The value of (\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} dx) is equal to

(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} dx = \int_{0}^{\frac{\pi}{2}} \left(\frac{x^2 \cos x}{1+e^x} + \frac{x^2 \cos x}{1+e^{-x}}\right) dx ) (As numerator is even function but denominator is odd function)

(= \int_{0}^{\frac{\pi}{2}} \frac{x^2 \cos x + e^x (x^2 \cos x)}{1+e^x} dx = \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx )

(= \left(x^2 \sin x\right){0}^{\frac{\pi}{2}} - \int{0}^{\frac{\pi}{2}} 2x \sin x dx ) (uv rule of integration)

(= \frac{\pi^2}{4} - \left[\left(-x \cos x\right){0}^{\frac{\pi}{2}} - \int{0}^{\frac{\pi}{2}} -\cos x dx\right] ) (uv rule of integration)

(= \frac{\pi^2}{4} - \left[0 + (\sin x)_{0}^{\frac{\pi}{2}}\right] = \frac{\pi^2}{4} - 1)

This is the required solution.