$\begin{aligned}
& I=\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x} \
& I=\int_0^\pi \frac{8(\pi-x) d x}{4 \cos ^2 x+\sin ^2 x}
\end{aligned}$
2I=8π∫0π4cos2x+sin2xdx
2I=8π×2∫0π/24+tan2xsec2xdx
$\begin{aligned}
& \mathrm{I}=8 \pi \int_0^{\infty} \frac{\mathrm{dt}}{4+\mathrm{t}^2}=8 \pi \times \frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^{\infty} \
& =4 \pi \times \frac{\pi}{2}=2 \pi^2
\end{aligned}$
option (1)