Answer
$$\sin x - \frac{{{{\sin }^3}x}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^3}x} dx \cr
& {\text{split off }}\cos x \cr
& = \int {{{\cos }^2}x} \cos xdx \cr
& {\text{pythagorean identity}} \cr
& = \int {\left( {1 - {{\sin }^2}x} \right)\cos x} dx \cr
& u = \sin x,{\text{ }}du = \cos xdx \cr
& = \int {\left( {1 - {u^2}} \right)} du \cr
& {\text{evaluate the integral}} \cr
& = u - \frac{{{u^3}}}{3} + C \cr
& {\text{replace }}u = \cos x \cr
& = \sin x - \frac{{{{\sin }^3}x}}{3} + C \cr} $$
