1. We need to evaluate $I=\int_0^\pi \frac{8x\,dx}{4\cos^2 x+\sin^2 x}.$

2. First simplify the denominator: $4\cos^2 x+\sin^2 x=4\cos^2 x+(1-\cos^2 x)=1+3\cos^2 x.$ So, $I=\int_0^\pi \frac{8x}{1+3\cos^2 x}\,dx.$

3. Use the standard symmetry property: If $f(x)=f(\pi-x)$, then $\int_0^\pi x f(x)\,dx=\frac{\pi}{2}\int_0^\pi f(x)\,dx.$

Here, $f(x)=\frac{8}{1+3\cos^2 x}.$ Now check symmetry: f(\pi-x)=\frac{8}{1+3\cos^2(\pi-x)}=rac{8}{1+3\cos^2 x}=f(x), since $\cos(\pi-x)=-\cos x$ and hence $\cos^2(\pi-x)=\cos^2 x$.

Therefore, $I=\frac{\pi}{2}\int_0^\pi \frac{8}{1+3\cos^2 x}\,dx.$ So, $I=4\pi\int_0^\pi \frac{dx}{1+3\cos^2 x}.$

4. Again use symmetry about $\pi/2$: $\int_0^\pi \frac{dx}{1+3\cos^2 x}=2\int_0^{\pi/2} \frac{dx}{1+3\cos^2 x}.$ Hence, $I=8\pi\int_0^{\pi/2} \frac{dx}{1+3\cos^2 x}.$

5. Evaluate $J=\int_0^{\pi/2} \frac{dx}{1+3\cos^2 x}.$ Use $t=\tan x$. Then $dx=\frac{dt}{1+t^2}, \qquad \cos^2 x=\frac{1}{1+t^2}.$ Thus, $1+3\cos^2 x=1+\frac{3}{1+t^2}=\frac{t^2+4}{1+t^2}.$ So,