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