Answer
$$4\arcsin \left( {\frac{x}{2}} \right) + x\sqrt {4 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {16 - 4{x^2}} } dx \cr
& {\text{Rewrite}} \cr
& = \int {\sqrt {4\left( {4 - {x^2}} \right)} } dx \cr
& = 2\int {\sqrt {4 - {x^2}} } dx \cr
& {\text{Let }}u = x,{\text{ and }}du = dx \cr
& = {\text{2}}\int {\sqrt {4 - {x^2}} } dx \cr
& {\text{Using trigonometric substitution}} \cr
& 2\sin \theta = x,{\text{ }}dx = 2\cos \theta d\theta \cr
& {\text{2}}\int {\sqrt {4 - {x^2}} } dx = {\text{2}}\int {\sqrt {4 - 4{{\sin }^2}\theta } } \left( {2\cos \theta } \right)d\theta \cr
& = {\text{2}}{\left( 2 \right)^2}\int {\sqrt {{{\cos }^2}\theta } } \left( {\cos \theta } \right)d\theta \cr
& = 8\int {{{\cos }^2}\theta d\theta } \cr
& {\text{Using the power reducing formula }}{\cos ^2}u = \frac{{1 + \cos 2u}}{2} \cr
& = 8\int {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr
& = 8\int {\left( {1 + \cos 2\theta } \right)} d\theta \cr
& {\text{Integrating}} \cr
& = 8\left( {\theta + \frac{1}{4}\sin 2\theta } \right) + C \cr
& = 8\left( {\theta + \frac{1}{2}\sin \theta \cos \theta } \right) + C \cr
& = 8\theta + 4\sin \theta \cos \theta + C \cr
& {\text{Write in terms of }}x \cr
& = 8\left( {\frac{1}{2}} \right)\arcsin \left( {\frac{x}{2}} \right) + 4\left( {\frac{x}{2}} \right)\left( {\frac{{\sqrt {4 - {x^2}} }}{2}} \right) + C \cr
& = 4\arcsin \left( {\frac{x}{2}} \right) + x\sqrt {4 - {x^2}} + C \cr} $$
