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 $$ \log (z) = \ln |z| + i \mathrm{Arg}z, \quad -\pi <\mathrm{Arg} z \leq \pi.$$
2026-04-08 02:10:18.1775614218
What is the real part of $\int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$?
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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.