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, $x=0\Rightarrow t=0$ $x=1\Rightarrow t=\frac{\pi }{6}$
$\therefore$ $I={\int }_0^{\pi /6}\frac{2t}{2\sin t}.2\cos tdt$
$={\int }_0^{\pi /6}\frac{2t}{\tan t}dt={\int }_0^{\pi /6}\frac{2x}{\tan x}dx$