Question

If $I={\int }_0^{\pi }\frac{x\sin 2x\cdot \sin \left(\frac{\pi }{2}\cos x\right)}{(2x-\pi )}dx,$ then the value of ${\pi }^{2}I=$

Answer 1

Given : $I={\int }_0^{\pi }\frac{x\sin 2x\cdot \sin \left(\frac{\pi }{2}\cos x\right)}{(2x-\pi )}dx\cdots \left(i\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)\sin 2x\cdot \sin \left(\frac{\pi }{2}\cos x\right)}{(\pi -2x)}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right):$
$\Rightarrow 2I={\int }_0^{\pi }\sin 2x\cdot \sin \left(\frac{\pi }{2}\cos x\right)dx\Rightarrow 2I=2{\int }_0^{\pi /2}\sin 2x\cdot \sin \left(\frac{\pi }{2}\cos x\right)dx$
[$\because f(\pi -x)=f\left(x\right)$ ]
Substitute $\frac{\pi }{2}\cos x=t,-\frac{\pi }{2}\sin xdx=dt$
$\Rightarrow I=\frac{8}{{\pi }^2}\underset{0}{\overset{\pi /2}{\int }}t\sin tdt\Rightarrow I=\frac{8}{{\pi }^2}[-t\cos t+\sin t{]}_0^{\pi /2}\Rightarrow I=\frac{8}{{\pi }^2}\therefore {\pi }^{2}I=8$