Question

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

Answer 1

\[\text{We have},\]
\[I = \int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[ \Rightarrow I = \int x\sqrt{\frac{\left( 1 - x \right)\left( 1 - x \right)}{\left( 1 + x \right)\left( 1 - x \right)}} \text{ dx }\]
\[ \Rightarrow I = \int x\frac{\left( 1 - x \right)}{\sqrt{1 - x^2}} \text{ dx }\]
\[ \Rightarrow I = \int\frac{x - x^2}{\sqrt{1 - x^2}} \text{ dx }\]
\[ \Rightarrow I = \int\frac{x - x^2 - 1 + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[ \Rightarrow I = \int\frac{- x^2 + 1}{\sqrt{1 - x^2}} dx + \int\frac{x - 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[ \Rightarrow I = \int\sqrt{1 - x^2} dx + \int\frac{x}{\sqrt{1 - x^2}} dx - \int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
\[ \Rightarrow I = \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2} \sin^{- 1} \left( x \right) + C_1 - \sqrt{1 - x^2} + C_2 - \sin^{- 1} \left( x \right) + C_3 ....................\left[ \because \int\frac{x}{\sqrt{1 - x^2}}\text{  dx }= - \sqrt{1 - x^2} + C_2 \right]\]
\[ \Rightarrow I = \sqrt{1 - x^2}\left( \frac{x}{2} - 1 \right) - \frac{1}{2} \sin^{- 1} \left( x \right) + C\]