Let $I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{\beta}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\pi / 3$. Then, $3 I^2=$