_0^ 1 / 2 x ^ -1 x 1-x^2 d x= - Vidyadip

MCQ

$\int_0^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=$

A
$\frac{1}{2}+\frac{\sqrt{3} \pi}{12}$
✓
$\frac{1}{2}-\frac{\sqrt{3} \pi}{12}$
C
$\frac{1}{2}+\frac{\sqrt{3 \pi}}{12}$
D
None of these

✓

Answer

Correct option: B.

$\frac{1}{2}-\frac{\sqrt{3} \pi}{12}$

(B)
Put $t =\sin ^{-1} x \Rightarrow dt =\frac{1}{\sqrt{1-x^2}} d x$
$\therefore \quad \int_0^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=\int_0^{\pi / 6} t \sin t dt$
$=[- t \cos t +\sin t ]_0^{\frac{\pi}{6}}$
$\begin{array}{l}=-\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \\ =\frac{1}{2}-\frac{\sqrt{3} \pi}{12}\end{array}$

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