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
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