Answer
$$ - \sqrt {16 - {y^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ydy}}{{\sqrt {16 - {y^2}} }}} \cr
& \left( {\bf{a}} \right){\text{using the substitution method}} \cr
& \,\,\,\,{\text{Let }}16 - {y^2} = u,\,\,\,\,\, - 2ydy = du,\,\,\,\,dy = - \frac{{du}}{{2y}} \cr
& \int {\frac{{ydy}}{{\sqrt {16 - {y^2}} }}} = \int {\frac{y}{{\sqrt u }}} \left( { - \frac{{du}}{{2y}}} \right) \cr
& = - \int {\frac{{du}}{{\sqrt u }}} = - \frac{1}{2}\int {{u^{ - 1/2}}du} \cr
& {\text{integrate using the power rule}} \cr
& = - \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = - \sqrt u + C \cr
& {\text{write in terms of }}y,{\text{ substitute }}16 - {y^2}{\text{ for }}u \cr
& = - \sqrt {16 - {y^2}} + C \cr
& \cr
& \left( {\bf{b}} \right){\text{using a trigonometric substitution}} \cr
& \,\,\,\,\,{\text{Let }}y = 4\sin \theta ,\,\,\,dy = 4\cos \theta d\theta \cr
& \,\,\,\,\,\int {\frac{{ydy}}{{\sqrt {16 - {y^2}} }}} = \int {\frac{{4\sin \theta \left( {4\cos \theta } \right)d\theta }}{{\sqrt {16 - {{\left( {4\sin \theta } \right)}^2}} }}} \cr
& = \int {\frac{{16\sin \theta \cos \theta }}{{\sqrt {16 - 16{{\sin }^2}\theta } }}d\theta } \cr
& = \int {\frac{{16\sin \theta \cos \theta }}{{4\sqrt {1 - {{\sin }^2}\theta } }}d\theta } \cr
& = 4\int {\frac{{\sin \theta \cos \theta }}{{\sqrt {{{\cos }^2}\theta } }}d\theta } \cr
& = 4\int {\sin \theta d\theta } \cr
& {\text{Integrate}} \cr
& = 4\left( { - \cos \theta } \right) + C \cr
& = - 4\cos \theta + C \cr
& {\text{where cos}}\theta = \frac{{\sqrt {16 - {y^2}} }}{4} \cr
& = - 4\left( {\frac{{\sqrt {16 - {y^2}} }}{4}} \right) + C \cr
& = - \sqrt {16 - {y^2}} + C \cr} $$
