Answer
$${\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {16 - {x^2}} \right)}^{1/2}}}}} \cr
& {\text{substitute }}x = 4\sin \theta \cr
& = \int {\frac{{4\cos \theta d\theta }}{{{{\left( {16 - 16{{\sin }^2}\theta } \right)}^{1/2}}}}} \cr
& = \int {\frac{{4\cos \theta d\theta }}{{4{{\left( {1 - {{\sin }^2}\theta } \right)}^{1/2}}}}} \cr
& {\text{pythagorean identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \int {\frac{{4\cos \theta d\theta }}{{4{{\left( {{{\cos }^2}\theta } \right)}^{1/2}}}}} \cr
& = \int {d\theta } \cr
& {\text{integrating}} \cr
& = \theta + C \cr
& x = 4\sin \theta ,{\text{ then}} \cr
& = {\sin ^{ - 1}}\left( {\frac{x}{4}} \right) + C \cr} $$