Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \sqrt{\frac{1-x}{1+x}}dx \\ =\int \sqrt{\frac{\left(1-x\right)\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}}dx \\ =\int \left(\frac{1-x}{\sqrt{1-x^2}}\right)dx \\ =\int \frac{dx}{\sqrt{1-x^2}}-\int \frac{xdx}{\sqrt{1-x^2}} \\ \mathrm{Putting}1-x^2=t \\ \Rightarrow -2xdx=dt \\ \Rightarrow xdx=-\frac{dt}{2} \\ \mathrm{Then}, \\ I=\int \frac{dx}{\sqrt{1-x^2}}+\frac{1}{2}\int \frac{dt}{\sqrt{t}} \\ ={\sin }^{-1}\left(x\right)+\frac{1}{2}\times 2\sqrt{t}+C \\ ={\sin }^{-1}\left(x\right)+\sqrt{1-x^2}+C \end{gathered}$$