Question

$I={\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$
$\therefore I={\pi }^2/k.$
what is k?

Answer 1

Let $I={\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$ ...(1)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
We have $I={\int }_0^{\pi /2}\frac{(\pi /2-x)\cos x\sin xdx}{{\sin }^{4}x+{\cos }^{4}x}.$
$2I=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx.$
$=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\tan x{\sec }^{2}xdx}{1+{\tan }^{4}x}=\frac{\pi }{2}.\frac{1}{2}{\int }_0^{\infty }\frac{dt}{1+t^2},$
putting ${\tan }^{2}x=t$
$2I=\frac{\pi }{4}{\left[{\tan }^{-1}t\right]}_0^{\infty }=\frac{\pi }{4}[\frac{\pi }{2}-0]=\frac{{\pi }^2}{8}.$
$\therefore I={\pi }^2/16.$