Question

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

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

Answer 1

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

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{x}^2+\ln \left(\frac{\pi +x}{\pi -x}\right)\cos xdx$

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{x}^2\cos xdx+{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\ln \left(\frac{\pi +x}{\pi -x}\right)\cos xdx$

$={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{x}^2\cos xdx+0$ ....... $[\because \ln \left(\frac{\pi +x}{\pi -x}\right)\cos x\text{is an odd function}]$
$=2{\int }_0^{\frac{\pi }{2}}{x}^2\cos xdx$ ....... $[\because x^2\cos x\text{is an even function}]$
$=2[x^2\sin x+2x\cos x-2\sin x{]}_0^{\frac{\pi }{2}}$
$=2[\frac{{\pi }^2}{4}-2]=\frac{{\pi }^2}{2}-4$.