Answer
$$\frac{{25\sqrt 5 }}{2}{\sin ^{ - 1}}\left( {\frac{x}{{\sqrt 5 }}} \right) - \frac{{5\sqrt 5 x\sqrt {5 - {x^2}} }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^3}\sqrt {5 - {x^2}} } dx \cr
& {\text{substitute }}x = \sqrt 5 \sin \theta ,{\text{ }}dx = \sqrt 5 \cos \theta d\theta \cr
& = \int {{{\left( {\sqrt 5 } \right)}^3}\sqrt {5 - {{\left( {\sqrt 5 \sin \theta } \right)}^2}} } \left( {\sqrt 5 \cos \theta } \right)d\theta \cr
& {\text{simplify}} \cr
& = 25\int {\sqrt {5 - {{\left( {\sqrt 5 \sin \theta } \right)}^2}} } \cos \theta d\theta \cr
& = 25\sqrt 5 \int {\sqrt {1 - {{\sin }^2}\theta } } \cos \theta d\theta \cr
& = 25\sqrt 5 \int {\sqrt {{{\cos }^2}\theta } \cos \theta } d\theta \cr
& = 25\sqrt 5 \int {{{\cos }^2}\theta } d\theta \cr
& = 25\sqrt 5 \int {\frac{{1 + \cos 2\theta }}{2}} d\theta \cr
& = 25\sqrt 5 \int {\left( {\frac{1}{2} - \frac{{\cos 2\theta }}{2}} \right)} d\theta \cr
& {\text{integrating}} \cr
& = 25\sqrt 5 \left( {\frac{\theta }{2} - \frac{{\sin 2\theta }}{4}} \right) + C \cr
& = 25\sqrt 5 \left( {\frac{\theta }{2} - \frac{{\sin \theta \cos \theta }}{2}} \right) + C \cr
& {\text{write in terms of }}x \cr
& = 25\sqrt 5 \left( {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 5 }}} \right) - \frac{1}{2}\left( {\frac{x}{{\sqrt 5 }}} \right)\left( {\frac{{\sqrt {5 - {x^2}} }}{{\sqrt 5 }}} \right)} \right) + C \cr
& = 25\sqrt 5 \left( {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{x}{{\sqrt 5 }}} \right) - \frac{{x\sqrt {5 - {x^2}} }}{{10}}} \right) + C \cr
& = \frac{{25\sqrt 5 }}{2}{\sin ^{ - 1}}\left( {\frac{x}{{\sqrt 5 }}} \right) - \frac{{5\sqrt 5 x\sqrt {5 - {x^2}} }}{2} + C \cr} $$
