Answer
$$\frac{1}{3}\sin \left( {3x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\cos 3x} dx \cr
& {\text{set }}u = 3x{\text{ then }}\frac{{du}}{{dx}} = 3,\,\,\,\,\,\,\,\,\frac{{du}}{3} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\cos 3x} dx = \int {\cos u} \left( {\frac{{du}}{3}} \right) \cr
& {\text{use multiple constant rule}} \cr
& = \frac{1}{3}\int {\cos u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\cos x} dx = \sin x + C \cr
& = \frac{1}{3}\sin u + C \cr
& {\text{write in terms of }}x.{\text{ Replacing }}3x{\text{ for }}u{\text{ gives}} \cr
& = \frac{1}{3}\sin \left( {3x} \right) + C \cr} $$
