Another interesting integral

Evaluate $\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + cos^4 x} \; dx$

If we substitute $u = \frac{\pi}{2} - x$, then we will get

I = -$\int_{\frac{\pi}{2}}^0 \frac{x \sin x \cos x}{\sin^4 x + cos^4 x} \; dx$

$= \int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2} - x\right) \sin x \cos x}{\sin^4 x + cos^4 x} \; dx$

$= \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + cos^4 x} \; dx - I$

or, $2I = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + cos^4 x} \; dx$

or, $I= \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin^4 x + cos^4 x} \; dx$

or, $I= \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x \sec^2 x}{\tan^4 x + 1} \; dx$

Substitute $v = \tan x$, and we have

$I= \frac{\pi}{4} \int_0^{\infty} \frac{v}{v^4 + 1} \; dv$

Substitute $w = v^2$, and we have

$I= \frac{\pi}{8} \int_0^{\infty} \frac{1}{w^2 + 1} \; dw$

$= \frac{\pi}{8} \left. \tan^{-1} w \right|_0^{\infty} \; dw$

$= \frac{\pi}{8} \cdot \frac{\pi}{2}$

$= \frac{\pi^2}{16}$