$\sin^4 x + \cos^4 x = (\sin^2 x +\cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \dfrac{1}{2}\sin^2 2x$
$I= \int \cos 2x (1 -\dfrac{1}{2}\sin^2 2x) dx = \int \cos 2x dx -\dfrac{1}{2}\int \cos 2x \sin^2 2x dx$
$= \dfrac{1}{4}\int \cos 2x d(2x) - \dfrac{1}{2}\int \sin^2 2x d(\sin 2x)$
$= \dfrac{1}{4}\sin 2x - \dfrac{1}{6}\sin^3 2x + C$