Question

The value of the integral $\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^2+\ln \left(\frac{\pi +x}{\pi -x}\right))\cos xdx$ is

• A. $0$
• B. $\frac{{\pi }^2}{2}-4$
• C. $\frac{{\pi }^2}{2}+4$
• D. $\frac{{\pi }^2}{2}$

Answer 1

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

$\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^2+\ln \left(\frac{\pi +x}{\pi -x}\right))\cos xdx$
$=\underset{-\pi /2}{\overset{\pi /2}{\int }}{x}^2\cos xdx+\underset{-\pi /2}{\overset{\pi /2}{\int }}\ln \left(\frac{\pi +x}{\pi -x}\right)\cos xdx$
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$=2\underset{0}{\overset{\pi /2}{\int }}{x}^2\cos xdx+0$
Applying successive by-part,
$=2[x^2\sin x+2x\cos x-2\sin x{]}_0^{\pi /2}$
$=2[\frac{{\pi }^2}{4}-2]=\frac{{\pi }^2}{2}-4$