Answer
$$ - 16\sqrt {16 - {x^2}} + \frac{1}{3}\left( {16 - {x^2}} \right)\sqrt {16 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {16 - {x^2}} }}} dx \cr
& {\text{Using substitution }}x = 4\sin \theta \cr
& x = 4\sin \theta ,{\text{ }}dx = 4\cos \theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {16 - {x^2}} }}} dx = \int {\frac{{{{\left( {4\sin \theta } \right)}^3}}}{{\sqrt {16 - {{\left( {4\sin \theta } \right)}^2}} }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{{{{\left( {4\sin \theta } \right)}^3}}}{{\sqrt {16 - 16{{\sin }^2}\theta } }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{{{{\left( {4\sin \theta } \right)}^3}}}{{\sqrt {16\left( {1 - {{\sin }^2}\theta } \right)} }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{{{{\left( {4\sin \theta } \right)}^3}}}{{\sqrt {16{{\cos }^2}\theta } }}} \left( {4\cos \theta } \right)d\theta \cr
& = \int {\frac{{64{{\sin }^3}\theta }}{{4\cos \theta }}} \left( {4\cos \theta } \right)d\theta \cr
& = 64\int {{{\sin }^3}\theta } d\theta \cr
& = 64\int {{{\sin }^2}\theta } \sin \theta d\theta \cr
& {\text{Use the pythagorean identity co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1 \cr
& = 64\int {\left( {1 - {{\cos }^2}\theta } \right)} \sin \theta d\theta \cr
& = 64\int {\sin \theta } d\theta - 64\int {{{\cos }^2}\theta } \sin \theta d\theta \cr
& = 64\int {\sin \theta } d\theta + 64\int {{{\cos }^2}\theta } \left( { - \sin \theta } \right)d\theta \cr
& {\text{Integrating}} \cr
& = - 64\cos \theta + 64\left( {\frac{{{{\cos }^3}\theta }}{3}} \right) + C \cr
& = - 64\cos \theta + \frac{{64}}{3}{\cos ^3}\theta + C \cr
& {\text{Where }}\cos {\text{ }}\theta = \frac{{\sqrt {16 - {x^2}} }}{4} \cr
& = - 64\left( {\frac{{\sqrt {16 - {x^2}} }}{4}} \right) + \frac{{64}}{3}{\left( {\frac{{\sqrt {16 - {x^2}} }}{4}} \right)^3} + C \cr
& = - 16\sqrt {16 - {x^2}} + \frac{{64}}{3}\left( {\frac{{{{\left( {\sqrt {16 - {x^2}} } \right)}^3}}}{{64}}} \right) + C \cr
& = - 16\sqrt {16 - {x^2}} + \frac{1}{3}\left( {16 - {x^2}} \right)\sqrt {16 - {x^2}} + C \cr} $$
