Question

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

• A. $\frac{{\pi }^2}{4}-2$
• B. $\frac{{\pi }^2}{4}+2$
• C. ${\pi }^2-e^{\frac{\pi }{2}}$
• D. ${\pi }^2+e^{\frac{\pi }{2}}$

Answer 1

The correct option is A $\frac{{\pi }^2}{4}-2$

$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{x^2\cos x}{1+e^x}dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}(\frac{x^2\cos x}{1+e^x}+\frac{x^2\cos x}{1+e^{-x}})dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{x^2\cos x+e^x{x}^2\cos x}{1+e^x}\right)dx$
$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{x}^2\cos xdx$
$\Rightarrow I=[x^2\sin x{]}_0^{\frac{\pi }{2}}-\underset{0}{\overset{\frac{\pi }{2}}{\int }}2x\sin xdx$
$\Rightarrow I=[x^2\sin x{]}_0^{\frac{\pi }{2}}-2[{[-x\cos x]}_0^{\frac{\pi }{2}}-\underset{0}{\overset{\frac{\pi }{2}}{\int }}(-\cos x\left)dx\right]$
$\Rightarrow I=\frac{{\pi }^2}{4}-2[-(0-0)+[\sin x{]}_0^{\frac{\pi }{2}}]$
$\therefore I=\frac{{\pi }^2}{4}-2$