Let $I=\int_{0}^{\pi} \frac{x \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \quad \text{...(i)}$
$\Rightarrow \quad I=\int_{0}^{\pi} \frac{(\pi-x) \sin ^{2 n}(\pi-x)}{\sin ^{2 n}(\pi-x)+\cos ^{2 n}(\pi-x)} d x$
$$
=\int_{0}^{\pi} \frac{(\pi-x) \sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \quad \text{...(ii)}
$$
Adding Eqs. (i) and (ii), we get
$$
2 I=\pi \int_{0}^{\pi} \frac{\sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \quad \text{...(iii)}
$$
Let $I_{1}=\int_{0}^{\pi} \frac{\sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x$
$$
\Rightarrow \quad I_{1}=2 \int_{0}^{\pi / 2} \frac{\sin ^{2 n} x}{\sin ^{2 n} x+\cos ^{2 n} x} d x \quad \text{...(iv)}
$$
$$
\Rightarrow I_{1}=2 \int_{0}^{\pi / 2} \frac{\sin ^{2 n}\left(\frac{\pi}{2}-x\right)}{\sin ^{2 n}\left(\frac{\pi}{2}-x\right)+\cos ^{2 n}\left(\frac{\pi}{2}-x\right)} d x
$$
$$
=2 \int_{0}^{\pi / 2} \frac{\cos ^{2 n} x}{\cos ^{2 n} x+\sin ^{2 n} x} d x \quad \text{...(v)}
$$
Adding Eqs. (iv) and (v), we get
$$
\begin{aligned}
&2 I_{1}=2 \int_{0}^{\pi / 2} 1 d x=2 \cdot \frac{\pi}{2}=\pi \\
&\Rightarrow \quad I_{1}=\frac{\pi}{2}
\end{aligned}
$$
Substituting the value of $I_{1}$ in $\mathrm{Eq}$. (iii), we get
$$
2 I=(\pi)\left(\frac{\pi}{2}\right) \Rightarrow I=\frac{\pi^{2}}{4}
$$