Question

$\int {\sin }^{-1}\sqrt{x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int {\sin }^{-1}\sqrt{x}dx \\ \mathrm{Putting}\sqrt{x}=\sin \theta \\ \Rightarrow x={\sin }^2\theta \\ \Rightarrow dx=2\sin \theta \cos \theta d\theta \\ \Rightarrow dx=\sin \left(2\theta \right)d\theta \\ \therefore I=\int \theta \sin \left(2\theta \right)d\theta \\ =\theta \left[\frac{-\cos 2\theta }{2}\right]-\int 1\left(\frac{-\cos 2\theta }{2}\right)d\theta \\ =-\frac{\theta \cos \left(2\theta \right)}{2}+\frac{1}{2}\int \cos \left(2\theta \right)d\theta \\ =-\frac{\theta \cos \left(2\theta \right)}{2}+\frac{1}{2}\left[\frac{\sin \left(2\theta \right)}{2}\right]+\mathrm{C} \\ =\frac{-{\sin }^{-1}\sqrt{x}\left(1-2{\sin }^2\theta \right)}{2}+\frac{1}{2}\left[\frac{2\sin \theta \cos \theta }{2}\right]+C \\ =\frac{-\sin \sqrt{x}\left(1-2x\right)}{2}+\frac{\sin \theta \sqrt{1-{\sin }^2\theta }}{2}+C \\ =\frac{-{\sin }^{-1}\sqrt{x}\left(1-2x\right)}{2}+\frac{\sqrt{x}\sqrt{1-x}}{2}+C \\ =-\frac{1}{2}{\sin }^{-1}\left(\sqrt{x}\right)\left(1-2x\right)+\frac{1}{2}\sqrt{\mathrm{x}-x^{\mathit{2}}}+C \end{gathered}$$