$$ \begin{aligned} \int _{0}^{\pi }\frac {\sin x\ln \sin \frac {x}2\ln \tan \frac {x}4}{3+\cos x}dx &=2\int _{0}^{\frac {\pi }2}\frac {\sin (2x)\ln \sin (x)\ln \tan \frac {x}2}{3+\cos 2x}\\ &=\int _{0}^{\frac {\pi }2}\frac {\sin x\cos x\ln \sin ^{2}x\ln \tan ^{2}\frac {x}2}{2+2\cos ^{2}x}dx\\ &=\frac {1}2\int _{0}^{\frac {\pi }2}\frac {\cos x(\ln (1-\cos x)+\ln (1+\cos x))(\ln (1-\cos x)-\ln (1+\cos x))}{1+\cos^2x}\sin xdx\\ &=\frac {1}2\int _{0}^{1}\frac {t(\ln ^{2}(1-t)-\ln ^{2}(1+t))}{1+t^2}dt\\ &=-\frac {1}2\int _{-1}^{1}\frac {t\ln ^2(1+t)}{1+t^2}\\ &=-\frac {1}2\Re \int _{-1}^{1}\frac {\ln ^{2}(1+t)}{t+i}dt\\ &=-\frac {1}2\Re \int _{0}^{2}\frac {\ln ^{2}s}{s-(1-i)}ds\\ &=-\frac {1}2\Re \left [\ln \left (1-\frac {(1+i)s}2\right )\ln ^{2}s+2\operatorname {Li} _2 \left (\frac {(1+i)s}2\right )\ln s-2\operatorname{Li} _{3}\left (\frac {(1+i)s}2\right ) \right ]_0^2\\ &=-\frac {1}2\Re \left (-\frac {\pi i}2\ln ^{2}2+2\operatorname {Li} _2 (1+i)\ln 2-2\operatorname{Li} _{3}\left (1+i\right ) \right )\\ &=-\frac {\pi ^{2}}{16}\ln 2+\frac {35}{64}\zeta (3)+\frac {\pi ^{2}\ln 2}{32}\\ &=\textcolor {blue}{\frac {35}{64}\zeta (3)-\frac {\pi ^{2}\ln 2}{32}}. \end{aligned} $$