Integral $\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$ $$I:=\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$$ 
I am trying to see nice solutions to this integral.  I tried the following
$$
I=\int_0^{\pi/4}\frac{\ln \sin x}{\cos 2x} dx-\int_0^{\pi/4} \frac{\ln \cos x }{\cos 2x}dx
$$
but am not sure how to work with this denominator of $\cos 2x$.  If this helps:
$$
\int_0^{\pi/4}\log \sin x \, dx=-\frac{1}{4}\big(2K+\pi \ln 2\big)
$$
where K is the Catalan constant (note I am using Borwein convention not mathematica of using a C to define this constant.)  It is given by
$$
K=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}=\beta(2) 
$$
where $\beta(2)$ is the  Dirichlet beta function.
However I cannot solve this integral either.  Thanks

$$
\begin{aligned}
\int_0^{\frac{\pi}{4}} \frac{\ln (\tan x)}{\cos 2 x} d x=&\int_0^{\frac{\pi}{4}} \frac{\ln \left(\tan \left(\frac{\pi}{4}-x\right)\right)}{\cos 2\left(\frac{\pi}{4}-x\right)} d x \\
=& \int_0^{\frac{\pi}{4}} \frac{\ln \left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)}{\sin 2 x} d x\\=&\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{\ln \left(\frac{1-\sin 2 x}{1+\sin 2 x}\right)}{\sin 2 x} d x\\=&\frac{1}{4}\left[\int_0^{\frac{\pi}{2}} \frac{\ln (1-\sin x)}{\sin x} d x-\int_0^\frac\pi2 \frac{\ln (1+\sin x)}{\sin x} d x\right]
\end{aligned}
$$
From my post,
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\ln (1+a \sin x)}{\sin x} d x &=\int_0^{\frac{\pi}{2}} \frac{\ln (1+a \cos x)}{\cos x} d x \\
&=\frac{\pi^2}{8}-\frac{\left(\cos ^{-1} a\right)^2}{2}
\end{aligned}
$$
Now we can conclude that
$$\int_0^{\frac{\pi}{4}} \frac{\ln (\tan x)}{\cos 2 x} d x =\frac{1}{4}\left(-\frac{3 \pi^2}{8^2}-\frac{\pi^2}{8}\right)=-\frac{\pi^2}{8} $$
