Answer
$${\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} \cr
& \left( {\bf{a}} \right){\text{ Integrate by tables}} \cr
& \,\,\,\,\,\,{\text{Rewrite the radicand }}9 - {x^2} \cr
& = \int {\frac{{\left( {1/3} \right)dx}}{{\sqrt {\frac{9}{9} - \frac{{{x^2}}}{9}} }}} \cr
& = \int {\frac{{\left( {1/3} \right)dx}}{{\sqrt {1 - \frac{{{x^2}}}{9}} }}} = \int {\frac{{\left( {1/3} \right)dx}}{{\sqrt {1 - {{\left( {\frac{x}{3}} \right)}^2}} }}} \cr
& {\text{Let }}u = \frac{x}{3},\,\,\,du = \frac{1}{3}dx.{\text{ Then}}{\text{,}} \cr
& \int {\frac{{\left( {1/3} \right)dx}}{{\sqrt {1 - {{\left( {\frac{x}{3}} \right)}^2}} }}} = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{Use the formula }}\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x + C} \cr
& \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} = {\sin ^{ - 1}}u + C \cr
& {\text{Replace back }}u = \frac{x}{3} \cr
& = {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr
& \cr
& \left( {\bf{b}} \right){\text{Integrate using trigonometric substitution}} \cr
& \,\,\,\,\,{\text{Let }}x = 3\sin \theta ,\,\,\,dx = 3\cos \theta d\theta \cr
& {\text{Write in terms of }}\theta \cr
& \,\,\,\,\int {\frac{{dx}}{{\sqrt {9 - {x^2}} }}} = \int {\frac{{3\cos \theta d\theta }}{{\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}} \cr
& \,\,\, = \int {\frac{{3\cos \theta d\theta }}{{\sqrt {9 - 9{{\sin }^2}\theta } }}} \cr
& \,\,{\text{Factor and simplify}} \cr
& \,\,\, = \int {\frac{{3\cos \theta d\theta }}{{\sqrt {9\left( {1 - {{\sin }^2}\theta } \right)} }}} \cr
& \,\,\, = \int {\frac{{3\cos \theta d\theta }}{{3\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& \,\,{\text{Use si}}{{\text{n}}^2}\theta + {\cos ^2}\theta = 1 \cr
& \,\,\, = \int {\frac{{3\cos \theta d\theta }}{{3\sqrt {{{\cos }^2}\theta } }}} \cr
& \,\,\, = \int {d\theta } \cr
& {\text{Integrating}} \cr
& = \theta + C \cr
& {\text{We know that }}x = 3\sin \theta ,\,\, then\,\,\,\theta = {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) \cr
& = {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr} $$
