For $n\in N,$$\underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx=$

$$\begin{gathered} \underset{0}{\overset{2\pi }{\int }}\frac{x{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx \\ =\frac{2\pi }{2}\underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx \\ \left[\underset{0}{\overset{a}{\int }}xf\left(x\right)dx=\frac{a}{2}\underset{0}{\overset{a}{\int }}f\left(x\right)dx if f(a-x)=f\left(x\right)\right] \\ and here f\left(x\right)=\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x} \\ f(2\pi -x)=\frac{{\sin }^{2n}\left(2\pi -x\right)}{{\sin }^{2n}\left(2\pi -x\right)+{\cos }^{2n}\left(2\pi -x\right)} \\ =\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}=f\left(x\right) \\ I=\pi \underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x} \\ =\pi 2\underset{0}{\overset{\pi }{\int }}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx \\ \left[\underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx\right] \\ =\pi \left(2\right).2\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{2n}x}{{\sin }^{2n}x+{\cos }^{2n}x}dx \\ =\left(4\pi \right)\left(\frac{\pi }{4}\right) \\ I={\pi }^2 \end{gathered}$$