Question

The integral $ \int_{0}^{1}{\frac{2{{\sin }^{-1}}\frac{x}{2}}{x}}dx $ equals

• A. $ \int_{0}^{\pi /6}{\frac{xdx}{\tan x}} $
• B. $ \int_{0}^{\pi /6}{\frac{2x}{\tan x}}dx $
• C. $ \int_{0}^{\pi /2}{\frac{2xdx}{\tan x}} $
• D. $ \int_{0}^{\pi /6}{\frac{xdx}{\sin x}} $
• E. $ \int_{0}^{\pi /6}{\frac{2x}{\sin x}}dx $

Answer 1

Answer: (B)

$ z=\int_{0}^{1}{\frac{2{{\sin }^{-1}}\frac{x}{2}}{x}}dx $ Put $ {{\sin }^{-1}}\frac{x}{2}=t $
$ \Rightarrow $ $ x=2\sin t $
$ \Rightarrow $ $ dx=2\cos tdt $
Also, $ x=0\Rightarrow t=0 $ $ x=1\Rightarrow t=\frac{\pi }{6} $
$ \therefore $ $ I=\int_{0}^{\pi /6}{\frac{2t}{2\sin t}}.2\cos t\,dt $
$=\int_{0}^{\pi /6}{\frac{2t}{\tan t}}dt=\int_{0}^{\pi /6}{\frac{2x}{\tan x}}dx $