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$\ds{I \equiv \int_{0}^{\pi}\ln^{2}\pars{\tan\pars{x \over 4}}\,\dd x
={\pi^{3} \over 4\phantom{^{3}}}:\ {\large ?}}$
\begin{align}
I&=4\int_{0}^{\pi/4}\ln^{2}\pars{\tan\pars{x}}\,\dd x
=4\int_{0}^{1}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x
=2\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x
\\[3mm]&=2\lim_{\mu \to 0}\partiald[2]{}{\mu}
\int_{0}^{\infty}{x^{\mu} \over x^{2} + 1}\,\dd x
=\lim_{\mu \to 0}\partiald[2]{}{\mu}
\int_{0}^{\infty}{x^{\pars{\mu - 1}/2} \over x + 1}\,\dd x
\end{align}
With $\ds{t \equiv {1 \over x + 1}\quad\imp\quad x = {1 \over t} - 1}$:
\begin{align}
I&=\lim_{\mu \to 0}\partiald[2]{}{\mu}
\int_{1}^{0}t\pars{1 - t}^{\pars{\mu - 1}/2}t^{\pars{1 - \mu}/2}\,
\pars{-\,{\dd t \over t^{2}}}
\\[3mm]&=\lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{1}t^{-\pars{1 + \mu}/2}
\pars{1 - t}^{\pars{\mu - 1}/2}\,\dd t
=\lim_{\mu \to 0}\partiald[2]{{\rm B}\pars{1/2 - \mu/2,1/2 + \mu/2}}{\mu}
\\[3mm]&=\lim_{\mu \to 0}\partiald[2]{}{\mu}
\bracks{\Gamma\pars{1/2 - \mu/2}\Gamma\pars{1/2 + \mu/2} \over \Gamma\pars{1}}
=\lim_{\mu \to 0}\partiald[2]{}{\mu}
\braces{\pi \over \sin\pars{\pi\bracks{1/2 + \mu/2}}}
\\[3mm]&=\pi\lim_{\mu \to 0}\partiald[2]{\sec\pars{\pi\mu/2}}{\mu}
=\pi\lim_{\mu \to 0}\bracks{%
{1 \over 4}\,\pi^{2}\sec^{3}\pars{\pi\mu \over 2}
+ {1 \over 4}\,\pi^{2}\sec\pars{\pi\mu \over 2}\tan^{2}\pars{\pi\mu \over 2}}
\\[3mm]&=\pi\pars{\pi^{2} \over 4}
\end{align}
$\ds{{\rm B}\pars{x,y}}$ and $\ds{\Gamma\pars{z}}$ are the Beta and Gamma Functions, respectively, and we used well known properties of them.
$$
\int_{0}^{\pi}\ln^{2}\pars{\tan\pars{x \over 4}}\,\dd x
= {\pi^{3} \over 4\phantom{^{3}}}
$$
Integrate by parts; then you get that
$$I(\alpha) = \left [\frac1{2 \alpha} \arctan{(\alpha \sin{2 x})} \log{(\tan{x})} \right ]_0^{\pi/4} - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\alpha \sin{2 x}}$$
The first term on the RHS is zero. To evaluate the integral, expand the arctan into a Taylor series and get
$$I(\alpha) = -\frac12 \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \alpha^{2 k} \int_0^{\pi/2} du \, \sin^{2 k}{u} = -\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k}$$
A little manipulation leads us to
$$\alpha I'(\alpha) +I(\alpha) = -\frac{\pi}{4} \sum_{k=0}^{\infty} (-1)^k \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k} = -\frac{\pi}{4} \frac1{\sqrt{1+\alpha^2}}$$
The LHS is just $[\alpha I(\alpha)]'$, so the solution is
$$I(\alpha) = -\frac{\pi}{4} \frac{\operatorname{arcsinh}(\alpha)}{\alpha} $$
Best Answer
Introduce variables $t$ and $y$ such that $t = \tan x = e^{-y}$, we have
$$\begin{align} \int_0^{\pi/4}\frac{\log\tan x}{\cos 2x} dx &= \int_0^1\frac{\log t}{\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}\\ &= -\int_0^\infty \frac{y}{1 - e^{-2y}} e^{-y} dy = -\int_0^\infty \sum_{k=0}^\infty y e^{-(2k+1)y} dy\\ &= -\sum_{k=0}^\infty\frac{1}{(2k+1)^2} = -\left[\sum_{k=1}^\infty\frac{1}{k^2} - \sum_{k=1}^\infty\frac{1}{(2k)^2}\right]\\ &= -\left(1-\frac14 \right)\zeta(2) = -\frac{\pi^2}{8} \end{align} $$