Question

Evaluate ${\int }_0^{\pi /2}\frac{{\cos }^{2}x}{{\cos }^{2}x+4{\sin }^{2}x}dx$

Answer 1

Let $I={\int }_0^{\pi /2}\frac{{\cos }^{2}x}{{\cos }^{2}x+4{\sin }^{2}x}dx$

$={\int }_0^{\pi /2}\frac{{\cos }^{2}a}{4-3{\cos }^{2}x}dx=-\frac{1}{3}{\int }_0^{\pi /2}\frac{4-3{\cos }^{2}x-4}{4-3{\cos }^{2}x}dx$

$=-\frac{1}{3}{\int }_0^{\pi /2}(1-\frac{4}{4-{\cos }^{2}x})dx=-\frac{1}{3}[x{]}_0^{\pi /2}+\frac{4}{3}{\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{4{\sec }^2\frac{x}{3}}$

$[\tan x=z\Rightarrow {\sec }^{2}xdx=dz,\because x=\frac{\pi }{2}\Rightarrow z=\infty andx=0\Rightarrow z=0]$

$\therefore I=-\frac{1}{3}\left[\frac{\pi }{2}\right]+\frac{4}{3}{\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{4(1+{\tan }^{2}x)-3}$

$=-\frac{\pi }{6}+\frac{4}{3}{\int }_0^{\infty }\frac{dz}{4+4z^2-3}=-\frac{\pi }{6}+\frac{4}{3\times 4}{\int }_0^{\infty }\frac{dz}{z^2+{\left(\frac{1}{2}\right)}^2}$

$=-\frac{\pi }{6}+\frac{1}{3}.2{\left[{\tan }^{-1}\frac{1}{\frac{1}{2}}\right]}_0^{\infty }=-\frac{\pi }{6}+\frac{1}{3}\times [{\tan }^{-1}2z{]}_0^{\infty }$

$=-\frac{\pi }{6}+\frac{2}{3}[{\tan }^{-1}\infty -{\tan }^{-1}0]=-\frac{\pi }{6}+\frac{2}{3}[\frac{x}{2}-0]$

$=-\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{6}$ Ans