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 $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$
 A: \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
A: Method. $(1)$, Use Mittag-Leffler's series for $\cot(x)$
$$\begin{align}
I&=\int_0^\frac{\pi}{2}x\ln(\sin(x))~dx=\frac{1}{2}\int_0^\frac{\pi}{2}\ln(\sin(x))~dx^2\\
\\
&=-\frac{1}2 \int_0^\frac{\pi}{2}\frac{x^2}{\tan(x)}~dx=-\frac{1}2\int_0^\frac{\pi}{2}x^2\left(\frac{1}x+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2} \right)~dx
\end{align}$$
Next, integrate them term by term and you will get the answer.
Method. $(2)$, Construct two bases, as mentioned by @Quanto:
$$\begin{align}
I_1&=\int_0^\frac{\pi}{2}x\left[\ln(\sin(x))+\ln(\cos(x))\right]dx=\int_0^\frac{\pi}{2}x\ln\left( \frac{1}{2}\sin(2x) \right)dx\\
\\
I_2&=\int_0^\frac{\pi}{2}x\left[\ln(\sin(x))-\ln(\cos(x))\right]dx=\int_0^\frac{\pi}{2}x\ln(\tan(x))dx
\end{align}$$
A: Even if there is an antiderivative (have a look here), you could write
$$\log(\sin(x))=\log(x)-\sum_{n=1}^\infty(-1)^{n+1}\, \frac{ 2^{2 n-1}\, B_{2 n} }{n \,(2 n)!}x^{2 n}$$ where appear Bernoulli numbers.
Integrate termwise
$$\int_0^{\frac \pi 2}x\,\log(x)=-\frac{1}{16} \pi ^2 (1+2 \log (2)-2 \log (\pi ))$$
$$\int_0^{\frac \pi 2} (-1)^{n+1}\, \frac{ 2^{2 n-1}\, B_{2 n} }{n \,(2 n)!}x^{2 n+1}=(-1)^{n+1}\, \frac{\pi ^{2 n+2} B_{2 n}}{16 n (n+1) (2 n)!}$$ The infinite summation gives
$$-\frac{1}{16} \left(7 \zeta (3)+\pi ^2-2 \pi ^2 \log (\pi )\right)$$ and then the result (with a sign difference)
A: Enforcing the substitution $x\mapsto\frac {\pi}2-x$, then we get
\begin{align*}
\int\limits_0^{\pi/2}x\log\sin x\,\mathrm dx & =\int\limits_0^{\pi/2}\left(\frac {\pi}2-x\right)\log\cos x\,\mathrm dx\\ & =\frac {\pi}2\int\limits_0^{\pi/2}\log\cos x\,\mathrm dx-\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx\\ & =-\frac {\pi^2}4\log 2-\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx
\end{align*}
Now, express $\cos x$ as
$$2\cos x=e^{ix}+e^{-ix}$$
And expand the logarithm into three separate terms. Since the integral is entirely real, the complex components will cancel out in the end.
\begin{align*}
\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx & =\int\limits_0^{\pi/2}x\log\left(e^{ix}+e^{-ix}\right)\,\mathrm dx-\log 2\int\limits_0^{\pi/2}x\,\mathrm dx\\ & =i\int\limits_0^{\pi/2}x^2\,\mathrm dx+\int\limits_0^{\pi/2}x\log\left(1+e^{-2ix}\right)\,\mathrm dx-\frac {\pi^2}8\log 2\\ & =\frac {\pi^3i}{24}-\frac {\pi^2}8\log 2+\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}n\int\limits_0^{\pi/2}xe^{-2nix}\,\mathrm dx
\end{align*}
Focusing on the remaining sum, we use integration by parts
\begin{align*}
\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}n\int\limits_0^{\pi/2}xe^{-2nix}\,\mathrm dx & =-\frac 14\sum\limits_{n=1}^{+\infty}\frac 1{n^3}-\frac {\pi i}4\sum\limits_{n=1}^{+\infty}\frac 1{n^2}-\frac 14\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}{n^3}\\ & =-\frac 7{16}\zeta(3)-\frac {\pi^3 i}{24}
\end{align*}
So the integral is
\begin{align*}
\int\limits_0^{\pi/2}x\log\sin x\,\mathrm dx=-\frac {\pi^2}4\log 2+\frac {\pi^2}8\log 2+\frac 7{16}\zeta(3)\color{blue}{=-\frac {\pi^2}8\log 2+\frac 7{16}\zeta(3)}
\end{align*}
