Mathematics

# Solve:$\displaystyle\int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^2}xdx}}{{{{\cos }^2}x + 4{{\sin }^2}x}}}$

##### SOLUTION
$I=\displaystyle \int_{0}^{\pi/2}{\dfrac{\cos^2x dx}{\cos^2x+4\sin^2 x}}=\displaystyle \int_{0}^{\pi/2}{\dfrac{dx}{1+4\tan^2x}}$
put $\tan x=t\Rightarrow \sec^2 xdx=dt$
$\Rightarrow dx=\dfrac{dt}{(1+t^2)}$
$I=\displaystyle \int_{0}^{\infty}{\dfrac{dt}{(1+t^2)(1+4t^2)}}$
$\dfrac{1}{(1+t^2)(1+4t^2)}=\dfrac{A}{(1+t^2)}+\dfrac{B}{(1+4t^2)}$
comparing coefficients :- $A+B=1$
$4A+B=0\Rightarrow B=-4A$
$A-4A=1$ as $A=-1/3\Rightarrow B=4/3$
$\therefore I=\displaystyle \int_{0}^{\infty}{dt\left[\dfrac{4}{3}\times \dfrac{1}{(1+4t^2)}-\dfrac{1}{3(1+t^2)}\right]dt}$
$I=\displaystyle \frac { 4 }{ 3 } \int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+4t^{ 2 }) } dt } --\frac { 1 }{ 3 } \int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+t^{ 2 }) } dt } } { \left. \frac { 1 }{ 2 } \times \frac { 4 }{ 3 } \tan ^{ -1 } (2t) \right] }_{ 0 }^{ \infty }-{ \left. \frac { 1 }{ 3 } \tan ^{ -1 }{ (t) } \right] }_{ 0 }^{ \infty }$
$I=\dfrac{2}{3}\left[\tan^{-1} \infty -\tan^{-1}(0)\right]-\dfrac{1}{3}\left[\tan^{-1}\infty -\tan^{-1}(0)\right]$
$I=\dfrac{1}{3}\left[\pi/2-0\right]=\dfrac{\pi}{6}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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lf $f(x)$ is a polynomial satisfying$f(x)f(\frac{1}{x})=f(x) +f(\frac{1}{x})$, and$f(3)=82$, then $\displaystyle \int\frac{f(x)}{x^{2}+1}dx=$
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