$\begin{align}  & \int{\frac{1}{\sqrt{4-{{x}^{2}}}}dx}=\int{\frac{1}{\sqrt{{{2}^{2}}-{{x}^{2}}}}dx}=\arcsin \frac{x}{2}+C \\ & \text{Alternatively Let }x=2\sin \theta  \\ & \frac{dx}{d\theta }=2\cos \theta ,\text{ }dx=2\cos \theta d\theta  \\ & \text{ }\int{\frac{1}{\sqrt{4-{{x}^{2}}}}dx}=\int{\frac{2\cos \theta }{\sqrt{4-4{{\sin }^{2}}\theta }}d\theta } \\ & \int{\frac{1}{\sqrt{4-{{x}^{2}}}}dx}=\int{\frac{2\cos \theta }{2\sqrt{1-{{\sin }^{2}}\theta }}d\theta } \\ & \int{\frac{1}{\sqrt{4-{{x}^{2}}}}dx}=\int{\frac{2\cos \theta }{2\sqrt{{{\cos }^{2}}\theta }}d\theta } \\ & \int{\frac{1}{\sqrt{4-{{x}^{2}}}}dx}=\int{\frac{\cos \theta }{\cos \theta }d\theta }=\int{d\theta }=\theta +C \\ & \text{since }x=2\sin \theta ,\text{ }\theta ={{\sin }^{-1}}\frac{x}{2} \\ & \int{\frac{1}{\sqrt{4-{{x}^{2}}}}dx}={{\sin }^{-1}}\frac{x}{2}+C \\\end{align}$