Question

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

Answer 1

\[\text{We have}, \]

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

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

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

\[ \Rightarrow dx = - 2 \cos\theta \sin\text{ θ  dθ }\]

\[ \Rightarrow dx = - \text{ sin}\left( 2\theta \right) \text{  dθ }\]

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

\[ = \int\sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} \left( - 2 \sin\theta \cos\theta \right) d\theta\]

\[ = \int\left( \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \right) \left( - 2 \times 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\cos\theta \right) \text{   dθ }\]

\[ = - 4\int \sin^2 \frac{\theta}{2} \times \cos\text{ θ  dθ }\]

\[ = - 4\int\left( \frac{1 - \cos\theta}{2} \right) \cos\text{ θ  dθ }\]

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

\[ = - 2\int\left\{ \cos\theta - \left( \frac{1 + \cos 2\theta}{2} \right) \right\}d\theta\]

\[ = - 2\int\cos \text{ θ  dθ } + \int\left( 1 + \cos 2\theta \right) d\theta\]

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

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

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

\[ = - 2\sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{1 - x}\sqrt{x} + C\]

\[ = - 2\sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x}\sqrt{1 - x} + C\]