Applying $\left(a+b-x\right)$ and adding, we get,

 $\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{dx}{1+{\tan }^{8}x}$
$\Rightarrow 2I=\pi {\int }_0^{\pi /2}\left(\frac{1}{1+{\tan }^{8}x}+\frac{1}{1+{\tan }^8\left(\pi -x\right)}\right)dx$
$\Rightarrow 2I=2\pi {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{dx}{1+{\tan }^{8}x}=2\pi {\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{8}x}{{\cos }^{8}x+{\sin }^{8}x}$

$\Rightarrow I=\pi {\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{8}x}{{\cos }^{8}x+{\sin }^{8}x}$
Applying $\left(a+b-x\right)$ and adding,

$2I=\pi {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{8}x+{\cos }^{8}x}{{\sin }^{8}x+{\cos }^{8}x}dx$
$\Rightarrow 2I=\pi {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}1dx=\pi \left(\frac{\pi }{2}\right)$
$\Rightarrow I=\frac{{\pi }^2}{4}\Rightarrow \left[I\right]=2$