And in the darkness bind them

Solution sketch :

Using Parseval's theorem applied to $f(x) = \ln \cos \dfrac{x}{2}$ * (see footnotes for extra stuff), we have

$\frac{1}{\pi} \int_0^{\pi/2} \ln^2 \cos x \ dx = \underbrace{\frac{1}{4} \left[ \frac{1}{\pi} \int_{-\pi}^{\pi} \ln^2 \cos \frac{x}{2} \ dx \right] = \frac{1}{4} \left[ 2\ln^22 + \sum_{n=1}^\infty \frac{1}{n^2} \right]}_{\text{Parseval's theorem}} = \frac{\ln^22}{2} + \frac{\pi^2}{24}$

Now, let $\displaystyle I = \int_0^{\pi/2} \ln^2 \cos x \ dx = \int_0^{\pi/2} \ln^2 \sin x \ dx = \frac{\pi\ln^22}{2} + \frac{\pi^3}{24}$ for convenience's sake.

On the one hand, using the double-angle formula we have

$\begin{aligned} \int_0^{\pi/2} (\ln \cos x + \ln \sin x)^2 \ dx &= \int_0^{\pi/2} (\ln \sin(2x) - \ln 2)^2 \ dx \\ &= I + \frac{3\pi \ln^22}{2} \end{aligned}$

On the other hand, simply expanding gives

$\begin{aligned} \int_0^{\pi/2} (\ln \cos x + \ln \sin x)^2 \ dx &= \int_0^{\pi/2} \ln^2 \cos x + 2 \ln \cos x \ln \sin x + \ln^2 \sin x \ dx \\ &= 2I + 2\int_0^{\pi/2} 2 \ln \cos x \ln \sin x \ dx \end{aligned}$

Equating the two we have

$\int_0^{\pi/2} 2 \ln \cos x \ln \sin x \ dx = \frac{3\pi \ln^22}{4} - \frac{I}{2}$

See https://brilliant.org/problems/logarithmtrignometryintegrals/ .