Answer
$$ - \frac{1}{a}\cos a\theta + \frac{1}{{3a}}{\cos ^3}a\theta + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}a\theta d\theta } \cr
& {\text{split off }}{\sin ^3}a\theta d\theta \cr
& = \int {{{\sin }^2}a\theta \sin a\theta d\theta } \cr
& {\text{identity }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = \int {\left( {1 - {{\cos }^2}a\theta } \right)\sin a\theta } d\theta \cr
& = \int {\left( {\sin a\theta - {{\cos }^2}a\theta \sin a\theta } \right)} d\theta \cr
& {\text{sum rule}} \cr
& = \int {\sin a\theta } d\theta - \int {{{\cos }^2}a\theta } \sin a\theta d\theta \cr
& u = \cos a\theta ,{\text{ }}du = - a\sin a\theta d\theta \cr
& = \int {\sin a\theta } d\theta + \frac{1}{a}\int {{u^2}} du \cr
& {\text{find antiderivatives}} \cr
& = - \frac{1}{a}\cos a\theta + \frac{1}{{3a}}{u^3} + C \cr
& {\text{write in terms of }}\theta ,{\text{ replace }}u = \cos a\theta \cr
& = - \frac{1}{a}\cos a\theta + \frac{1}{{3a}}{\cos ^3}a\theta + C \cr} $$
