Integral exercise asked in the math group

Exercise:

Compute, $\int_{0}^{\pi }{\frac{{{x}^{2}}\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)}{2x-\pi }dx}$

Solution: Let $I=\int_{0}^{\pi }{\frac{{{x}^{2}}\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)}{2x-\pi }dx}$

So $I=\int_{0}^{\pi }{\frac{{{\left( \pi -x \right)}^{2}}\sin 2\left( \pi -x \right)\sin \left( \frac{\pi }{2}\cos \left( \pi -x \right) \right)}{2\left( \pi -x \right)-\pi }dx}$

$=\int_{0}^{\pi }{\frac{{{\left( \pi -x \right)}^{2}}\sin \left( 2\pi -2x \right)\sin \left( -\frac{\pi }{2}\cos x \right)}{\pi -2x}dx}=-\int_{0}^{\pi }{\frac{{{\left( \pi -x \right)}^{2}}\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)}{2x-\pi }dx}$

$=-\int_{0}^{\pi }{\frac{\left( {{\pi }^{2}}-2x\pi +{{x}^{2}} \right)\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)}{2x-\pi }dx}$

$=-\int_{0}^{\pi }{\frac{\pi \left( \pi -2x \right)\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)+{{x}^{2}}\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)}{2x-\pi }dx}$

$=\int_{0}^{\pi }{\pi \sin 2x\sin \left( \frac{\pi }{2}\cos x \right)dx-\int_{0}^{\pi }{\frac{{{x}^{2}}\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)}{2x-\pi }dx}}$

$\Rightarrow 2I=\pi \int_{0}^{\pi }{\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)dx}\Leftrightarrow I=\frac{\pi }{2}\int_{0}^{\pi }{\sin 2x\sin \left( \frac{\pi }{2}\cos x \right)dx}$

Let $u=\frac{\pi }{2}\cos x\Leftrightarrow du=-\frac{\pi }{2}\sin x\,dx$ and $\sin 2x=2\sin x\cos x$

So $I=\frac{\pi }{2}\int_{0}^{\pi }{2\sin x\cos x\sin \left( \frac{\pi }{2}\cos x \right)dx}=\pi \int_{0}^{\pi }{\sin x\cos x\sin \left( \frac{\pi }{2}\cos x \right)dx}$

$=-\pi \int_{\frac{\pi }{2}}^{-\frac{\pi }{2}}{\frac{2}{\pi }\times \frac{2}{\pi }u\sin udu=\frac{4}{\pi }\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{u\sin u\,du}}=\frac{8}{\pi }$

Let $U=u\,\And \,dV=\sin udu\Leftrightarrow dU=du\,\And \,V=-\cos u$

$\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{u\sin u\,du}=-\left( u\cos u \right)_{\frac{-\pi }{2}}^{\frac{\pi }{2}}+\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\cos udu}=0+\left( \sin u \right)_{\frac{-\pi }{2}}^{\frac{\pi }{2}}=1+1=2$