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

I'm still struggling with this integral, but I guess the following result may have a chance to be helpful:
\begin{align*}
\int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx
&= \frac{11 \pi^5}{1440} + \frac{\pi^3}{24} \log^2 2 + \frac{\pi}{2}\zeta(3) \log 2 \tag{1} \\
&\approx 4.2671523609840988652 \cdots.
\end{align*}
To prove this, let us consider the following identity
$$ \int_{0}^{\frac{\pi}{2}} \cos^{z}x \cos wx \, dx = \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}}.$$
You can find the proof of this identity at here. Thus it follows that
$$ \int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx = - \left. \frac{\partial^4}{\partial z^2 \partial w^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)}. $$
Performing a bunch of calculations, we obtain $(1)$. Similar idea shows that
$$ \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx = \left. \frac{\partial^2}{\partial z^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)} = \frac{\pi^3}{24} + \frac{\pi}{2}\log 2. \tag{2} $$

Indeed, starting from the identity
$$ \log^2 \left( \frac{\sin 2x}{2} \right) = \log^2 \cos x + \log^2 \sin x + 2\log \cos x \log \sin x, $$
I obtained
\begin{align*}I
&= -\frac{7}{8}\int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx
   + \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx
   - \frac{3\pi^2}{32} \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx \\
&\quad -\frac{\log 2}{8}\int_{0}^{\pi} x^2 \log \sin x \, dx
   + \frac{\pi^3}{48} \log^2 2 \\
&\approx 0.077821979372293864338\cdots.
\end{align*}
From the identity
$$ \log \sin x = -\log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nx}{n}, $$
we obtain
$$\int_{0}^{\pi} x^2 \log \sin x \, dx = -\frac{\pi}{2} \zeta (3) - \frac{\pi^3}{3} \log 2.  \tag{3}$$
Putting $(1)$, $(2)$ and $(3)$ together, I was able to derive
\begin{align*}I
&= -\frac{61 \pi^5}{5760} - \frac{3\pi}{8} \zeta (3) \log 2 -\frac{\pi^3}{48} \log^2 2
   + \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx.
\end{align*}
I'm not sure if this formula will be helpful, since the last remaining integral seems to defy my techniques.
