1. Let $I=\int_0^{2\pi}\frac{x\sin^8 x}{\sin^8 x+\cos^8 x}\,dx.$

We use the standard symmetry property for integrals over $[0,a]$: $\int_0^a x f(x)\,dx=\int_0^a (a-x)f(a-x)\,dx.$

Here, take $a=2\pi$ and $f(x)=\frac{\sin^8 x}{\sin^8 x+\cos^8 x}.$

Then $I=\int_0^{2\pi}(2\pi-x)\frac{\sin^8(2\pi-x)}{\sin^8(2\pi-x)+\cos^8(2\pi-x)}dx.$

2. Simplify the trigonometric terms: $\sin(2\pi-x)=-\sin x \implies \sin^8(2\pi-x)=\sin^8 x,$ $\cos(2\pi-x)=\cos x \implies \cos^8(2\pi-x)=\cos^8 x.$

So, $I=\int_0^{2\pi}(2\pi-x)\frac{\sin^8 x}{\sin^8 x+\cos^8 x}\,dx.$

3. Add the two expressions for $I$: $2I=\int_0^{2\pi}\left[x+(2\pi-x)\right]\frac{\sin^8 x}{\sin^8 x+\cos^8 x}dx$ $=2\pi\int_0^{2\pi}\frac{\sin^8 x}{\sin^8 x+\cos^8 x}dx.$

Hence, $I=\pi\int_0^{2\pi}\frac{\sin^8 x}{\sin^8 x+\cos^8 x}dx.$

4. Now evaluate $J=\int_0^{2\pi}\frac{\sin^8 x}{\sin^8 x+\cos^8 x}dx.$

Use the substitution $x\mapsto \frac{\pi}{2}-x$. Then over any full period, $\frac{\sin^8 x}{\sin^8 x+\cos^8 x}+\frac{\cos^8 x}{\sin^8 x+\cos^8 x}=1.$

So if $J=\int_0^{2\pi}\frac{\sin^8 x}{\sin^8 x+\cos^8 x}dx,$ then also $J=\int_0^{2\pi}\frac{\cos^8 x}{\sin^8 x+\cos^8 x}dx.$

Adding, $2J=\int_0^{2\pi}1\,dx=2\pi,$ which gives $J=\pi.$

5. Therefore, $I=\pi J=\pi\cdot \pi=\pi^2.$

So the value of the integral is $\boxed{\pi^2}.$

6. Comparing with the options:

• A: $4\pi$
• B: $2\pi$
• C: $\pi^2$
• D: $2\pi^2$

Thus the correct option is C.