How to show that $\int_{0}^{\pi/4}{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx={\ln(2^2)-\ln^2(2)\over 8}?$ Given that:

$$\int_{0}^{\pi/4}{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx={\ln(2^2)-\ln^2(2)\over 8}\tag1$$

I am not sure how to go about to begin tackling this problem of proving $(1)$. 
 A: Let $$I=\int{\ln(\cos x)\ln(\sin x)\over \sin(2x)\tan(2x)}\mathrm dx\tag1$$  we have
$$I=\int \frac{\ln(\cos x)\ln(\sin x) \cos(2x)\:dx}{\sin^2(2x)}$$  Using Integration by Parts taking $u=\ln(\cos x)\ln(\sin x) $  and $v=\frac{\cos (2x)}{\sin^2(2x)}$ 
Now  we have $$\int v dx=\frac{-1}{2 \sin(2x)}$$
$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}+\int \left(\frac{d}{dx}(\ln(\cos x)\ln(\sin x))\right) \times \frac{-1}{2 \sin(2x)}\:dx$$
so
$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)} -\frac{1}{2}\int \frac{\ln(\sin x)\tan x \:dx}{\sin(2x)}+\frac{1}{2}\int\frac{\ln(\cos x)\cot x \:dx}{\sin(2x)}$$  hence
$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}-\frac{1}{4}\int \ln(\sin x)\sec^2 x \: dx+\frac{1}{4}\int \ln(\cos x)\csc^2 x\:dx$$
But
$$\int\ln(\sin x)\sec^2 x \:dx=\ln(\sin x)\tan x-x+C$$  and
$$ \int\ln(\cos x)\csc^2 x \:dx=-\ln(\cos x)\cot x-x+C$$ so Finally
$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}-\frac{1}{4}\left(\ln(\sin x)\tan x-x\right)+\frac{1}{4}\left(-\ln(\cos x)\cot x-x\right)$$  So
$$I=\frac{-\ln(\sin x)\ln(\cos x)}{2\sin(2x)}-\frac{1}{4}\left(\ln(\sin x)\tan x+\ln(\cos x)\cot x \right)$$
Now apply limits from $0$ to $\frac{\pi}{4}$ and using $$\lim_{x \to 0^+}x \:\ln x=0$$  we get
$$I=\frac{\ln (2^2)-\ln^2 (2)}{8}$$
