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.
 A: Fourier series provide an alternative and simple way. Over $(0,\pi)$ we have
$$ \log\sin x = -\log 2-\sum_{m\geq 1}\frac{\cos(2m x)}{m} \tag{1}$$
pointwise and uniformly over any compact subset of $(0,\pi)$. Similarly, over $(0,\pi/2)$
$$ \log\cos x = -\log 2 + \sum_{n\geq 1}\frac{\cos(2n x)}{n}(-1)^{n+1} \tag{2} $$
holds pointwise and uniformly over compact subsets. Since for any $m,n\in\mathbb{N}^+$
$$ \int_{0}^{\pi/2}\cos(2n x)\cos(2mx)\,dx =\frac{\pi}{4}\delta(m,n)\tag{3}$$
we have
$$ \int_{0}^{\pi/2}\log(\sin\theta)\log(\cos\theta)\,d\theta = \frac{\pi}{2}\log^2 2-\frac{\pi}{4}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\color{red}{\frac{\pi}{2}\log^2 2-\frac{\pi^3}{48}}.\tag{4} $$

Yet another way is to consider that the integral is
$$ \lim_{\alpha,\beta\to 0^+}\frac{\partial^2}{\partial a\partial b}\int_{0}^{\pi/2}(\sin\theta)^{\alpha}(\cos\theta)^{\beta}\,d\theta $$
where the last integral is a value of the Beta function.
A: 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) $
A: Hint:
First, establish
$$\int_0^{\pi/2} \ln(2\sin x)\ln(2\cos x)dx
=-\frac12\int_0^{\pi/2} \ln^2(2\sin x)dx
$$
and the use the known $\int_0^{\pi/2} \ln^2(2\sin x)dx=\frac{\pi}{4}\zeta(2)$.
A: In fact
\begin{eqnarray}
&&\int_0^{\pi/2}\ln(\sin x)\ln(\cos x)dx\\
&=&\frac12\int_0^{\pi/2}\bigg[\ln(\sin x)+\ln(\cos x)\bigg]^2dx-\frac12\int_0^{\pi/2}\bigg[\ln^2(\sin x)+\ln^2(\cos x)\bigg]dx\\
&=&\frac12\int_0^{\pi/2}\ln^2(\sin x\cos x)dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\
&=&\frac12\int_0^{\pi/2}\ln^2(\frac12\sin 2x)dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\
&=&\frac14\int_0^{\pi}(\ln2-\ln\sin x)^2dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\
&=&\frac12\int_0^{\pi/2}(\ln2-\ln\sin x)^2dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\
&=&\frac12\bigg(\frac\pi2\ln^22-2\ln 2\int_0^{\pi/2}\ln\sin xdx+\int_0^{\pi/2}\ln^2\sin xdx\bigg)-\frac12\int_0^{\pi/2}\ln^2(\sin x)dx\\
&=&\frac{\pi}{2}\ln^2 2-\frac{\pi^3}{48}.
\end{eqnarray}
Here the famous results
$$\int_0^{\pi/2}\ln(\sin x)dx=-\frac\pi2\ln2,\int_0^{\pi/2}\ln^2(\sin x)dx=\frac{1}{24} \left(\pi^3 + 3\pi \ln^2(4)\right) $$
are used.
