Mathematics - Definite Integration Question with Solution | TestHub

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