Curious integral, $8\pi\int_{0}^{\pi/2}\cos^2x{\ln^2(\tan^2 x)\over [\pi^2+\ln^2(\tan^2 x)]^2}dx=\ln 2-{1\over 4}\zeta(2)$ How to show that,
$$8\pi\int_{0}^{\pi/2}\cos^2x\cdot{\ln^2(\tan^2 x)\over [\pi^2+\ln^2(\tan^2 x)]^2}\mathrm dx=\ln 2-{1\over 4}\zeta(2)$$
This integral is an extract from this paper of Olivier Oloa line $2.14$
I have not ideas where to begin, any help!

By letting $x=\arctan u$ the original integral is converted into
$$ 32\pi \int_{0}^{+\infty}\frac{\log^2(u)}{(1+u^2)^2\left(\pi^2+\log^2 u\right)^2}\,du\stackrel{u\mapsto e^{\pi x}}{=}32\int_{\mathbb{R}}\frac{e^{\pi x}x^2}{(1+e^{2\pi x})^2(1+x^2)^2}\,dx $$
and by symmetry the RHS collapses into
$$ 16 \int_{0}^{+\infty}\frac{x^2}{\cosh(\pi x)(1+x^2)^2}\,dx = 8 \int_{\mathbb{R}}\frac{x^2}{\cosh(\pi x)(1+x^2)^2}\,dx$$
which can be easily evaluated through the Fourier transform, since $\frac{1}{\cosh(\pi x)}$ is essentially a fixed point for $\mathscr{F}$ and $\mathscr{F}\left(\frac{x^2}{(1+x^2)^2}\right)(s)$ is related to the Laplace distribution.
