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