TITLE: How can we show that $\int_{0}^{1}\cos(\ln x)\cdot{\mathrm dx\over 1+x^2}={\pi\over 4}\cdot{1\over \cosh\left({\pi\over 2}\right)}?$
QUESTION [17 upvotes]: Proposed:

$$\int_{0}^{1}\cos(\ln x)\cdot{\mathrm dx\over 1+x^2}={\pi\over 4}\cdot{1\over \cosh\left({\pi\over 2}\right)}\tag1$$

My try:
$x=\tan u\implies dx=\sec^2 u du$, then $(1)$ becomes
$$\int_{0}^{\pi/4}\cos(\ln\tan u)\mathrm du\tag2$$
Recall of $$\int_{0}^{\pi/2}\ln\sin x\mathrm dx=\int_{0}^{\pi/2}\ln\cos x\mathrm dx=-{\pi\over 2}\ln(2)\tag3$$

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
\int_{0}^{1}\cos\pars{\ln\pars{x}}\,{\dd x \over 1 + x^{2}} & =
\Re\int_{0}^{1}\expo{\ic\ln\pars{x}}\,{\dd x \over 1 + x^{2}} =
\Re\int_{0}^{1}{x^{\ic} \over 1 + x^{2}}\,\dd x =
\Re\int_{0}^{1}{x^{\ic} - x^{\ic + 2} \over 1 - x^{4}}\,\dd x
\\[5mm] & =
{1 \over 4}\,\Re\int_{0}^{1}{x^{\ic/4 -3/4} - x^{\ic/4 - 1/4} \over 1 - x}\,\dd x
\\[5mm] & =
{1 \over 4}\,\Re\pars{%
\int_{0}^{1}{1 - x^{\ic/4 - 1/4} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{\ic/4 - 3/4} \over 1 - x}\,\dd x}
\\[5mm] & =
{1 \over 4}\,\Re\pars{H_{\ic/4 - 1/4} - H_{\ic/4 - 3/4}}\qquad
\pars{~H_{z}:\ Harmonic\ Number~}
\\[5mm] & =
{1 \over 4}\
\overbrace{\Re\pars{H_{\ic/4 - 1/4} - H_{-\ic/4 - 1/4}}}^{\ds{=\ 0}}\ -\
{1 \over 4}\
\underbrace{\Re\pars{\pi\cot\pars{\pi\bracks{{3 \over 4} - {\ic \over 4}}}}}
_{\ds{\substack{\ds{=\ -\,{\pi \over \cosh\pars{\pi/2}}\\
{\tiny after\ a\ tedious\ algebraic\ task}}}}}
\\[5mm] & =
\bbx{{1 \over 4}\,\pi\,\mrm{sech}\pars{\pi \over 2}}
\end{align}