2}{3}(t^3)_0^1+(t)_0^1$$
$$=\dfrac{1}{5}-\dfrac{2}{3}+1=\dfrac{3-10+15}{15}=\dfrac{8}{15}$$
Using Trapezoidal rule and following table $$\int_0^8 {f(x)}dx$$ is equal to
$$x$$
0
2
4
6
8
$$f(x)$$
2
5
10
17
26
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$$184$$
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$$92$$
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$$46$$
0%
$$-36$$
$$\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}$$ equals
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$$\dfrac {\pi}{3}$$
0%
$$\dfrac {2\pi}{3}$$
0%
$$\dfrac {\pi}{6}$$
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$$\dfrac {\pi}{12}$$
Explanation
$$\displaystyle \int \frac {d}{1+x^2}=\tan^{-1}x=F(x)$$
By second fundamental theorem of calculus, we obtain
$$\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}=F(\sqrt 3)-F(1)$$
$$=\tan^{-1}\sqrt 3-\tan^{-1}1$$
$$\displaystyle =\frac {\pi}{3}-\frac {\pi}{4}=\frac {\pi}{12}$$
$$\displaystyle\int _{ 1 }^{ 3 }{ \dfrac { \cos { \left( \log { x } \right) } }{ x } dx } $$ is equal to
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$$1$$
0%
$$\cos { \left( \log { 3 } \right) } $$
0%
$$\sin { \left( \log { 3 } \right) } $$
0%
$$\dfrac { \pi }{ 4 } $$
Explanation
$$\displaystyle \int_{1}^{3} \frac{cos(logx)}{x}dx $$
Substituting $$ log\,x = t $$
$$\displaystyle dt = \frac{1}{x}dx $$
$$\displaystyle \therefore \int_{t_{1}}^{t_{2}} cos\,t\,dt $$
Lim $$: t = log\,1 = 0 $$ to $$ t = log\,3 $$
$$\displaystyle \therefore \int_{0}^{log3} cos\,t\,dt \Rightarrow [sint]_{0}^{log3} = sin\,log3-0 $$
$$\displaystyle \therefore \int_{1}^{3} \frac{cos(log\,x)}{x}dx = sin\,log3 $$
The value of $$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \tan ^{ n+1 }{ x } \right) dx } +\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \tan ^{ n+1 }{ \left( \frac { x }{ 2 } \right) } \right) dx } $$ is
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0%
$$\cfrac{1}{n}$$
0%
$$\cfrac{n+2}{2n+1}$$
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$$\cfrac{2n-1}{n}$$
0%
$$\cfrac{2n-3}{3n-2}$$
Explanation
Given $$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \tan ^{ n+1 }{ x } \right) dx } +\cfrac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \tan ^{ n+1 }{ \left( \cfrac { x }{ 2 } \right) } \right) dx } $$
In second integral, put $$t=\cfrac{x}{2} \Rightarrow$$ $$dx=2dt$$
$$\Rightarrow$$ Also, when $$x=0$$ then $$t=0$$
When $$x=\dfrac{\pi}2$$ then $$t=\dfrac{\pi}4$$
Then $$I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } \right) } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ t } \right) } dt$$
$$I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } \right) } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) } dx\quad (\because \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(y)dy } )$$
$$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } +\tan ^{ n-1 }{ x } \right) } dx$$
$$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) .\left( \tan ^{ 2 }{ x+1 } \right) } dx$$
$$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) .\left( \sec ^{ 2 }{ x } \right) } dx$$
Put $$t=\tan{x}$$
$$\Rightarrow dt=\sec ^{ 2 }{ x } dx$$
Also when $$x=0$$ then $$t=0$$
when $$x=\pi /4$$, then $$t=1$$
$$I=\displaystyle \int _{ 0 }^{ 1 }{ { t }^{ n-1 }dt } $$
$$={ \left[ \cfrac { { t }^{ n } }{ n } \right] }_{ 0 }^{ 1 }=\cfrac { 1 }{ n } $$
What is $$\displaystyle \int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x } }{ 1+{ x }^{ 2 } } } dx$$ equal to?
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$$\cfrac { { \pi }^{ 2 } }{ 8 } $$
0%
$$\cfrac { { \pi }^{ 2 } }{ 32 } $$
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$$\cfrac { \pi }{ 4 } $$
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$$\cfrac { \pi }{ 8 } $$
Explanation
$$I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { \tan ^{ -1 }{ x } }{ 1+{ x }^{ 2 } } } dx$$
Put $$\tan ^{ -1 }{ x } =p\Rightarrow \dfrac { dx }{ 1+{ x }^{ 2 } } =dp$$
When $$x=0, p=0$$ and $$x=1,p=\dfrac{\pi}{4}$$
$$ \Rightarrow I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ pdp }=\dfrac{p^{2}}{2} =\dfrac { { \pi }^{ 2 } }{ 32} $$
Hence, B is correct.
What is $$\displaystyle\int _{ 1 }^{ 2 }{ \ln { x } dx } $$ equal to?
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$$\ln { 2 } $$
0%
$$1$$
0%
$$\ln { \left( \dfrac { 4 }{ e } \right) } $$
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$$\ln { \left( \dfrac { e }{ 4 } \right) } $$
Explanation
Let, $$y=\ln x$$
then, $$ x={ e^{ y } }$$
Now at, $$ x=1,y=0$$
$$ x=2,y=\ln 2$$
$$\Rightarrow dx={ e }^{ y }.dy$$
Putting the values, we get
$$\displaystyle \int_{1}^{2} \ln x dx=\int _{ 0 }^{ \ln2 }{ { e }^{ y }.y.dy }$$
$$ =\displaystyle { \left[ y.{ e }^{ y } \right] }_{ 0 }^{\ln2 }-\int _{ 0 }^{\ln2 }{ { e }^{ y }dy }$$
$$ =2\ln 2-1$$
$$=2\ln 2-\ln(e)$$
$$ =\ln\left(\dfrac { 4 }{ e }\right)$$
Hence, C is correct.
If $$\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}$$, then $$b = $$
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0%
$$\tan ^{-1}\, \left (\displaystyle \frac{1}{3} \right )$$
0%
$$\displaystyle \frac{\sqrt 3}{2}$$
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$$\sqrt 3$$
0%
$$1$$
Explanation
$$\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}$$
$$\Rightarrow [\tan^{-1}x]_0^b=[\tan^{-1}x]_b^{\infty}$$
$$\Rightarrow \tan^{-1}b-0=\dfrac{\pi}{2}-\tan^{-1}b$$
$$\Rightarrow \tan^{-1}b=\dfrac{\pi}{4}$$
$$\Rightarrow b=\tan\dfrac{\pi}{4}=1$$
Select and write the most appropriate answer from the given alternatives for question :
If $$\displaystyle \int^k_0 4x^3dx=16$$, then the value of $$k$$ is _____.
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0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Given:
$$\displaystyle \int^K_0 4x^3dx=16$$
$$4\left(\dfrac{x^4}{4}\right)^K_0=16$$
$$(x^4)^k_0=16$$
$$k^4=16$$
$$k^4=2^4$$
$$\boxed{k=2}$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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