_0^1 ^ -1 (x 2 ) x d x= - Vidyadip

MCQ

$\int_0^1 \frac{\sin ^{-1}\left(\frac{x}{2}\right)}{x} d x=$

✓
$\int_0^{\frac{\pi}{6}} \frac{x d x}{\tan x}$
B
$\int_0^{\frac{\pi}{6}} \frac{2 x}{\tan x} d x$
C
$\int_0^{\frac{\pi}{2}} \frac{2 x d x}{\tan x}$
D
$\int_0^{\frac{\pi}{6}} \frac{x d x}{\sin x}$

✓

Answer

Correct option: A.

$\int_0^{\frac{\pi}{6}} \frac{x d x}{\tan x}$

(A)
Put $\sin ^{-1}\left(\frac{x}{2}\right)= t \Rightarrow x=2 \sin t \Rightarrow d x=2 \cos tdt$
$\therefore \int_0^1 \frac{\sin ^{-1}\left(\frac{x}{2}\right)}{x} d x$
$=\int_0^{\frac{\pi}{6}} \frac{t}{(2 \sin t)}(2 \cos t d t) \\ =\int_0^{\frac{\pi}{6}} \frac{t}{\tan t} d t$
$=\int_0^{\frac{\pi}{6}} \frac{x}{\tan x} d x \quad \ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( t ) dt \right]$

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