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!
 A: Substituting $\tan(x)^2=t$, splitting the integral into the intervals $(0,1)$ and $(1,\infty)$, then substituting $t=e^{-u}$ in the first integral, and $t=e^{u}$ in the second, adding both integrals and observing the symmetry of the resulting integrand whch allows to extend the integral from $-\infty$ to $\infty$ leads to the integral which I provided in my comment:
$$I=\frac{1}{\pi} \int_{-\infty }^{\infty } \frac{1}{\left(\frac{u}{\pi }+\frac{\pi }{u}\right)^2 \cosh \left(\frac{u}{2}\right)} \, du\tag{1}$$
That time I had got stuck at this point.
But in the meantime the next steps have been outlined by Jack D'Aurizio. Here I carry out those steps explicitly.


*

*Replacing the $\cosh$ term by its Fourier transform, i.e.


$$\frac{1}{\cosh \left(\frac{u}{2}\right)}= \int_{-\infty }^{\infty } \text{sech}(\pi  t) \exp (-i t u) \, dt\tag{2}$$


*Interchanging the order of integration we can perform the $u$-integral:


$$\frac{1}{\pi} \int_{-\infty }^{\infty } \frac{\exp (-i t u)}{\left(\frac{u}{\pi }+\frac{\pi }{u}\right)^2} \, du = \frac{1}{2} \pi  e^{-\pi  \left| t\right| } (1-\pi  \left| t\right| )\tag{3}$$


*Now the final $t$-integral gives


$$\int_{-\infty }^{\infty } \frac{1}{2} \left(\pi  e^{-\pi  \left| t\right| } (1-\pi  \left| t\right| )\right) \text{sech}(\pi  t) \, dt\\=\int_0^{\infty } \pi  e^{-\pi  t} (1-\pi  t) \text{sech}(\pi  t) \, dt = \log (2)-\frac{\pi ^2}{24}\tag{4}$$
Observing finally that $\zeta (2)=\frac{\pi ^2}{6}$ we have found the result to be proved. QED.
A: 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.
