Answer
\[I=\frac{\sin^{-1}x}{2}+\frac{x\sqrt{1-x^2}}{2}+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int \sqrt{1-x^2}dx\]
Make a substitution $ \: x=\sin \theta\;\;\;...(1)$
\[\Rightarrow dx=\cos\theta \: d\theta\]
\[I=\int \sqrt{1-\sin^2 \theta}\cos \theta\:d\theta\]
\[I=\int\cos^2 \theta\:d \theta\]
\[I=\int\frac{1+\cos2 \theta}{2}d \theta\]
\[I=\frac{ \theta}{2}+\frac{\sin 2 \theta}{4}\]
\[I=\frac{\theta}{2}+\frac{2\sin\theta\cos\theta}{4}\]
Using (1)
\[I=\frac{\sin^{-1}x}{2}+\frac{x\sqrt{1-x^2}}{2}+C\]
Where $C$ is constant of integration
Hence,
\[I=\frac{\sin^{-1}x}{2}+\frac{x\sqrt{1-x^2}}{2}+C\]
