Q. Let $I_{1} = \int_{a}^{\pi - a}{xf(\sin{x)\ dx,\ I_{2} = \int_{a}^{\pi - a}{f\ (\sin{x)\ dx,}}}}$ then $I_{2}$ is equal to
A) $\frac{\pi}{2}I_{1}$
B) $\pi I_{1}$
C) $\frac{2}{\pi}I_{1}$
D) $2I_{1}$
Explanation
(c)
Given,
$I_{1} = \int_{a}^{\pi - a}{x\ f(\sin{x)\ dx}}$
and $I_{2} = \int_{a}^{\pi - a}{\ f(\sin{x)\ dx}}$
Now, $I_{1} = \int_{a}^{\pi - a}{x\ f(\sin{x)\ dx}}$
$= \int_{a}^{\pi - a}{(\pi - x)f\lbrack\sin{(\pi - x)\rbrack\ dx}}$
$= \int_{a}^{\pi - a}{(\pi - x)f{(sin}{x)\ dx}}$
$= \int_{a}^{\pi - a}{\pi\ f(\sin{x)dx - I_{1}}}$
$\Rightarrow 2I_{1} = \pi\ I_{2}\ \Rightarrow I_{2} = \frac{2}{\pi}I_{1}\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.