Imaginary part of $\int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$ and $\int_{0}^{\pi/2} \frac{\log \cos x}{x^2}\:\mathrm{d}x$

Question

I have found the following new result connecting two rational log-cosine integrals.

Proposition. \begin{align} \displaystyle & {\Im} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x = \frac{\pi^2}{16} - \frac{\ln 2}{4} + \frac{\pi}{8} \int_{0}^{\pi/2} \frac{\log \cos x}{x^2}\:\mathrm{d}x \end{align} where $\displaystyle \log (z)$ denotes the principal value of the logarithm defined for $z \neq 0$ by \begin{align} \displaystyle \log (z) = \ln |z| + i \: \mathrm{arg}z, \quad -\pi <\mathrm{arg} z \leq \pi. \nonumber \end{align}

How would you prove it?

Answer 1

Numeric confirmation

From Mathematica. Left hand side:

Right hand side:

$$ f(z) = \frac{\ln \cos x}{x^2} $$

$$ \text{Im }f(z) = \frac{z^2}{z^2+\ln^{2} \left(-2 \cos z \right)} $$