TITLE: Integral $\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$
QUESTION [21 upvotes]: $$I:=\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx=-\frac{\pi^2}{8}.$$ 
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

REPLY [11 votes]: Using
$$
\sin^2(x)=\frac{1-\cos(2x)}{2}\qquad\text{and}\qquad
\cos^2(x)=\frac{1+\cos(2x)}{2}\tag{1}
$$
and
$$
\frac12\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+\dots\tag{2}
$$
we get
$$
\begin{align}
\log(\tan(x))
&=\frac12\log\left(\sin^2(x)\right)-\frac12\log\left(\cos^2(x)\right)\\
&=\frac12\log\left(\frac{1-\cos(2x)}{2}\right)-\frac12\log\left(\frac{1+\cos(2x)}{2}\right)\\
&=-\frac12\log\left(\frac{1+\cos(2x)}{1-\cos(2x)}\right)\\
&=-\cos(2x)-\frac{\cos^3(2x)}{3}-\frac{\cos^5(2x)}{5}-\dots\tag{3}
\end{align}
$$
Using
$$
\begin{align}
\int_0^{\pi/4}\cos^{2n}(2x)\,\mathrm{d}x
&=\frac12\int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x\\
&=\frac\pi4\binom{2n}{n}4^{-n}\tag{4}
\end{align}
$$
we get
$$
\begin{align}
\int_0^{\pi/4}\frac{\log(\tan(x))}{\cos(2x)}\mathrm{d}x
&=-\frac\pi4\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{2n+1}\\
&=-\frac\pi4\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}\color{#C00000}{1}^{2n+1}}{2n+1}\\
&=-\frac\pi4\arcsin(\color{#C00000}{1})\\
&=-\frac{\pi^2}8\tag{5}
\end{align}
$$