Question

Evaluate the definite integral ${\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}xdx}{{\cos }^{2}x+4{\sin }^{2}x}$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{{\cos }^{2}x+4{\sin }^{2}x}dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{{\cos }^{2}x+4(1-{\cos }^{2}x)}dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{{\cos }^{2}x+4-4{\cos }^{2}x}dx$
$\Rightarrow I=\frac{-1}{3}{\int }_0^{\frac{\pi }{2}}\frac{4-3{\cos }^{2}x-4}{4-3{\cos }^{2}x}dx$
$\Rightarrow I=\frac{-1}{3}{\int }_0^{\frac{\pi }{2}}\frac{4-3{\cos }^{2}x}{4-3{\cos }^{2}x}dx+\frac{1}{3}{\int }_0^{\frac{\pi }{2}}\frac{4}{4-3{\cos }^{2}x}dx$
$\Rightarrow I=\frac{-1}{3}{\int }_0^{\frac{\pi }{2}}1dx+\frac{1}{3}{\int }_0^{\frac{\pi }{2}}\frac{4{\sec }^{2}x}{4{\sec }^{2}x-3}dx$
$\Rightarrow I=\frac{-1}{3}[x{]}_0^{\frac{\pi }{2}}+\frac{1}{3}{\int }_0^{\frac{\pi }{2}}\frac{4{\sec }^{2}x}{4(1+{\tan }^{2}x)-3}dx$
$\Rightarrow I=-\frac{\pi }{6}+\frac{2}{3}{\int }_0^{\frac{\pi }{2}}\frac{2{\sec }^{2}x}{1+4{\tan }^{2}x}dx$ .............. (1)
Consider, ${\int }_0^{\frac{\pi }{2}}\frac{2{\sec }^{2}x}{1+4{\tan }^{2}x}dx$
Let $2\tan x=t\Rightarrow 2{\sec }^{2}xdx=dt$
When $x=0,t=0$ and when $x=\frac{\pi }{2},t=\infty$
$\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{2{\sec }^{2}x}{1+4{\tan }^{2}x}dx={\int }_0^{\infty }\frac{dt}{1+t^2}$
$=[{\tan }^{-1}t{]}_0^{\infty }$
$=\left[{\tan }^{-1}\right(\infty )-{\tan }^{-1}(0\left)\right]=\frac{\pi }{2}$
Therefore, from (1), we obtain
$I=-\frac{\pi }{6}+\frac{2}{3}\left[\frac{\pi }{2}\right]=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$