If $$I_1=\int_{0}^{\frac{\pi}{2}}\frac{(\ln(\tan x))^2}{1-\sin 2x}dx$$ and $$I_2=\int_{0}^{\frac{\pi}{2}}(\ln(1-\sin x))(\cot x)dx$$ then evaluate $$\left|\frac{I_1}{I_2}\right|$$
My Attempt
$$I_1=\int_{0}^{\frac{\pi}{2}}\frac{(\ln(\tan x))^2}{1-\sin 2x}dx=2\int_{0}^{\frac{\pi}{4}}\frac{(\ln(\tan x))^2}{1-\sin 2x}dx$$Taking $t=\tan x$, the integral transforms to$$I_1=2\int_{0}^{1}\frac{(\ln t)^2}{(1-t)^2}dt$$Putting $t=1-\sin x$ in $I_2$ we have $$I_2=\int_{0}^{1}\frac{\ln t}{1-t}dt$$
What to do after this