Question

$\int \sqrt{\frac{1+x}{1-x}}dx$ is/are equal to (where $C$ is integration constant)

• A. $-{\cos }^{-1}x-\sqrt{1-x^2}+C$
• B. ${\cos }^{-1}x+\sqrt{1-x^2}+C$
• C. ${\sin }^{-1}x-\sqrt{1-x^2}+C$
• D. $\sqrt{1-x^2}+{\sin }^{-1}x+C$

Answer 1

Answer: (A), (C)

${\sin }^{-1}x-\sqrt{1-x^2}+C$

$I=\int \sqrt{\frac{1+x}{1-x}}dx$
Let $x=\cos 2\theta$
$\Rightarrow dx=-2\sin 2\theta d\theta$
$\Rightarrow I=\int \sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}(-2\sin 2\theta )d\theta$
$=-2\int \frac{\cos \theta }{\sin \theta }\left(2\cos \theta \sin \theta \right)d\theta$
$=-4\int {\cos }^2\theta d\theta =-2\int (1+\cos 2\theta )d\theta =-2\int d\theta -2\int \cos 2\theta d\theta$
$=-2\theta -\sin 2\theta +C$
$=-\sqrt{1-x^2}-{\cos }^{-1}x+C$
Replacing ${\cos }^{-1}x$ with $\frac{\pi }{2}-{\sin }^{-1}x,$ we get
$=-\sqrt{1-x^2}-\frac{\pi }{2}+{\sin }^{-1}x+C$
$={\sin }^{-1}x-\sqrt{1-x^2}+C$