The value of 2 , ,dx 1 - 4 x^2 is - Vidyadip

MCQ

The value of $\int {\frac{{2\,\,dx}}{{\sqrt {1 - 4{x^2}} }}} $ is

A
${\tan ^{ - 1}}(2x) + c$
B
${\cot ^{ - 1}}(2x) + c$
C
${\cos ^{ - 1}}(2x) + c$
✓
${\sin ^{ - 1}}(2x) + c$

✓

Answer

Correct option: D.

${\sin ^{ - 1}}(2x) + c$

d
(d) $I = \int {\frac{{2dx}}{{\sqrt {1 - 4{x^2}} }}} $.

Put $2x = \sin \theta $ ==> $2dx = \cos \theta \,\,d\theta $
$ \Rightarrow I = \int {\frac{{\cos \theta \,\,d\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }} = \int {\frac{{\cos \theta }}{{\cos \theta }}d\theta = \int {d\theta + c = \theta + c} } } $.
Therefore, $I = {\sin ^{ - 1}}(2x) + c.$

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