$$\begin {aligned} &\quad I:=\int _{0}^{\frac {\pi }2}\ln ^2\left (\sin x\right )dx\\ &\mathrm{This \ integral\ can\ be\ calculated\ by\ differentiating\ Beta\ function.}\\ &\quad \begin {aligned} I&=\frac {1}{2\cdot 2^2}\left .\frac {d^2}{dx^2}B\left (x,\frac {1}2\right )\right |_{x=\frac {1}2}\\ &=\frac {1}{8}\left (\frac {\pi ^{3}}3+4\pi \ln^2 2\right )\\ &=\frac {\pi ^{3}}{24}+\frac {\pi \ln ^{2}2}2. \end {aligned}\\ &\mathrm{On\ the\ other\ hand,}\\ &\quad \begin {aligned} I&=\int _{0}^{1}\ln ^{2}\left (\frac {2t}{1+t^{2}}\right )\cdot \frac {2}{1+t^{2}}dt\quad \left (t=\tan \frac {x}2\right )\\ &=2\int _{0}^{1}\left (\ln 2+\ln t-\ln (1+t^{2})\right )^2\frac {dt}{1+t^{2}}\\ &=2\int _{0}^{1}\left (\ln ^{2}2+\ln ^{2}t+\ln ^{2}(1+t^{2})+2\ln (2)\ln t-2\ln (t)\ln (1+t^{2})-2\ln (2)\ln (1+t^{2})\right )\frac {dt}{1+t^{2}}\\ &=2\left (\frac {\pi \ln ^{2}2}4+2\beta (3)+\underbrace{\int _{0}^{1}\frac {\ln ^{2}(1+t^{2})}{1+t^2}dt}_{J}-2\ln 2\beta (2)-2\underbrace{\int _{0}^{1}\frac {\ln (t)\ln (1+t^{2})}{1+t^{2}}dt}_{K}-2\ln 2\underbrace{\int _{0}^{1}\frac {\ln (1+t^{2})}{1+t^{2}}dt}_L\right )\\&=\frac {\pi \ln ^{2}2}2+\frac {\pi ^{3}}{8}+2J-4\ln (2)\beta (2)-4K -4\ln (2)L. \end {aligned}\\ &\quad \begin {aligned} J&=\int _{0}^{\frac {\pi }4}\ln ^{2}\left (1+\tan ^{2}x\right )dx\\ &=4\int _{0}^{\frac {\pi }4}\ln ^{2}\cos xdx.\\ K&=\frac {3\pi ^{3}}{32}+\frac {\pi \ln ^{2}2}8-\ln (2)\beta (2)-2\Im \operatorname{Li} _{3}(1+i).\\ L&=\frac {\pi \ln 2}2-\beta (2). \end {aligned}\\ &\mathrm{Solving\ for\ 2J:}\\ &\quad \begin {aligned} 2J&=I-\frac {\pi \ln ^{2}2}2-\frac {\pi ^{3}}8+2\ln (2)\beta (2)+4K+4\ln (2)L\\ &=\frac {\pi ^{3}}{24}+\frac {\pi \ln ^{2}2}2-\frac {\pi \ln ^{2}2}2-\frac {\pi ^{3}}8+4\ln (2)\beta (2)+\frac {3\pi ^{3}}{8}+\frac {\pi \ln ^{2}2}2-4\ln (2)\beta (2)-8\Im \operatorname{Li} _{3}(1+i)+2\pi \ln ^{2}2 -4\ln (2)\beta (2)\\ &=\frac {7\pi ^{3}}{24}+\frac {5\pi \ln ^{2}2}2-4\ln (2)\beta (2)-8\Im \operatorname{Li} _{3}(1+i) \end {aligned}\\ &\mathrm{Now\ we \ get \ the\ answer}\\ &\quad \begin {aligned} \int _{0}^{\frac {\pi }4}\ln ^{2}(\sin x)dx &=\int _{0}^{\frac {\pi }2}\ln ^{2}(\sin x)dx-\int _{0}^{\frac {\pi }4}\ln ^{2}(\cos x)dx\\ &=I-\frac {1}4J\\ &=\frac {\pi ^{3}}{24}+\frac {\pi \ln ^{2}2}2-\frac {1}8\left (\frac {7\pi ^{3}}{24}+\frac {5\pi \ln ^{2}2}2-4\ln (2)\beta (2)-8\Im \operatorname{Li} _{3}(1+i) \right )\\ &=\textcolor {blue}{\frac {\pi ^{3}}{192}+\frac {3\pi \ln ^{2}2}{16}+\frac {\ln (2)\beta (2)}2+\Im \operatorname{Li} _{3}(1+i) }. \end {aligned}　 \end {aligned}$$