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