Mathematics

$$\displaystyle\int _{ 0 }^{ 1 }{ \dfrac { 2\sin ^{ -1 }{ \dfrac { x }{ 2 }  }  }{ x }  } dx$$ is equal to


ANSWER

$$\displaystyle\int _{ 0 }^{ \pi /6 }{ \cfrac { 2x }{ \tan { x } } } dx$$


SOLUTION
$$I=\displaystyle\int^1_0\dfrac{2\sin^{-1}\dfrac{x}{2}}{x}dx$$
put $$\sin^{-1}\dfrac{x}{2}=t$$
$$\Rightarrow \dfrac{x}{2}=\sin t$$
$$\Rightarrow x=2\sin t$$
$$\therefore dx=2\cos t dt$$
As $$x+0, t\rightarrow 0$$
$$x\rightarrow 1, t\rightarrow \dfrac{\pi}{6}$$
$$=\displaystyle\int^{\dfrac{\pi}{6}}_0\dfrac{2t}{2\sin t}2\cos tdt$$
$$=\displaystyle\int^{\dfrac{\pi}{6}}_0\dfrac{2t}{\tan t}dt$$
$$=\displaystyle\int^{\dfrac{\pi}{6}}_0\dfrac{2x}{\tan x}dx$$.
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Single Correct Medium Published on 17th 09, 2020
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