Question

\[\int \sin^{- 1} \sqrt{x}\ dx\]

Answer 1

\[\text{We have}, \]

\[I = \int \sin^{- 1} \sqrt{x} dx\]

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

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

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

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

\[ \therefore I = \int \theta \text{ sin } \left( 2\theta \right)d\theta\]

\[ = \theta\left[ \frac{- \text{ cos }2\theta}{2} \right] - \int1\left( \frac{- \text{ cos }2\theta}{2} \right)d\theta\]

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

\[ = - \frac{\theta \text{ cos} \left( 2\theta \right)}{2} + \frac{1}{2}\left[ \frac{\text{ sin} \left( 2\theta \right)}{2} \right] + C\]

\[ = \frac{- \sin^{- 1} \sqrt{x} \left( 1 - 2 \text{ sin}^2 \theta \right)}{2} + \frac{1}{2}\left[ \frac{2 \sin \theta \cos \theta}{2} \right] + C\]

\[ = \frac{- \sin \sqrt{x}\left( 1 - 2x \right)}{2} + \frac{\sin \theta\sqrt{1 - \sin^2 \theta}}{2} + C\]

\[ = \frac{- \sin^{- 1} \sqrt{x} \left( 1 - 2x \right)}{2} + \frac{\sqrt{x} \sqrt{1 - x}}{2} + C\]

\[ = - \frac{1}{2} \sin^{- 1} \left( \sqrt{x} \right) \left( 1 - 2x \right) + \frac{1}{2}\sqrt{x - x^2} + C\]