Question

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

Answer 1

Let $I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{x^2\cos x}{1+e^x}dx$

Using ${\int }_{-a}^{a}f\left(x\right)dx={\int }_0^a\left(f\right(x)+f(-x\left)\right)dx$ , we get

$I={\int }_0^{\frac{\pi }{2}}\frac{x^2\cos x(1+e^x)dx}{(1+e^x)}=-{\int }_0^{\frac{\pi }{2}}{x}^2\cos xdx$

using product rule, we get

$I=x^2{\int }_0^{\frac{\pi }{2}}\cos xdx-{\int }_0^{\frac{\pi }{2}}\left[\right(\frac{d{x}^2}{dx})\int \cos xdx]dxI=x^2{\left[\sin x\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}2x\sin xdxI=\frac{{\pi }^2}{4}-2[x{\int }_0^{\frac{\pi }{2}}\sin xdx-{\int }_0^{\frac{\pi }{2}}[\frac{dx}{dx}\int \sin xdx]dx]I=\frac{{\pi }^2}{4}-2[0+{\int }_0^{\frac{\pi }{2}}\cos xdx]I=\frac{{\pi }^2}{4}-2\left(1\right)=\frac{{\pi }^2}{4}-2$