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

The value of ∫_(-π/2)^(π/2) (x²cos x)/(1+eˣ) dx is equal to

$I = \int_{-\pi/2}^{\pi/2} \dfrac{x^2\cos x}{1+e^x}\,dx$

$\text{symmetric limits } \left(-\tfrac{\pi}{2},\ \tfrac{\pi}{2}\right)$

$\text{To find: } I$

$I' = \int_{-\pi/2}^{\pi/2} \dfrac{x^2\cos x}{1+e^{-x}}\,dx$

$x\to -x:\quad I' = \int_{-\pi/2}^{\pi/2} \dfrac{x^2\cos x}{1+e^{x}}\,dx = I$

$I + I' = \int_{-\pi/2}^{\pi/2} x^2\cos x\left(\dfrac{1}{1+e^x} + \dfrac{1}{1+e^{-x}}\right)dx$

$\dfrac{1}{1+e^{-x}} = \dfrac{e^x}{1+e^x}\ \Rightarrow\ \dfrac{1}{1+e^x}+\dfrac{e^x}{1+e^x} = 1$

$2I = \int_{-\pi/2}^{\pi/2} x^2\cos x\,dx = 2\int_{0}^{\pi/2} x^2\cos x\,dx$

$\int x^2\cos x\,dx = x^2\sin x + 2x\cos x - 2\sin x$

$\int_0^{\pi/2} = \dfrac{\pi^2}{4} - 2$

$2I = 2\left(\dfrac{\pi^2}{4} - 2\right) \Rightarrow I = \dfrac{\pi^2}{4} - 2$

$\boxed{I = \dfrac{\pi^2}{4} - 2 \ \text{— option } (1)}$