Question

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

Answer 1

\[\text{ Let I }= \int \sin^{- 1} \sqrt{x} dx\]
\[ = \int \frac{\sqrt{x} . \sin^{- 1} \sqrt{x}}{\sqrt{x}}dx\]
\[\text{ Let} \sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]
\[ \therefore I = \int t_{II} . \sin^{- 1} t_I dt\]
\[ = \sin^{- 1} t\int t\ dt - \int\left\{ \frac{d}{dt}\left( \sin^{- 1} t \right)\int t\ dt \right\}dt\]
\[ = 2 \left[ \sin^{- 1} t . \frac{t^2}{2} - \int \frac{1}{\sqrt{1 - t^2}} \times \frac{t^2}{2}dt \right]\]
\[ = \sin^{- 1} t . t^2 - \int \frac{t^2}{\sqrt{1 - t^2}}dt\]
\[ = \sin^{- 1} t . t^2 + \int\left( \frac{1 - t^2 - 1}{\sqrt{1 - t^2}} \right)dt\]
\[ = \sin^{- 1} t . t^2 + \int \sqrt{1 - t^2} dt - \int \frac{dt}{\sqrt{1 - t^2}}\]
\[ = \sin^{- 1} t . t^2 + \frac{t}{2}\sqrt{1 - t^2} + \frac{1}{2} \sin^{- 1} t - \sin^{- 1} t + C\]
\[ = \sin^{- 1} t . t^2 + \frac{t}{2}\sqrt{1 - t^2} - \frac{1}{2} \sin^{- 1} t + C\]
\[ = x . \sin^{- 1} \sqrt{x} + \frac{\sqrt{x}}{2} \sqrt{1 - x} - \frac{1}{2} \sin^{- 1} \left( \sqrt{x} \right) + C \left( \because \sqrt{x} = t \right)\]
\[ = \frac{\left( 2x - 1 \right) \sin^{- 1} \sqrt{x}}{2} + \frac{\sqrt{x - x^2}}{2} + C\]