$\int \cos^4 (2t)dt$
$u = 2t$ $\frac {1}{2}\int \cos^4 u du$ $\frac {1}{2}\int \frac{(1+ \cos^2u)^2}{2} du$
$\frac {1}{8}\int 1 - \cos^2 2u du$
$ \frac{1}{8} \int \sin^2 u du$
From here I am not sure what to do. Nothing seems to simplify the problem, only complicate it.