$$\begin {aligned} \int _{0}^{\frac {\pi }4}\frac {x\ln \tan x}{\sin 2x}dx &=\frac {1}2\int _{0}^{1}\frac {\arctan (t)\ln (t)}{t}dt\quad \quad \left (t=\tan x\right )\\ &=\frac {1}4\left [\arctan (t)\ln ^{2}(t)\right ]_0^1-\frac {1}4\int _{0}^1\frac {\ln ^{2}t}{1+t^{2}}dt\\ &=-\frac {\Gamma (3)\beta (3)}4\\ &=\textcolor {blue}{-\frac {\pi ^{3}}{64}}. \end {aligned}$$