How to evaluate $\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$ Find the value of
$$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$$
We have the information that
$$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{\pi^2}{8}\ln^2(2)-\dfrac{\pi^4}{192}$$

This is not quite a complete answer but goes a good way towards showing that the idea of @kalpeshmpopat is not so far off the mark - if we want to answer the question that was orginally asked.
First, numerical investigation indicates that the correct integral is
$$I=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)dx=\dfrac{(\pi\ln{2})^2}{8}-\dfrac{\pi^4}{192}.$$
Now, as @kalpeshmpopat points out, a simple substitution, together with the facts that $\cos(\frac{\pi}{2}-x)=\sin(x)$ and vice-versa, shows that
$$\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)dx=\int_0^{\pi/2}(\frac{\pi}{2}-x)\ln(\sin x)\ln(\cos x)dx.$$
Thus, if we add these two together we get
$$\displaystyle\int_0^{\pi/2} \frac{\pi}{2} \ln(\sin x)\ln(\cos x)dx=2I.$$
All that remains to show is that
$$\displaystyle\int_0^{\pi/2} \frac{\pi}{2} \ln(\sin x)\ln(\cos x)dx =
\frac{1}{96} \pi ^2 \left(6 \log ^2(4)-\pi^2\right),$$
which Mathematica can do.  It's getting late, but my guess on this last integral would be to expand $\ln(\cos(x))$ into a power series (which is easy, since we know $\ln(1+y)$) and try to integrate $x^n \ln(\sin(x))$.
