How to compute $\int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x$? By my post, I had found that
$$
\begin{aligned}
\frac{\pi^{3}}{8}&= \int_{0}^{\infty} \frac{\ln ^{2} x}{x^{2}+1} d x \\&\stackrel{x\mapsto\tan x}{=} \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\tan x) d x \\
&= \int_{0}^{\frac{\pi}{2}}[\ln (\sin x)-\ln (\cos x)]^{2} d x \\
&= \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\sin x) d x-2 \int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x +\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\cos x) d x \\
&=2\left[\frac{\pi^{2}}{24}\left(\pi^{2}+3 \ln ^{2} 4\right)\right]-2 I,
\end{aligned}
$$
the last line comes from the post.
Rearranging yields the result
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x \\
&=\frac{1}{2}\left(\frac{\pi^{3}}{12}+\pi \ln^{2} 2-\frac{\pi^{3}}{8}\right) \\
&=-\frac{\pi^{3}}{48}+\frac{\pi}{2} \ln ^{2} 2
\end{aligned}
$$
Is there any other solution? Comments and solutions are highly appreciated.

If you are familiar with beta function and feymman's trick then
$$ B(x,y) = 2\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \cos^{2y-1}(t) \ dt $$
$$ \frac{\partial }{\partial x} B(x,y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t)\ln(\sin t) \cos^{2y-1}(t) \ dt$$
$$ \frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} B(x,y) \right) = 8\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \ln(\sin t ) \ln(\cos t) \cos^{2y-1}(t) \ dt $$
and we know that  $$ B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$
so differentiate and put $ x =\frac{1}{2} , y = \frac{1}{2} $
$$\frac{\partial^2 B}{\partial x\partial y}=(\psi(x)-\psi(x+y))(\psi(y)-\psi(x+y))B(x,y)+B(x,y)(-\psi'(x+y)))$$
$$ \psi \left(\frac{1}{2} \right) = -2\ln 2 - \gamma $$
$$ \psi(1) = -\gamma $$
$$ \psi^{(1)}(1) = \frac{\pi^2}{6} $$
$$ B\left( \frac{1}{2} , \frac{1}{2} \right) = \frac{\Gamma \left( \frac{1}{2} \right)^2}{\Gamma(1)} = \pi $$
thus you'll have the integral $ = \frac{\pi}{8} \left(4\ln(2)^2 - \frac{\pi^2}{6} \right) $
