Answer
$${\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt {16 - {x^2}} }}} dx \cr
& {\text{Refer to the triangle below}} \cr
& \sin \theta = \frac{x}{4} \cr
& x = 4\sin \theta ,{\text{ }}dx = 4\cos \theta d\theta ,{\text{ }} \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{\sqrt {16 - {x^2}} }}} dx = \int {\frac{1}{{\sqrt {16 - {{\left( {4\sin \theta } \right)}^2}} }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{1}{{\sqrt {16 - 16{{\sin }^2}\theta } }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{1}{{\sqrt {16\left( {1 - {{\sin }^2}\theta } \right)} }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{1}{{\sqrt {16{{\cos }^2}\theta } }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{1}{{4\cos \theta }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {d\theta } \cr
& {\text{Integrating}} \cr
& = \theta + C \cr
& x = 4\sin \theta \Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + C \cr
& = {\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + C \cr} $$
