Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$ Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
Furthermore, $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x&=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x)-\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x \\
&=-\frac{\pi^2}{4} \ln 2-\left(-\frac{\pi^2}{8} \ln 2+\frac{7}{16}\zeta(3)\right) \\
&=-\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3)
\end{aligned}
$$
and $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$

\begin{align}J&=\int_0^{\frac{\pi}{2}}x\ln(\sin x)dx\\
K&=\int_0^{\frac{\pi}{2}}x\ln(\cos x)dx\\
J+K&=\int_0^{\frac{\pi}{2}}x\ln(\sin x\cos x)dx\\
&\overset{y=\frac{\pi}{2}-x}=\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-y\right)\ln(\sin y\cos y)dy\\
&\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln(\sin y\cos y)dy\\
&=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin(2y)}{2}\right)dy\\
&\frac{\pi}{4}\underbrace{\int_0^{\frac{\pi}{2}}\ln\left(\sin(2y)\right)dy}_{z=2y}-\frac{\pi^2}{8}\ln 2\\
&=\frac{\pi}{8}\int_0^\pi \ln(\sin z)dz-\frac{\pi^2}{8}\ln 2\\
&=\frac{\pi}{8}\int_0^{\frac{\pi}{2}} \ln(\sin z)dz+\frac{\pi}{8}\underbrace{\int_{\frac{\pi}{2}}^\pi \ln(\sin z)dz}_{x=\pi-z}-\frac{\pi}{8}\ln 2\\
&=\frac{\pi}{4}\underbrace{\int_0^{\frac{\pi}{2}} \ln(\sin x)dx}_{=-\frac{\pi}{2}\ln 2}-\frac{\pi^2}{8}\ln 2=\boxed{-\frac{\pi^2}{4}\ln 2}\\
J-K&=\int_0^{\frac{\pi}{2}}x\ln(\tan x)dx\\
&\overset{t=\tan x}=\int_0^\infty \frac{\ln t\arctan t}{1+t^2}dt
\end{align}
And see:
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$
https://math.stackexchange.com/a/4500094/186817
