Q. The integral $\int_{0}^{1}{\frac{2\sin^{- 1}\frac{x}{2}}{x}dx}$ equals
A) $\int_{0}^{\pi/6}{\frac{x}{\tan x}dx}$
B) $\int_{0}^{\pi/6}{\frac{2x}{\tan x}dx}$
C) $\int_{0}^{\pi/2}{\frac{2x}{\tan x}dx}$
D) $\int_{0}^{\pi/6}{\frac{x}{\sin x}dx}$
Explanation
(b)
Let $I = \int_{0}^{1}{\frac{2\sin^{- 1}\frac{x}{2}}{x}dx}$
Put ${\sin^{- 1}\frac{x}{2} = t \Rightarrow \frac{x}{2} = sin\ t}\$
$\Rightarrow x = 2\sin I$
$\Rightarrow dx = 2\cos{t\ dt}$
$\therefore I = \int_{0}^{\frac{\pi}{6}}{\frac{2t}{2\sin t}.2\cos{t\ dt}}$
$= \int_{0}^{\frac{\pi}{6}}{\frac{2t}{\tan t}dt}$
$= \int_{0}^{\frac{\pi}{6}}{\frac{2x}{\tan x}dx}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.