$\int_{}^{}{\cos{\left\{ 2 an^{- 1}\sqrt{\frac{1 - x}{1 + x}} ight\} dx}}$ is equal to

Question

$\int_{}^{}{\cos{\left\{ 2	an^{- 1}\sqrt{\frac{1 - x}{1 + x}} ight\} dx}}$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Let $I = \int_{}^{}{\cos{\left\{ 2	an^{- 1}\sqrt{\frac{1 - x}{1 + x}} ight\} dx}}$
Put $x = \cos	heta$, then
$I = \int_{}^{}{\cos{\left\{ 2	an^{- 1}\sqrt{\frac{1 - \cos	heta}{1 + \cos	heta}} ight\} dx}}$
$= \int_{}^{}{\cos{\left\{ 2	an^{- 1}\left( 	an\frac{	heta}{2} ight) ight\} dx}}$
$= \int_{}^{}{\cos	heta}dx = \int_{}^{}xdx = \frac{x^{2}}{2} + c$

Q. $\int_{}^{}{\cos{\left\{ 2\tan^{- 1}\sqrt{\frac{1 - x}{1 + x}} \right\} dx}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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