Evaluate $\int_0^{\pi/4} \frac {\sin x} {x \cos^2 x} \mathrm d x$ I'm trying to evaluate:

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

Mathematica gives the numerical approximation:
$0.959926156626593638859649248036004150970933774605514278777212260466184427508$
I cannot find a closed form though. Any ideas would be appreciated. 
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$\ds{\int_{0}^{\pi/4}{\sin\pars{x} \over x\cos^{2}\pars{x}}\,\dd x:\ {\large ?}}$

\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.

