Prove $\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx=\frac{\sqrt{2}-2}{4} \sqrt{\pi}~ \zeta \left( \frac{1}{2} \right)$ Wolfram Alpha evaluates this integral numerically as 
$$\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx=0.379064 \dots$$
Its value is apparently
$$\frac{\sqrt{2}-2}{4} \sqrt{\pi}~ \zeta \left( \frac{1}{2} \right)=0.37906401072\dots$$

How would you solve this integral?

Obviously, we can make a substitution $t=x^2$
\begin{align}
\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx&=\frac{1}{2} \int_0^{\infty} \frac{\sqrt{t}}{\cosh^2 (t)} dt\\[10pt]
&=\int_0^{\infty} \frac{\sqrt{t}}{\cosh (2t)+1} dt\\[10pt]
&=\frac{1}{2 \sqrt{2}}\int_0^{\infty} \frac{\sqrt{u}}{\cosh (u)+1} du
\end{align}
We could use geometric series since $\cosh (u) \geq 1$, but I don't know how it will help.

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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{2} \over \cosh^{2}\pars{x^{2}}}\,\dd x}\,\,\
\stackrel{x\ \to\ x^{1/2}}{=}\,\,\,\
2\int_{0}^{\infty}{x^{1/2}\expo{2x} \over \pars{\expo{2x} + 1}^{2}}\,\dd x
\\[5mm] = &\
-\int_{x = 0}^{x \to \infty}x^{1/2}\,\dd\pars{{1 \over \expo{2x} + 1}} =
\half\int_{0}^{\infty}{x^{-1/2}\expo{-2x} \over 1 + \expo{-2x}}\,\dd x
\\[5mm] \stackrel{2x\ \to x}{=}\,\,\,&
{\root{2} \over 4}\int_{0}^{\infty}{x^{-1/2}\expo{-x} \over 1 + \expo{-x}}\,\dd x
\\[5mm] = &
{\root{2} \over 4}\sum_{n = 0}^{\infty}\pars{-1}^{n}
\int_{0}^{\infty}x^{-1/2}\expo{-\pars{n + 1}x}\,\dd x
\\[3mm] \stackrel{\pars{n + 1}x\ \to x}{=}\,\,\,&
{\root{2} \over 4}\sum_{n = 0}^{\infty}
{\pars{-1}^{n} \over \pars{n + 1}^{1/2}}\ \overbrace{%
\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x}^{\ds{\Gamma\pars{\half}\ =\ \root{\pi}}}
\\[5mm] = &\
{\root{2} \over 4}\,\root{\pi}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{1/2}}
\end{align}

With the identity
$\ds{\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{s}} =
     \pars{1 - 2^{1 - s}}\zeta\pars{s}}$:
\begin{align}
\bbox[#ffd,10px]{\int_{0}^{\infty}{x^{2} \over \cosh^{2}\pars{x^{2}}}\,\dd x} & =
{\root{2} \over 4}\,\root{\pi}\pars{1 - 2^{1 - 1/2}}\zeta\pars{\half}
\\[5mm] & =
\bbox[10px,border:1px groove navy]{{\root{2} - 2 \over 4}\,\root{\pi}\zeta\pars{\half}} \approx 0.3791
\end{align}
