MCQ
$\int {\frac{{2\,\,dx}}{{\sqrt {1 - 4{x^2}} }}} $ =
- A${\tan ^{ - 1}}(2x) + c$
- B${\cot ^{ - 1}}(2x) + c$
- C${\cos ^{ - 1}}(2x) + c$
- ✓${\sin ^{ - 1}}(2x) + c$
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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