${\int }_0^{\frac{\pi }{2}} \frac{x\tan x{\sec }^{2}x}{{\tan }^{4}x}dx=$

$I={\int }_0^{\frac{\pi }{2}} \frac{x\tan x{\sec }^{2}x}{{\tan }^{4}x}dx$
$I={\int }_0^{\frac{\pi }{2}} \frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\to \left(1\right)$
By (a-x) property
$\begin{array}{l}I={\int }_0^{\frac{\pi }{2}} \frac{\left(\frac{\pi }{2}-x\right)\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\to \left(2\right)\\ \left(1\right)+\left(2\right)\Rightarrow 2I={\int }_0^{\frac{\pi }{2}} \frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\\ Put{\sin }^{2}x=t\Rightarrow 2\sin x\cos xdx=dt\\ x=0\Rightarrow t=0\mathrm{\&}x=\frac{\pi }{2}\Rightarrow t=1\\ I=\frac{\pi }{8}{\int }_0^1 \frac{1}{2t^2-2t+1}dt\\ \Rightarrow I=\frac{\pi }{8}{\int }_0^1 \frac{1}{{\left(t-\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}dt\\ \Rightarrow I={\left[\frac{\pi }{8}{\tan }^{-1}(2t-1)\right]}_0^1=\frac{{\pi }^2}{16}\end{array}$