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 $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$

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)
