Answer
$$\frac{{{{\cos }^6}x}}{6} - \frac{{{{\cos }^4}x}}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}x{{\cos }^3}x} dx \cr
& {\text{split off }}{\sin ^3}x \cr
& = \int {{{\sin }^2}x{{\cos }^3}x} \sin xdx \cr
& {\text{identity }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^3}x} \sin xdx \cr
& = \int {\left( {{{\cos }^3}x - {{\cos }^5}x} \right)} \sin xdx \cr
& {\text{substitute }}u = \cos x,{\text{ }}du = - \sin xdx \cr
& = \int {\left( {{u^3} - {u^5}} \right)} \left( { - du} \right) \cr
& = \int {\left( {{u^5} - {u^3}} \right)du} \cr
& {\text{find the antiderivatives by the power rule}} \cr
& = \frac{{{u^6}}}{6} - \frac{{{u^4}}}{4} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \cos x \cr
& = \frac{{{{\cos }^6}x}}{6} - \frac{{{{\cos }^4}x}}{4} + C \cr} $$
