Question

${\int }_0^1\frac{2{\sin }^{-1}\frac{x}{2}}{x}dx$ is equal to

• 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. None of these

Answer 1

Answer: (B)

${\int }_0^{\pi /6}\frac{2x}{\tan x}dx$

$I={\int }_0^1\frac{2{\sin }^{-1}\frac{x}{2}}{x}dx$

put ${\sin }^{-1}\frac{x}{2}=t$

$\Rightarrow \frac{x}{2}=\sin t$

$\Rightarrow x=2\sin t$

$\therefore dx=2\cos tdt$

As $x+0,t\to 0$

$x\to 1,t\to \frac{\pi }{6}$

$={\int }_0^{\frac{\pi }{6}}\frac{2t}{2\sin t}2\cos tdt$

$={\int }_0^{\frac{\pi }{6}}\frac{2t}{\tan t}dt$

$={\int }_0^{\frac{\pi }{6}}\frac{2x}{\tan x}dx$.