Question

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

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

Answer 1

Answer: (B)

$\underset{-\pi /2}{\overset{\pi /2}{\int }}\left\{x^2+\ln \left(\frac{\pi +x}{\pi -x}\right)\right\}\cos xdx$
$=\underset{-\pi /2}{\overset{\pi /2}{\int }}{x}^2\cos xdx+\underset{-\pi /2}{\overset{\pi /2}{\int }}\ln \left(\frac{\pi +x}{\pi -x}\right)\cos xdx$
$=2\underset{0}{\overset{\pi /2}{\int }}{x}^2\cos xdx$
$=2{\left[x^2\sin x+2x\cos x-2\sin x\right]}_0^{\pi /2}$
$=2\left[\frac{{\pi }^2}{4}-2\right]=\frac{{\pi }^2}{2}-4$.