The value of

$\int_{0}^{2 \pi} \frac{x \sin ^{8} x}{\sin ^{8} x+\cos ^{8} x} d x$

$=\int_{0}^{\pi}\left[\frac{x \sin ^{8} x}{\sin ^{8} x+\cos ^{8} x}+\frac{(2 \pi-x) \sin ^{8} x}{\sin ^{8} x+\cos ^{8} x}\right] d x$

$\left[\because \int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x\right]$

$=\int_{0}^{\pi} \frac{2 \pi \sin ^{8} x}{\sin ^{8} x+\cos ^{8} x} d x$

$=2 \pi \int_{0}^{\pi / 2}\left[\frac{\sin ^{8} x}{\sin ^{8} x+\cos ^{8} x}+\frac{\cos ^{8} x}{\sin ^{8} x+\cos ^{8} x}\right] d x$

$=2 \pi \int_{0}^{\pi / 2} 1 d x=2 \pi \times \frac{\pi}{2}=\pi^{2}$