Question

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

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

Answer 1

Answer: (B)

$\frac{{\pi }^2}{2}-4$

$I={\int }_{-\pi /2}^{\pi /2}[x^2+log\left(\frac{\pi -x}{\pi +x}\right)]cosxdx$

As,${\int }_{-a}^{a}f\left(x\right)dx=0,$when$f(-x)=-f\left(x\right)$

$\therefore I={\int }_{-\pi /2}^{\pi /2}{x}^{2}cosxdx+0=2{\int }_0^{\pi /2}\left(x^{2}cosx\right)dx$

$=2\left\{\right(x^{2}sinx{)}_0^{\pi /2}-{\int }_0^{\pi /2}\left(x^{3}cosx\right)\}$

$2[\frac{{\pi }^2}{4}-2\left\{\right(-xcosx{)}_0^{\pi /2}-{\int }_0^{\pi /2}1.(-cosx)dx\}]$

$=2[\frac{{\pi }^2}{4}-2(sinx{)}_0^{\pi /2}]=2[\frac{{\pi }^2}{4}-2]=(\frac{{\pi }^2}{2}-4)$