Q. $\int_{}^{}{\cos{\left\lbrack 2\cot^{- 1}\sqrt{\frac{1 - x}{1 + x}} \right\rbrack dx}}$ is equal to
A) $\frac{1}{2}x^{2} + c$
B) $\frac{1}{2}\sin\left\lbrack 2\cot^{- 1}\sqrt{\frac{1 - x}{1 + x}} \right\rbrack + c$
C) $- \frac{1}{2}x^{2} + c$
D) $\frac{1}{2}x + c$
Explanation
(c)
Let $I = \int_{}^{}{\cos{\left\lbrack 2\cot^{- 1}\sqrt{\frac{1 - x}{1 + x}} \right\rbrack dx}}$
$= \int_{}^{}{\cos{\left\lbrack 2\tan^{- 1}\sqrt{\frac{1 + x}{1 - x}} \right\rbrack dx}}$
$= \int_{}^{}{\cos{\left\lbrack \cos^{- 1}{( - x)} \right\rbrack dx}}\ \left\lbrack \because 2\tan^{- 1}x = \cos^{- 1}\left( \frac{1 - x^{2}}{1 + x^{2}} \right) \right\rbrack$
$= - \frac{x^{2}}{2} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.