The value of ∫ _ - (π)/(2) ^ (π)/(2) x ^ 2 cos x 1 + e ^ x dx is...

Let $I = \mathop{∫}_{- \dfrac{x}{2}}^{\dfrac{x}{2}} \dfrac{{x}^{2} \cos x}{1 + {e}^{x}} dx$
Replacing $x$ by $- x$
$I = \mathop{∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}} ( \dfrac{{x}^{2} \cos x}{1 + {e}^{- x}} ) dx$
$⇒ 2I = \mathop{∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}} ( \dfrac{1}{1 + {e}^{x}} + \dfrac{1}{1 + {e}^{- x}} ){x}^{2} cosxdx$
$⇒ 2I = \mathop{∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}} {x}^{2} cosxdx$, (even function)
$I$ $= \mathop{∫}_{0}^{\dfrac{\pi}{2}}{x}^{2}cos xdx = {\left({x}^{2}sin x\right)}_{0}^{\dfrac{\pi}{2}} - 2\mathop{∫}_{0}^{\dfrac{\pi}{2}}x . sin xdx$
(Using integration by parts, taking ${x}^{2} = u, cosxdx = dv$ )
$= \dfrac{{\pi}^{2}}{4} - 2\left[- {\left(x . cos x\right)}_{0}^{\dfrac{\pi}{2}} + \mathop{∫}_{0}^{\dfrac{\pi}{2}}1 . cos xdx\right] = \dfrac{{\pi}^{2}}{4} - 2\left[0 + 1\right] = \left(\dfrac{{\pi}^{2}}{4} - 2\right)