Question

Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]

Answer 1

\[\text{ Let I} = \int\frac{dx}{\sqrt{1 - x^2}}\]

\[\text{ Let  x }= \sin \theta\]

\[ \Rightarrow dx = \cos \theta\]

\[ \therefore I = \int\frac{\cos \theta}{\cos \theta}d\theta\]

\[ = \int d\theta\]

\[ = \theta + C\]

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