Show that $\int\limits_0^{\frac{\pi}2}4\cos^2(x)\log^2(\cos x)~\mathrm dx=-\pi\log 2+\pi\log^2 2-\frac{\pi}2+\frac{\pi^3}{12}$

You can first substitute $x=\arctan u$: \begin{align} 4\int_0^{\pi/2} \cos^2 (x) \log^2 \cos x\,dx&=\int_0^\infty \frac{\log^2(1+u^2)}{(1+u^2)^2} \,du \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \int_0^\infty \frac{1}{(1+u^2)^{2-\beta}}\,du \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{1}{\Gamma(2-\beta)}\int_0^\infty \int_0^\infty \nu^{1-\beta} e^{-\nu(1+u^2)}\,d\nu\, du \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{1}{\Gamma(2-\beta)}\int_0^\infty e^{-\nu}\nu^{1-\beta} \int_0^\infty e^{-\nu u^2}\,du\, d\nu \\ &= \frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{\sqrt{\pi}}{\Gamma(2-\beta)}\int_0^\infty e^{-\nu}\nu^{1/2 -\beta} d\nu \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{\sqrt{\pi}\,\Gamma\left(3/2-\beta\right)}{\Gamma(2-\beta)} \end{align}

After doing the computations, we can find that this is equal to

\begin{align} \frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{\sqrt{\pi}\,\Gamma\left(3/2-\beta\right)}{\Gamma(2-\beta)} &= \sqrt{\pi}\frac{\Gamma(3/2)}{\Gamma(2)}\left(\psi^{(0)}(3/2)^2-2\psi^{(0)}(2)\psi^{(0)}(3/2)+\psi^{(1)}(3/2)+\psi^{(0)}(2)^2-\psi^{(1)}(2)\right) \\ &=\frac{\pi}{4}\left(\frac{\pi^2}{3}+4\log^2-4\log 2 -2\right) \\ &=\frac{\pi^3}{12}+\pi\log^2 2 -\pi\log 2 -\frac{\pi}{2} \end{align}