こういう系,積分範囲が違うともっと真面目に多重対数の計算しなきゃいけなくて大変
$$\begin {aligned}
\int _{0}^{\frac {\pi }2}\ln \sin x\ln \tan xdx&=
-\int _{0}^{\frac {\pi }2}\ln \cos x\ln \tan xdx\\
&=\frac {1}2\int _{0}^{\infty }\frac {\ln (1+t^{2})\ln t}{1+t^{2}}dt\quad \quad (\tan x\mapsto t)\\
&=\frac {1}2\Re\left (\int _{0}^{\infty }\frac {\ln (1+t^{2})\left (\ln (t)+\frac {\pi i}2\right )}{1+t^2}dt\right )\\
&=\frac {1}2\Re \left (\int _{0}^{\infty }\frac {\ln (1-s^{2})\ln s}{1-s^{2}}\frac {ds}i\right )\quad \quad \\
&=-\frac {\pi }2\int _{1}^{\infty }\frac {\ln s}{1-s^{2}}ds\\
&=-\frac {\pi }2\int _{0}^{1}\frac {\ln s}{1-s^{2}}ds\\
&=-\frac {\pi }4\int _{0}^{1}\left (\frac {\ln s}{1-s}+\frac {\ln s}{1+s}\right )ds\\
&=-\frac {\pi }4\left (\left [\ln s\ln (1+s)\right ]_0^1+\int _{0}^{1}\left (\frac {\ln (1-s)}s-\frac {\ln (1+s)}s\right )ds\right )\\
&=-\frac {\pi }4\left [-\operatorname{Li} _{2}(s) +\operatorname{Li} _{2}(-s) \right ]_0^1\\
&=-\frac {\pi }4\left (-\frac {\pi ^{2}}6-\frac {\pi ^{2}}{12}\right )\\
&=\textcolor {blue}{\frac {\pi ^{3}}{16}}.
\end {aligned}
$$