Mathematics - Definite Integration Question with Solution | TestHub

Let$I={\displaystyle \underset{-\frac{x}{2}}{\overset{\frac{x}{2}}{\int }}\frac{x^2\cos x}{1+e^x}dx}$
Replacing$x$by$-x$
$I={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{x^2\cos x}{1+e^{-x}}\right)dx}$
$\Rightarrow 2I={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right)x^2\cos xdx}$
$\Rightarrow 2I={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{x}^2\cos xdx}$, (even function)
$I$$= \underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^2\cos x dx={\left(x^2\sin x\right)}_0^{\frac{\pi }{2}}-2\underset{0}{\overset{\frac{\pi }{2}}{\int }}x.\sin x dx$
(Using integration by parts, taking$x^2=\mathrm{u},\cos xdx=dv$)
$=\frac{{\pi }^2}{4}-2\left[-{\left(x.\cos x\right)}_0^{\frac{\pi }{2}}+ \underset{0}{\overset{\frac{\pi }{2}}{\int }}1.\cos xdx\right]=\frac{{\pi }^2}{4}-2\left[0+1\right]=\left(\frac{{\pi }^2}{4}-2\right)$