${\cos ^4}2x$

Let $I = \int {{{\cos }^4}2x\,dx} = \int {{{\left( {\cfrac{{1 + \cos 4x}}{2}} \right)}^2}dx}$

$= \cfrac{1}{4}\int {\left( {1 + {{\cos }^2}4x + 2\cos 4x} \right)dx}$

$= \cfrac{1}{4}\int {\left[ {1 + \cfrac{{1 + \cos 8x}}{2} + 2\cos 4x} \right]dx}$

$= \cfrac{3}{8}\int {dx} + \cfrac{1}{8}\int {\cos 8x\,dx} + \cfrac{1}{2}\int {\cos 4x\,dx} \cdot$

$= \cfrac{3}{8}x + \cfrac{1}{{64}}\sin 8x + \cfrac{1}{8}\sin 4x + C$