Question

\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\sqrt{\frac{1 - x}{1 + x}} dx\]
\[ = \int\sqrt{\frac{\left( 1 - x \right)\left( 1 - x \right)}{\left( 1 + x \right)\left( 1 - x \right)}} dx\]
\[ = \int\left( \frac{1 - x}{\sqrt{1 - x^2}} \right) dx\]
\[ = \int\frac{dx}{\sqrt{1 - x^2}} - \int\frac{x dx}{\sqrt{1 - x^2}}\]
\[\text{ Putting }1 - x^2 = t\]
\[ \Rightarrow \text{ - 2x dx } = dt\]
\[ \Rightarrow \text{ x dx }= - \frac{dt}{2}\]
\[\text{ Then, } \]
\[I = \int\frac{dx}{\sqrt{1 - x^2}} + \frac{1}{2}\int\frac{dt}{\sqrt{t}}\]
\[ = \sin^{- 1} \left( x \right) + \frac{1}{2} \times 2\sqrt{t} + C\]
\[ = \sin^{- 1} \left( x \right) + \sqrt{1 - x^2} + C\]