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
$$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{\pi^2}{8}\ln^2(2)-\dfrac{\pi^4}{192}$$

Presented below is a complete solution that evaluates the integral to the following close-form
\begin{align}
&\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ dx\\
=& -\frac\pi2 \text{Li}_4(\frac12)-\frac{3\pi}8\ln2\ \zeta(3)
+\frac{\pi^5}{320}+\frac{\pi^3}{16}\ln^22-\frac\pi{48}\ln^42\tag1
\end{align}
To derive (1), evaluate the integrals below sequentially, with $\int_0^{\pi/2}\ln^2(2\sin x)dx=\frac{\pi^3}{24}$ and $\int_0^{\pi/2}x^2 \ln^2(2\cos x) dx=\frac{11\pi^5}{1440}$
\begin{align}
&\int_0^{\pi/2}x^2
\overset{x\to\frac\pi2-x}{\ln^2(2\sin x)}\ dx 
 =\pi\int_0^{\pi/2}x
\ln^2(2\sin x)\ dx-\frac{\pi^5}{360}\\
\\
&\int_0^{\pi/2}x^2 
\ln^2(2\sin 2x)\overset{2x\to x}{dx} \\
 =& \ \frac18\int_0^{\pi/2}(\pi^2-2\pi x+2x^2 )
\ln^2(2\sin x)\ dx=\frac{13\pi^5}{2880}\\
\\
&\int_0^{\pi/2}x^2\ln(2\sin x)\ln(2\cos x)\ dx\\
=& \ \frac12 \int_0^{\pi/2}[x^2 
\ln^2(2\sin 2x)-x^2 \ln^2(2\sin x)-x^2 \ln^2(2\cos x)]\ dx\\
=& - \frac{\pi^5}{5760}-\frac\pi2 \int_0^{\pi/2}x
\ln^2(2\sin x)\ dx\
\end{align}
Writing out both sides of the last integral and utilizing below to yield the close-form (1)
\begin{align}
\int_0^{\pi/2}x\ln(2\sin x)dx=\frac7{16}\zeta(3),\>\>\>
\int_0^{\pi/2}x^2\ln(2\sin x)dx=\frac{3\pi}{16}\zeta(3)\\
\end{align}
$$\int_0^{\pi/2}x\ln^2(\sin x)dx 
=  \text{Li}_4(\frac12)
-\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42
$$
