Letting $x\mapsto \frac{\pi}{2} -x$ converts the integral
$\displaystyle I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x \tag*{} $
Using the identity $ \displaystyle \tan x=\frac{\sin 2 x}{1+\cos 2 x} $ , we get
$$\displaystyle I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}}\left[\frac{\sin 2 x}{\cos 2 x} \ln (\sin 2 x)-\frac{\sin 2 x}{\cos 2 x} \ln (1+\cos 2 x)\right] d x \tag*{} $$
Letting $2x\mapsto x$ yields
$\displaystyle I=\frac{\pi}{8}\left[\underbrace{\int_{0}^{\pi} \frac{\sin x}{\cos x} \ln (\sin x)}_{J} d x-\underbrace{\int_{0}^{\pi} \frac{\sin x}{\cos x} \ln (1+\cos x) d x}_{K}\right]\tag*{} $
As $ \displaystyle J \stackrel{x \rightarrow \pi-x}{=}-J \Rightarrow J=0 $ , therefore $\displaystyle I=-\frac{\pi}{8} K. $
By my post ,
$\displaystyle \begin{aligned}K \stackrel{y=\cos x}{=}& \int_{-1}^{1} \frac{\ln (1+y)}{y} d y=\frac{\pi^{2}}{4}\end{aligned}\tag*{} $
Now we can conclude that
$\displaystyle \boxed{I=-\frac{\pi^{3}}{32}}\tag*{} $
Request for elegant solutions. Your suggestion and alternative methods are warmly welcome!
Best Answer
Substitute $t=\tan^2x$, along with $\tan 2x =\frac{2\tan x}{1-\tan^2x}$\begin{align} & I = \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x = \frac\pi8 \int_0^\infty\frac{\ln t}{1-t^2}dt =\frac\pi8\left(-\frac{\pi^2}4\right)=-\frac{\pi^3}{32} \end{align} where $\int_0^\infty \frac {\ln t}{1-t^2}dt=-\frac{\pi^2}4 $