Answer
$$\frac{3}{8}\theta + \frac{1}{8}\sin 4\theta + \frac{1}{{64}}\sin 8\theta + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^4}2\theta } d\theta \cr
& {\text{applying }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr
& = \int {{{\left( {{{\cos }^2}2\theta } \right)}^2}} d\theta \cr
& {\text{trigonometric identity co}}{{\text{s}}^2}x = \frac{{1 + \cos 2x}}{2} \cr
& = \int {{{\left( {\frac{{1 + \cos 4\theta }}{2}} \right)}^2}} d\theta = \frac{1}{4}\int {{{\left( {1 + \cos 4\theta } \right)}^2}} d\theta \cr
& {\text{expand}} \cr
& = \frac{1}{4}\int {\left( {1 + 2\cos 4\theta + {{\cos }^2}4\theta } \right)} d\theta \cr
& {\text{trigonometric identity co}}{{\text{s}}^2}x = \frac{{1 + \cos 2x}}{2} \cr
& = \frac{1}{4}\int {\left( {1 + 2\cos 4\theta + \frac{{1 + \cos 8\theta }}{2}} \right)} d\theta \cr
& = \frac{1}{4}\int {\left( {1 + 2\cos 4\theta + \frac{1}{2} + \frac{{\cos 8\theta }}{2}} \right)} d\theta \cr
& = \frac{1}{4}\int {\left( {\frac{3}{2} + 2\cos 4\theta + \frac{{\cos 8\theta }}{2}} \right)} d\theta \cr
& {\text{integrating}} \cr
& = \frac{1}{4}\left( {\frac{3}{2}\theta + \frac{2}{4}\sin 4\theta + \frac{1}{{16}}\sin 8\theta } \right) + C \cr
& = \frac{3}{8}\theta + \frac{2}{{16}}\sin 4\theta + \frac{1}{{64}}\sin 8\theta + C \cr
& = \frac{3}{8}\theta + \frac{1}{8}\sin 4\theta + \frac{1}{{64}}\sin 8\theta + C \cr} $$
