Evaluate $\int_0^{\pi/4} \frac {\sin x} {x \cos^2 x} \mathrm d x$

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

\begin{align} &\color{#00f}{\large% \int_{0}^{\pi/4}{\sin\pars{x} \over x\cos^{2}\pars{x}}\,\dd x}= \int_{0}^{\pi/4}{\sec\pars{x}\tan\pars{x} \over x}\,\dd x =\int_{x=0}^{x=\pi/4}\,{\dd\bracks{\sec\pars{x} -1}\over x} \\[3mm]&={4 \over \pi}\,\pars{\root{2} - 1} +\int_{0}^{\pi/4}{\sec\pars{x} -1 \over x^{2}}\,\dd x \\[3mm]&={4 \over \pi}\,\pars{\root{2} - 1} +\sum_{n = 1}^{\infty}\pars{-1}^{n}{E_{2n} \over \pars{2n}!} \int_{0}^{\pi/4}x^{2n - 2}\,\dd x \\[3mm]&=\color{#00f}{\large{4 \over \pi}\,\pars{\root{2} - 1} +\sum_{n = 1}^{\infty}\pars{-1}^{n}{E_{2n} \over \pars{2n}!\pars{2n - 1}} \pars{\pi \over 4}^{2n - 1}} \end{align} where $\ds{E_{n}}$ is an Euler Number.