Q. If $\int_{0}^{\pi}{x\ f}\left( \sin x \right)dx = A\int_{0}^{\pi/2}{f(\sin{x)dx,\ then\ A}}$ is equal to
A) 0
B) $\pi$
C) $\frac{\pi}{4}$
D) $2\pi$
Explanation
(b)
Let $I = \int_{0}^{\pi}{x\ f(\sin{x)dx\ \ }}$…(i)
$= \int_{0}^{\pi}{(\pi - x)f\left\lbrack \sin(\pi - x) \right\rbrack dx}$
$\Rightarrow I = \int_{0}^{\pi}{(\pi - x)f(\sin{x)dx\ }}$…(ii)
On adding Eqs. (i) and (ii), we get
$2I = \int_{0}^{\pi}{\pi f(\sin{x)dx = 2\pi\int_{0}^{\pi/2}{f(\sin{x)dx}}}}$
$\Rightarrow I = \pi\int_{0}^{\pi/2}{f(\sin{x)dx = A\int_{0}^{\pi/2}{f(\sin{x)dx}}}}$
$\therefore A = \pi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.