Twist on log of sine and cosine integral $\int_{0}^{\frac{\pi}{2}}x\ln(\sin x)\ln(\cos x)dx$ I ran across this integral and have not been able to evaluate it.
$$\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\ln(\cos(x))dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$$
I had some ideas.  Perhaps some how arrive at $\displaystyle\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{4}}=\frac{{\pi}^{4}}{192}$.
and $\displaystyle \ln(2)\int_{0}^{\frac{\pi}{2}}x\ln(2)dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}$
by using the identity $\displaystyle\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}=-x\ln(\sin(x))-x\ln(2)$
and/or $\displaystyle \ln(\cos(x))=-\ln(2)-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}$
I have used the first one to evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$,  so I thought perhaps it could be used in some manner here.
I see some familiar things in the solution, but how to get there?.
Does anyone have any clever ideas?.
Thanks.

Here is an approach that avoids using the Fourier series expansions for $\ln (\cos x)$ and $\ln (\sin x)$.
Consider
$$I = \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx.$$
Enforcing a substitution of $x \mapsto \pi/2 - x$ gives
\begin{align*}
I &= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln \left [\sin \left (\frac{\pi}{2} - x \right ) \right ] \ln \left [\cos \left (\frac{\pi}{2} - x \right ) \right ] \, dx\\
&= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln (\cos x) \ln (\sin x) \, dx\\
&= \frac{\pi^2}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx - \pi \int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx\\
& \qquad + \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx,
\end{align*}
or
$$\int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx.$$
The integral appearing on the right can be found by differentiating the beta function. As
$$\text{B}(x,y) = 2 \int_0^{\pi/2} \cos^{2x - 1} \theta \sin^{2y - 1} \theta \, d\theta,$$
we see that
$$\int_0^{\pi/2} \ln (\sin \theta) \ln (\cos \theta) \, d\theta = \frac{1}{8} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2},$$
giving
$$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{32} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2}.$$
Since
$$\partial_x \text{B}(x,y) = \text{B}(x,y) [\psi(x) - \psi (x + y)] \quad \text{and} \quad \partial_y \text{B}(x,y) = \text{B}(x,y) [\psi(y) - \psi (x + y)],$$
where $\psi (z)$ is the digamma function, we have
$$\partial_x \partial_y \text{B}(x,y) = \text{B}(x,y) \left [ \left \{\psi(x) - \psi (x + y) \right \} \left \{\psi(y) - \psi (x + y) \right \} - \psi^{(1)}(x + y) \right ].$$
Thus
\begin{align*}
\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx &= \frac{\pi}{32} \text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) \left [ \left \{ \psi \left (\frac{1}{2} \right ) - \psi (1) \right \}^2 - \psi^{(1)}(1) \right ].
\end{align*}
Since
$$\text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) = \frac{\Gamma (1/2) \Gamma (1/2)}{\Gamma (1)} = \pi,$$
and
$$\psi^{(1)}(1) = \zeta (2) = \frac{\pi^2}{6} \quad \text{and} \quad \psi \left (\frac{1}{2} \right ) - \psi (1) = - 2\ln (2),$$
we have
$$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi^2}{8} \ln^2 (2) - \frac{\pi^4}{192},$$
as expected.
