$\mathrm{ln}\left(\mathrm{sin}x\right)=-\mathrm{ln}2-\sum _{n=1}^{\mathrm{\infty}}\frac{\mathrm{cos}\left(2nx\right)}{n}$ We can use the Fourier series and integrate by parts to obtain $I\equiv {\int}_{0}^{\frac{\pi}{4}}x[-\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)]dx=\frac{{\pi}^{2}}{32}\mathrm{ln}\left(2\right)+{\frac{14}{\sum}}_{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{2}}[\frac{\pi}{2}\mathrm{sin}\left(\frac{\pi}{2}n\right)-\frac{1}{n}(1-\mathrm{cos}\left(\frac{\pi}{2}n\right))]$ $\mathrm{sin}\left(\frac{\pi}{2}n\right)$ is non-zero and alternating for odd n, while $\mathrm{cos}\left(\frac{\pi}{2}n\right)$ is non-zero and alternating for even n. Therefore, $I=\frac{{\pi}^{2}}{32}\mathrm{ln}\left(2\right)+\frac{\pi}{8}\sum _{k=0}^{\mathrm{\infty}}\frac{{(-1)}^{k}}{{(2k+1)}^{2}}-\frac{1}{4}\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{n}^{3}}-\frac{1}{32}\sum _{k=1}^{\mathrm{\infty}}\frac{{(-1)}^{k-1}}{{k}^{3}}$ The first series is Catalan's constant G, the second one is $\zeta \left(3\right)$ and the third one is $\eta \left(3\right)=\frac{34}{\zeta}\left(3\right)$ (with the Riemann zeta function $\zeta$ and the Dirichlet eta function $\eta$), so we obtain $I=\frac{{\pi}^{2}}{32}\mathrm{ln}\left(2\right)+\frac{\pi}{8}G-\frac{35}{128}\zeta \left(3\right)$ and your integral is −I.