Question

\[\int\sqrt{\frac{x}{1 - x}} dx\]  is equal to

विकल्प

• \[\sin^{- 1} \sqrt{x} + C\]
• \[\sin^{- 1} \left\{ \sqrt{x} - \sqrt{x \left( 1 - x \right)} \right\} + C\]
• \[\sin^{- 1} \left\{ \sqrt{x \left( 1 - x \right)} \right\} + C\]
• \[\sin^{- 1} \sqrt{x} - \sqrt{x \left( 1 - x \right)} + C\]

Answer 1

\[\sin^{- 1} \sqrt{x} - \sqrt{x \left( 1 - x \right)} + C\]

 

 

\[\text{Let }I = \int\sqrt{\frac{x}{1 - x}}dx\]

\[\text{Putting }\sqrt{x} = \sin \theta\]

\[ \Rightarrow x = \sin^2 \theta\]

\[ \Rightarrow dx = 2 \sin \theta \cos \theta d\theta\]

\[ \Rightarrow dx = \sin \left( 2\theta \right) d\theta\]

\[ \therefore I = \int\sqrt{\frac{\sin^2 \theta}{1 - \sin^2 \theta}} \times \sin \left( 2\theta \right) \cdot d\theta\]

\[ = \int\frac{\sin \theta}{\cos \theta} \times 2 \sin \theta \cdot \cos \theta d\theta\]

\[ = \int2 \sin^2 \theta \cdot d\theta\]

\[ = \int\left( 1 - \cos 2\theta \right)d\theta\]

\[ = \theta - \frac{\sin \left( 2\theta \right)}{2} + C\]

\[ = \theta - \frac{2 \sin \theta \cos \theta}{2} + C \]

\[ = \theta - \sin \theta \sqrt{1 - \sin^2 \theta} + C\]

\[ = \sin^{- 1} \sqrt{x} - \sqrt{x} \sqrt{1 - x} + C ...........\left( \because \theta = \sin^{- 1} \sqrt{x} \right)\]

\[ = \sin^{- 1} \sqrt{x} - \sqrt{x\left( 1 - x \right)} + C\]