Question

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

Answer 1

Let $I=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\sin x\cos x}{{\cos }^{4}x}}{\frac{{\sin }^{4}x}{{\cos }^{4}x}+\frac{{\cos }^{4}x}{{\cos }^{4}x}}dx$

$=\underset{0}{\overset{\pi /2}{\int }}\frac{\frac{\sin x}{{\cos }^{3}x}}{{\tan }^{4}x+1}dx$

$=\underset{0}{\overset{\pi /2}{\int }}\frac{\tan x{\sec }^{2}x}{{\tan }^{4}x+1}dx$

Put ${\tan }^{2}x=u$, then, $2\tan x{\sec }^{2}xdx=du$.

$I=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\frac{du}{u^2+1}$

$=\frac{1}{2}{\left[{\tan }^{-1}u\right]}_0^{\pi /2}$

$=\frac{1}{2}{\left[{\tan }^{-1}\left({\tan }^{2}x\right)\right]}_0^{\pi /2}$

$=\frac{1}{2}\left[{\tan }^{-1}\left({\tan }^2\frac{\pi }{2}\right)-{\tan }^{-1}\left({\tan }^{2}0\right)\right]$

$=\frac{1}{2}\left[{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(0\right)\right]$

$=\frac{1}{2}\left[\frac{\pi }{2}-0\right]$

$=\frac{\pi }{4}$