Answer
$$ - 8\sqrt {4 - {w^2}} + \frac{{8{{\left( {4 - {w^2}} \right)}^{3/2}}}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{w^3}}}{{\sqrt {4 - {w^2}} }}} dw \cr
& {\text{substitute }}w = 2\sin \theta ,{\text{ }}dw = 2\cos \theta d\theta \cr
& \int {\frac{{{w^3}}}{{\sqrt {4 - {w^2}} }}} dw = \int {\frac{{{{\left( {2\sin \theta } \right)}^3}}}{{\sqrt {4 - {{\left( {2\sin \theta } \right)}^2}} }}} \left( {2\cos \theta } \right)d\theta \cr
& = \int {\frac{{8{{\sin }^3}\theta }}{{2\sqrt {{{\cos }^2}\theta } }}} \left( {2\cos \theta } \right)d\theta \cr
& = 8\int {{{\sin }^3}\theta } d\theta \cr
& = 8\int {\sin \theta } {\sin ^2}\theta d\theta \cr
& {\text{use }}{\sin ^2}\theta = 1 - {\cos ^2}\theta \cr
& = 8\int {\sin \theta } \left( {1 - {{\cos }^2}\theta } \right)d\theta \cr
& = 8\int {\left( {\sin \theta + {{\cos }^2}\theta \left( { - \sin \theta } \right)} \right)} d\theta \cr
& {\text{find the antiderivative}} \cr
& = 8\left( { - \cos \theta + \frac{{{{\cos }^3}\theta }}{3}} \right) + C \cr
& = - 8\cos \theta + \frac{{8{{\cos }^3}\theta }}{3} + C \cr
& {\text{write in terms of }}w \cr
& = - 8\sqrt {4 - {w^2}} + \frac{{8{{\left( {4 - {w^2}} \right)}^{3/2}}}}{3} + C \cr} $$
