Answer
$\frac{1}{3}\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} - \sqrt {1 - {x^2}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {1 - {x^2}} }}dx} ,{\text{ }}x = \sin \theta \cr
& {\text{Let }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& {\text{Substituting}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {1 - {x^2}} }}dx} = \int {\frac{{{{\left( {\sin \theta } \right)}^3}}}{{\sqrt {1 - {{\left( {\sin \theta } \right)}^2}} }}\left( {\cos \theta } \right)d\theta } \cr
& = \int {\frac{{{{\sin }^3}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}\left( {\cos \theta } \right)d\theta } \cr
& {\text{Use the identity si}}{{\text{n}}^2}\theta + {\cos ^2}\theta = 1 \cr
& = \int {\frac{{{{\sin }^3}\theta }}{{\sqrt {{{\cos }^2}\theta } }}\left( {\cos \theta } \right)d\theta } \cr
& = \int {\frac{{{{\sin }^3}\theta }}{{\cos \theta }}\left( {\cos \theta } \right)d\theta } \cr
& = \int {{{\sin }^3}\theta d\theta } \cr
& {\text{Split }}{\sin ^3}\theta d\theta {\text{ as }}{\sin ^2}\theta \sin \theta d\theta \cr
& = \int {{{\sin }^2}\theta \sin \theta d\theta } \cr
& {\text{Rewrite }}{\cos ^2}t{\text{ using }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \int {\left( {1 - {{\cos }^2}\theta } \right)\sin \theta d\theta } \cr
& {\text{Let }}u = \cos \theta ,{\text{ }}du = - \sin \theta d\theta \cr
& = - \int {\left( {1 - {u^2}} \right)du} \cr
& = \int {\left( {{u^2} - 1} \right)du} \cr
& {\text{Integrate}} \cr
& = \frac{{{u^3}}}{3} - u + C \cr
& {\text{Write in terms of }}\theta ,{\text{ }}u = \cos \theta \cr
& = \frac{1}{3}{\cos ^3}\theta - \cos \theta + C \cr
& {\text{Write in terms of }}x,{\text{ use the triangle shown below}} \cr
& \cos \theta = \sqrt {1 - {x^2}} ,{\text{ then}} \cr
& = \frac{1}{3}{\left( {\sqrt {1 - {x^2}} } \right)^3} - \sqrt {1 - {x^2}} + C \cr
& = \frac{1}{3}\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} - \sqrt {1 - {x^2}} + C \cr} $$
