Question

The value of the integral $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}(x^2+\log \frac{\pi -x}{\pi +x})\cos xdx$

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

Answer 1

Answer: (B)

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\left[x^2+\log \left(\frac{\pi -x}{\pi +x}\right)\right]\cos xdx$
As, $\underset{-a}{\overset{a}{\int }}f(x)dx=0$, when $f(-x)=-f(x)$
$\therefore I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{x}^2\cos xdx+0=2\underset{0}{\overset{\pi /2}{\int }}\left(x^2\cos x\right)dx$
$=2\left\{{\left(x^2\sin x\right)}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}2x\cdot \sin xdx\right\}$
$=2\left[\frac{{\pi }^2}{4}-2\left\{(-x\cdot \cos x{)}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}1\cdot (-\cos x)dx\right\}\right]$
$=2\left[\frac{{\pi }^2}{4}-2(\sin x{)}_0^{\pi /2}\right]=2\left[\frac{{\pi }^2}{4}-2\right]=\left(\frac{{\pi }^2}{2}-4\right)$