Answer
$$\frac{1}{2}\theta + \frac{1}{{16}}\sin 8\theta + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^2}4\theta } d\theta \cr
& {\text{identity co}}{{\text{s}}^2}x = \frac{{1 + \cos 2x}}{2} \cr
& = \int {\frac{{1 + \cos 8\theta }}{2}} d\theta \cr
& = \int {\left( {\frac{1}{2} + \frac{{\cos 8\theta }}{2}} \right)} d\theta \cr
& = \int {\frac{1}{2}} d\theta + \int {\frac{{\cos 8\theta }}{2}} d\theta \cr
& = \int {\frac{1}{2}} d\theta + \frac{1}{8}\int {\frac{{8\cos 8\theta }}{2}} d\theta \cr
& {\text{find antiderivatives}} \cr
& = \frac{1}{2}\theta + \frac{1}{{16}}\sin 8\theta + C \cr} $$
