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 yet another partial answer, and a verification of some other claims.
Using $(4)$ and $(8)$ from this answer, we get
$$
\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=\frac\pi2\log(2)^2-\frac{\pi^3}{48}\tag{1}
$$
Here is a way to extend kalpeshmpopat's suggestion about substituting $x\mapsto\frac\pi2-x$. Note that $g(x)=f(\sin(x))f(\cos(x))$ is even as a function of $x-\frac\pi4$; that is, $g(\frac\pi2-x)=g(x)$. Thus, if we multiply by an odd function of $x-\frac\pi4$, the integral over $[0,\frac\pi2]$ will be $0$. 
Therefore,
$$
\int_0^{\pi/2}\left(\frac\pi4-x\right)\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=0\tag{2}
$$
Using $(1)$ and $(2)$, we get
$$
\begin{align}
\int_0^{\pi/2}x\log(\sin(x))\log(\cos(x))\,\mathrm{d}x
&=\frac\pi4\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&=\frac\pi4\left(\frac\pi2\log(2)^2-\frac{\pi^3}{48}\right)\\
&=\frac{\pi^2}{8}\log(2)^2-\frac{\pi^4}{192}\tag{3}
\end{align}
$$
We also have
$$
\int_0^{\pi/2}\left(\frac\pi4-x\right)^3\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=0\tag{4}
$$
Which, along with $(1)$ and $(3)$, implies that
$$
\begin{align}
\int_0^{\pi/2}x^3\log(\sin(x))\log(\cos(x))\,\mathrm{d}x
&=\frac{3\pi}{4}\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&-\frac{3\pi^2}{16}\int_0^{\pi/2}x\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&+\frac{\pi^3}{64}\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&=\frac{3\pi}{4}\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&-\frac{\pi^4}{64}\log(2)^2+\frac{\pi^6}{1536}\tag{5}
\end{align}
$$
Equation $(5)$ supports math110's claim that if we know $I_2$, we know $I_3$.
