Answer
$$\frac{1}{2}x + \frac{1}{{12}}\sin 6x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^2}3x} dx \cr
& {\text{identity }}{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2},{\text{ }}\theta = 3x \cr
& = \int {\frac{{1 + \cos 2\left( {3x} \right)}}{2}} dx \cr
& = \int {\frac{{1 + \cos 6x}}{2}} dx \cr
& = \int {\left( {\frac{1}{2} + \frac{{\cos 6x}}{2}} \right)} d\theta \cr
& {\text{sum rule}} \cr
& = \int {\frac{1}{2}} dx + \frac{1}{2}\int {\cos 6x} dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}x + \frac{1}{2}\left( {\frac{1}{6}\sin 6x} \right) + C \cr
& = \frac{1}{2}x + \frac{1}{{12}}\sin 6x + C \cr} $$
