Question

Evaluate: $\int \frac{1}{\sqrt{1-x^2}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{\sqrt{1-x^2}} \\ \mathrm{Let}x=\sin \theta \\ \Rightarrow dx=\cos \theta \\ \therefore I=\int \frac{\cos \theta }{\cos \theta }d\theta \\ =\int d\theta \\ =\theta +C \\ ={\sin }^{-1}x+\mathrm{C}\left(\because x=\sin \theta \right) \end{gathered}$$