Question

Evaluate ${\int }_0^{\pi }\frac{x^2\sin 2x.\sin \left(\frac{\pi }{2}\cos x\right)dx}{\left(2x-\pi \right)}$

Answer 1

$\underset{0}{\overset{\pi }{\int }}\frac{n\sin 2x.\sin \left(\pi /2\cos x\right)}{2x-\pi }dx$

$I=\underset{0}{\overset{\pi }{\int }}\frac{\left(\pi -x\right)\sin \left(2\pi -2x\right)\sin \left(\frac{\pi }{2}\cos \left(\pi -x\right)\right)dx}{2\left(\pi -x\right)-\pi }$

$=\underset{0}{\overset{\pi }{\int }}\frac{\left(\pi -x\right)\left(-\sin 2x\right)\sin \left(\pi /2\left(-\cos x\right)\right)dx}{-2x+\pi }$

= $=\underset{0}{\overset{\pi }{\int }}\frac{\left(x-\pi \right)\sin 2x\sin \left(\frac{\pi }{2}\cos x\right)}{\pi -2x}dx$

$2I=\underset{0}{\overset{\pi }{\int }}\frac{\left(2x-\pi \right)\sin 2x\sin \left(\frac{\pi }{2}\cos x\right)dx}{2x-\pi }$

$=\underset{0}{\overset{\pi }{\int }}2\sin x\cos x.\sin \left(\frac{\pi }{2}\cos x\right)dx$

$\frac{\pi }{2}\cos x=t\Rightarrow -\sin dx=\frac{2dt}{\pi }$

$=2\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{-2dt}{\pi }\left(\frac{2t}{\pi }\right)\sin t.dt$

$=\frac{4}{{\pi }^2}{\int }_{-\pi /2}^{\pi /2}+\sin t.dt=\frac{8}{{\pi }^2}{\int }_0^{\pi /2}t\sin tdt$

$=\frac{8}{{\pi }^2}\left(\sin \frac{\pi }{2}-\sin 0\right)=\frac{8}{{\pi }^2}$