Let
$$
\begin{aligned}
& I=\int_0^{\pi / 2} \frac{16 x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =\int_0^{\pi / 2} \frac{16\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x \\
& =\int_0^{\pi / 2} \frac{(8 \pi-16 x) \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
& 2 I=8 \pi \int_0^{\pi / 2} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =4 \pi \int_0^{\pi / 2} \frac{2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& \therefore \quad I=2 \pi \int_0^{\pi / 2} \frac{2 \tan x \sec ^2 x}{\tan ^4 x+1} d x \\
& =2 \pi \int_0^{\pi / 2} d\left\{\tan ^{-1}\left(\tan ^2 x\right)\right\} \\
& =2 \pi\left[\tan ^{-1}\left(\tan ^2 x\right)\right]_0^{\pi / 2}=2 \pi\left[\frac{\pi}{2}-0\right] \\
&
\end{aligned}
$$

$$
\therefore \quad I=\pi^2
$$