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

Introduce variables $t$ and $y$ such that $t = \tan x = e^{-y}$, we have

$$\begin{align} \int_0^{\pi/4}\frac{\log\tan x}{\cos 2x} dx &= \int_0^1\frac{\log t}{\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}\\ &= -\int_0^\infty \frac{y}{1 - e^{-2y}} e^{-y} dy = -\int_0^\infty \sum_{k=0}^\infty y e^{-(2k+1)y} dy\\ &= -\sum_{k=0}^\infty\frac{1}{(2k+1)^2} = -\left[\sum_{k=1}^\infty\frac{1}{k^2} - \sum_{k=1}^\infty\frac{1}{(2k)^2}\right]\\ &= -\left(1-\frac14 \right)\zeta(2) = -\frac{\pi^2}{8} \end{align} $$

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 $$ \begin{align} \int_0^{\pi/4}\frac{\log(\tan(x))}{\cos(2x)}\mathrm{d}x &=-\frac\pi4\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{2n+1}\\ &=-\frac\pi4\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}\color{#C00000}{1}^{2n+1}}{2n+1}\\ &=-\frac\pi4\arcsin(\color{#C00000}{1})\\ &=-\frac{\pi^2}8\tag{5} \end{align} $$

Let $\tan x=e^{-t}$, then $$\cos 2x=\cos^2x-\sin^2x=\frac{1-e^{-2t}}{1+e^{-2t}}$$ and $$dx=-\frac{e^{-t}\ dt}{1+e^{-2t}}.$$ Therefore $$\begin{align} \int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} dx&=-\int_\infty^{0} \frac{\ln e^{-t}}{\dfrac{1-e^{-2t}}{1+e^{-2t}}}\cdot \frac{e^{-t}\ dt}{1+e^{-2t}}\\ &=-\int_0^{\infty}\frac{te^{-t}}{1-e^{-2t}} dt\tag1 \end{align}$$ Equation $(1)$ can be solved by using IBP. Let $u=t\;\rightarrow du=dt$ and $$\begin{align} dv&=\frac{e^{-t}}{1-e^{-2t}} dt\\ v&=\int\frac{e^{-t}}{1-e^{-2t}} dt\\ &=-\int\frac{d\left(e^{-t}\right)}{1-e^{-2t}}\quad\Rightarrow\quad y=e^{-t}\\ &=-\int\frac{dy}{1-y^2}\\ &=-\frac12\left(\int\frac{dy}{1-y}+\int\frac{dy}{1+y}\right)\\ &=\frac12\ln\left(1-y\right)-\frac12\ln\left(1+y\right)\\ &=\frac12\ln\left(1-e^{-t}\right)-\frac12\ln\left(1+e^{-t}\right). \end{align}$$ Hence $$\begin{align} -\int_0^{\infty}\frac{te^{-t}}{1-e^{-2t}} dt&=\left[\frac t2\ln\left(1-e^{-t}\right)+\frac t2\ln\left(1+e^{-t}\right)\right]_0^{\infty}\\&\frac12\int_0^{\infty}\ln\left(1-e^{-t}\right)\ dt-\frac12\int_0^{\infty}\ln\left(1+e^{-t}\right)\ dt\\ &=\frac12\int_0^{\infty}\ln\left(1-e^{-t}\right)\ dt-\frac12\int_0^{\infty}\ln\left(1+e^{-t}\right)\ dt\\ &=\frac12\int_0^1\ln\left(1-y\right)\ dy-\frac12\int_0^{1}\ln\left(1+y\right)\ dy. \end{align}$$ Since $|y|<1$, we can use Maclaurin series for natural logarithm. $$\begin{align} -\int_0^{\infty}\frac{te^{-t}}{1-e^{-2t}} dt &=\frac12\int_0^1\ln\left(1-y\right)\ dy-\frac12\int_0^{1}\ln\left(1+y\right)\ dy\\ &=-\frac12\int_0^1\sum_{n=1}^\infty\frac{y^n}{n}\ dy-\frac12\int_0^{1}\sum_{n=1}^\infty(-1)^{n+1}\frac{y^n}{n}\ dy\\ &=-\frac12\sum_{n=1}^\infty\frac{1}{n^2}-\frac12\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n^2}\\ &=-\frac12\zeta(2)-\frac12\eta(2)\\ &=-\frac12\zeta(2)-\frac14\zeta(2)\\ &=\Large\color{blue}{-\frac{\pi^2}{8}}, \end{align}$$ where $\zeta(s)$ is Riemann zeta function, $\eta(s)$ is Dirichlet eta function, and $\zeta(2)=\dfrac{\pi^2}{6}$.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{\pi/4}{\ln\pars{\tan\pars{x}} \over \cos\pars{2x}}\,\dd x =-\,{\pi^{2} \over 8}:\ {\large ?}}$.

\begin{align} I&=\half\int_{0}^{\pi/2}{\ln\pars{\tan\pars{x/2}} \over \cos\pars{x}}\,\dd x =\half\int_{0}^{1}{\ln\pars{t} \over \pars{1 - t^{2}}/\pars{1 + t^{2}}}\, {2\,\dd t \over 1 + t^{2}} =\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t \\[3mm]&=\int_{0}^{1}{\ln\pars{t^{1/2}} \over 1 - t}\,\half\,t^{-1/2}\,\dd t ={1 \over 4}\int_{0}^{1}{t^{-1/2}\ln\pars{t} \over 1 - t}\,\dd t =-\,{1 \over 4}\lim_{s \to -1/2}\partiald{}{s} \int_{0}^{1}{1 - t^{s} \over 1 - t}\,\dd t \end{align}

However, $\ds{\int_{0}^{1}{1 - t^{s} \over 1 - t}\,\dd t = \Psi\pars{s + 1} + \gamma}$. See ${\bf 6.3.22}$ in this link. $\ds{\Psi\pars{z}}$ and $\ds{\gamma}$ are the Digamma Function and the Euler-Mascheroni Constant, respectively. Then, $$ I = -\,{1 \over 4}\,\Psi'\pars{\half} $$

With Euler Reflection Formula ${\bf 6.4.7}$, $\ds{\Psi'\pars{\half} =\left.-\,\half\,\pi\cot'\pars{\pi z}\right\vert_{\,z\ =\ \half} =\half\,\pi^{2}\csc^{2}\pars{\pi \over 2} = {\pi^{2} \over 2}}$

$$ \color{#00f}{\large% I = \int_{0}^{\pi/4}{\ln\pars{\tan\pars{x}} \over \cos\pars{2x}}\,\dd x = -\,{\pi^{2} \over 8}} $$