$\underset{0}{\overset{\pi /2}{\int }}\frac{16\sin x.\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx=$

$$\begin{gathered} \begin{array}{l}{\int }_0^{\pi /4} \frac{16\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx\\ =16{\int }_0^{\pi /2} \frac{\sin x\cos x}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x}dx\\ =16{\int }_0^{\pi /2} \frac{\sin x\cos x}{1-\frac{1}{2}2{\sin }^{2}x{\cos }^{2}x}dx\\ =16{\int }_0^{\pi /2} \frac{\sin x\cos x}{1-\frac{{\sin }^{2}2x}{2}}\\ =16{\int }_0^{\pi /2} \frac{2\sin x\cos x}{2-{\sin }^{2}2x}dx\\ =16{\int }_0^{\pi /2} \frac{\sin 2x}{1+{\cos }^{2}2x}dx\\ \text{ Let }\mathrm{t}=\cos 2\mathrm{x}\\ \mathrm{dt}=-2\sin 2xdt\\ =-\frac{16}{2}{\int }_1^{-1} \frac{dt}{1+t^2}=2\frac{16}{2}{\int }_0^1 \frac{dt}{1+t^2} \\ =16\left[{\tan }^{-1}{t}_0^1\right.\\ =16\left[{\tan }^{-1}1-{\tan }^{-1}0\right]=16\left[\frac{\pi }{4}-0\right]\\ =4\mathrm{\pi }\\ \end{array} \end{gathered}$$