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प्रश्न
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
उत्तर
\[\text{ Let I} = \int\frac{dx}{\sqrt{1 - x^2}}\]
\[\text{ Let x }= \sin \theta\]
\[ \Rightarrow dx = \cos \theta\]
\[ \therefore I = \int\frac{\cos \theta}{\cos \theta}d\theta\]
\[ = \int d\theta\]
\[ = \theta + C\]
\[ = \sin^{- 1} x + C \left( \because x = \sin \theta \right)\]
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