Question

Evaluate: $\int \sqrt{\frac{1-x}{1+x}}dx$.

Answer 1

$=\int \sqrt{\frac{1-x}{1+x}}dx$

$=\int \sqrt{\frac{1-x}{1+x}\times \frac{1-x}{1-x}}dx$

$=\int \frac{1-x}{\sqrt{1-x^2}}dx$

$=\int \frac{1}{\sqrt{1-x^2}}dx-\int \frac{x}{\sqrt{1-x^2}}dx$

$={\sin }^{-1}x+\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$={\sin }^{-1}x+\frac{1}{2}\left(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right)$

$={\sin }^{-1}x+\frac{1}{2}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C$

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

$={\sin }^{-1}x+\sqrt{1-x^2}+C$. where $C$ is constant.