$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^2+\log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x\right) \mathrm{d} x=$

The integral is separated into even and odd components:

$I = \int_{-\pi/2}^{\pi/2} x^2 \, \mathrm{d}x + \int_{-\pi/2}^{\pi/2} \log\left(\frac{\pi-x}{\pi+x}\right) \cos x \, \mathrm{d}x$

Since $x^2$ is even, the first integral evaluates to twice the integral from $0$ to $\pi/2$:

$2 \int_0^{\pi/2} x^2 \, \mathrm{d}x = 2 \left[ \frac{x^3}{3} \right]_0^{\pi/2} = \frac{\pi^3}{12}$

The second integrand is odd, as $\log\left(\frac{\pi+x}{\pi-x}\right) \cos x = -\log\left(\frac{\pi-x}{\pi+x}\right) \cos x$, leading to cancellation over symmetric limits. Hence, the integral is zero.

Combining the results yields $I = \frac{\pi^3}{12} + 0 = \frac{\pi^3}{12}$.