Question

Let $f\left(x\right)=\frac{\sin 2x\cdot \sin \left(\frac{\pi }{2}\cos x\right)}{2x-\pi }$

Then answer the following question.

${\int }_0^{\pi }{x}^{2}f\left(x\right)dx$

• A. $0$
• B. $\frac{8}{\pi }$
• C. $\frac{16}{{\pi }^2}$
• D. $\frac{4}{{\pi }^2}$

Answer 1

The correct option is B $\frac{8}{\pi }$

Since ${\int }_0^{\pi }f\left(x\right)dx={\int }_0^{\pi /2}(f\left(x\right)+f(\pi -x))dx$

$\therefore$ $\int x^{2}f\left(x\right)dx=\pi {\int }_0^{\pi /2}\sin 2x\sin \left(\frac{\pi }{2}\cos x\right)dx$

$=2\pi {\int }_0^{1}t\sin \left(\frac{\pi }{2}t\right)dx$ (putting $\cos x=t$)

$=2\pi {[t\cos \frac{\pi }{2}t\times \frac{2}{\pi }]}_0^1-2\pi {\int }_0^1\left(\cos \frac{\pi }{2}t\right)(-\frac{2}{\pi })dt$

$=0+\frac{2}{\pi }\left(2\pi \right){\left(\sin \frac{\pi }{2}t\right)}_0^1\times \frac{2}{\pi }$

$=\frac{8}{\pi }$