The value of the integral \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^2 + ln(\frac{\pi + x}{\pi - x})) cosx dx is:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^2 + ln(\frac{\pi + x}{\pi - x}))cosx dx \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 cosx dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ln(\frac{\pi + x}{\pi - x}) cosx dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 cosx dx + 0 [\because ln(\frac{\pi + x}{\pi - x})cosx is an odd function] = 2\int_{0}^{\frac{\pi}{2}} x^2 cosx dx [\because x^2 cosx is an even function] = 2[x^2 sinx + 2xcosx - 2sinx]_{0}^{\frac{\pi}{2}} = 2[\frac{\pi^2}{4} - 2] = \frac{\pi^2}{2} - 4