Q. The value of $\int_{}^{}\frac{2\ dx}{\sqrt{1 - 4x^{2}}}$ is
A) $\tan^{- 1}{(2x) + c}$
B) $\cot^{- 1}{(2x) + c}$
C) $\cos^{- 1}{(2x) + c}$
D) $\sin^{- 1}{(2x) + c}$
Explanation
(d)
Let $I = \int_{}^{}\frac{2dx}{\sqrt{1 - 4x^{2}}}$
Put $2x = \sin{\theta\ \Rightarrow 2dx = \cos{\theta d\theta}}$
$\therefore I = \int_{}^{}\frac{\cos{\theta d\theta}}{1 - \sin^{2}\theta}$
$= \int_{}^{}\frac{\cos\theta}{\cos\theta}d\theta = \int_{}^{}{1\ d\theta}$
$\Rightarrow I = \theta + c$
$\therefore I = \sin^{- 1}{(2x) + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.