Question

If $I=\underset{0}{\overset{\pi }{\int }}\frac{x^2\sin 2x\sin \left(\frac{\pi }{2}\cos x\right)dx}{2x-\pi }$, then the value of $\frac{\pi }{2}I$ is

Answer 1

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

Now, add both equations
$2I=\underset{0}{\overset{\pi }{\int }}\pi \sin 2x\sin (\frac{\pi }{2}\cos x)dx$
$\Rightarrow 2I=2\pi \underset{0}{\overset{\pi }{\int }}\sin x\cos x\sin (\frac{\pi }{2}\cos x)dx$
Assume $\frac{\pi }{2}\cos x=t\Rightarrow -\frac{\pi }{2}\sin xdx=dt$

$\Rightarrow I=\underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}-\frac{4t}{\pi }\sin tdt\Rightarrow I=\frac{4}{\pi }\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}t\sin tdt\Rightarrow I=\frac{8}{\pi }[-t\cos t+\sin t{]}_0^{\frac{\pi }{2}}\Rightarrow I=\frac{8}{\pi }$

Hence, $\frac{\pi }{2}\times I=4$