Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {5{x^3}\sqrt {1 - {x^2}} } dx \cr
& {\text{substitute }}x = \sin \theta ,{\text{ }}\sqrt {1 - {x^2}} = \cos \theta ,\,\,\,dx = \cos \theta d\theta \,\,\,\,\theta = {\sin ^{ - 1}}x \cr
& \,\,\,x = 1 \to \theta = {\sin ^{ - 1}}\left( 1 \right) = \pi /2 \cr
& \,\,\,x = 0 \to \theta = {\sin ^{ - 1}}\left( 0 \right) = 0 \cr
& \cr
& {\text{write the integral in terms of }}\theta \cr
& \int_0^1 {5{x^3}\sqrt {1 - {x^2}} } dx = \int_0^{\pi /2} {5{{\sin }^3}\theta \cos \theta } \cos \theta d\theta \cr
& = \int_0^{\pi /2} {5{{\sin }^3}\theta {{\cos }^2}\theta } d\theta \cr
& = 5\int_0^{\pi /2} {{{\sin }^2}\theta {{\cos }^2}\theta \sin \theta } d\theta \cr
& \cr
& {\text{use the pythagorean identity co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1 \cr
& = 5\int_0^{\pi /2} {\left( {1 - {{\cos }^2}\theta } \right){{\cos }^2}\theta \sin \theta } d\theta \cr
& = 5\int_0^{\pi /2} {\left( {{{\cos }^2}\theta - {{\cos }^4}\theta } \right)\sin \theta } d\theta \cr
& \cr
& {\text{Integrate}} \cr
& = - 5\left[ {\frac{{{{\cos }^3}\theta }}{3} - \frac{{{{\cos }^5}\theta }}{5}} \right]_0^{\pi /2} \cr
& \cr
& {\text{evaluate the limits}} \cr
& = - 5\left[ {\frac{{{{\cos }^3}\left( {\pi /2} \right)}}{3} - \frac{{{{\cos }^5}\left( {\pi /2} \right)}}{5}} \right] + 5\left[ {\frac{{{{\cos }^3}\left( 0 \right)}}{3} - \frac{{{{\cos }^5}\left( 0 \right)}}{5}} \right] \cr
& = - 5\left[ 0 \right] + 5\left[ {\frac{1}{3} - \frac{1}{5}} \right] \cr
& = \frac{5}{3} - 1 \cr
& = \frac{2}{3} \cr} $$
