Let

$I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{2-\cos 2x}\left(\frac{3}{\pi }+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right)dx$

$\Rightarrow I=\frac{3}{\pi }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{\left(2-\cos 2x\right)}dx+{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\log \left(\frac{4+\sin x}{4-\sin x}\right)\frac{1}{\left(2-\cos 2x\right)}dx$

$\Rightarrow I=\frac{3}{\pi }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{\left(2-\cos 2x\right)}dx+0$

Since, $\log \left(\frac{4+\sin x}{4-\sin x}\right)\frac{1}{\left(2-\cos 2x\right)}$ is an odd function and ${\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{l}2{\int }_0^{a}f\left(x\right)dx \mathrm{if} f\left(-x\right)=f\left(x\right)\\ 0 \mathrm{if} f\left(-x\right)=-f\left(x\right)\end{array}\right.$.

$\Rightarrow I=\frac{6}{\pi }{\int }_0^{\frac{\pi }{4}}\frac{1}{\left(2-\cos 2x\right)}dx$

$\Rightarrow I=\frac{6}{\pi }{\int }_0^{\frac{\pi }{4}}\frac{1}{\left(2-\left({\displaystyle \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}\right)\right)}dx$

$\Rightarrow I=\frac{6}{\pi }{\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}x}{\left({\displaystyle 1+3{\tan }^{2}x}\right)}dx$

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$\Rightarrow I=\frac{6}{\pi }{\int }_0^1\frac{1}{\left({\displaystyle 1+3t^2}\right)}dt$

$\Rightarrow I=\frac{2}{\pi }{\int }_0^1\frac{1}{\left({\displaystyle \frac{{\displaystyle 1}}{3}+t^2}\right)}dt$

$\Rightarrow I=\frac{2\sqrt{3}}{\pi }{\left[\tan \left(\sqrt{3}t\right)\right]}_0^1$

$\Rightarrow I=\frac{2\sqrt{3}}{\pi }\left[\tan \left(\sqrt{3}\right)-0\right]$

$\Rightarrow I=\frac{2\sqrt{3}}{\pi }\times \frac{\mathrm{\pi }}{3}$

$\Rightarrow I=\frac{2}{\sqrt{3}}$

$\Rightarrow 3I^2=4$