What is the real part of $\int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$? This is a new integral that I propose to evaluate in closed form:
$$ {\mathfrak{R}} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$$
where $\Re$ denotes the real part and $\log (z)$ denotes the principal value of the logarithm defined for $z \neq 0$ by 
$$ \log (z)  = \ln |z| + i \mathrm{Arg}z, \quad -\pi <\mathrm{Arg} z \leq \pi.$$
 A: $$\color{blue}{\Re\int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)}dx = \frac{\pi }{8}\left( {1 - \ln 2 - \gamma } \right)}$$

For a proof, I will continue from $$\tag{4} - 8\Re I + 2\pi J = \pi \ln 2 - \pi$$ in this question.
where $$J=\int_0^\infty  {\frac{1}{{{x^2} + {\pi ^2}}}\left[ { - \frac{x}{{1 - {e^{ - x}}}} + \ln ({e^x} - 1)} \right]dx}$$
Integration by part shows
$$J = -\pi \int_0^\infty  {\arctan x\frac{{{e^{\pi x}}x}}{{{{\left( {{e^{\pi x}} - 1} \right)}^2}}}dx} $$
Invoke the Binet second formula:
$$\ln\Gamma(z) = (z-\frac{1}{2})\ln z - z + \frac{\ln(2\pi)}{2}+2z\int_0^\infty \frac{\arctan t}{e^{2\pi t z} - 1} dt$$
The value of $J$ immdiately follows from differentiating with respect to $z$, and then set $z = 1/2$:
$$J = -\frac{\gamma}{2}$$
I hope someone can explain reminiscence of this result to another question.
A: I don't think a closed form exists after computing that integral numerically in Mathematica, and looking up in the Inverse Symbolic Calculator. It is approximately equal to: $-0.10617124113817\ldots$ If you need more digits just ask.
