Answer
$$\left( {4{x^4} - 12{x^2} + 6} \right)\sin 2x + \left( {8{x^3} - 12x} \right)\cos 2x + C$$
Work Step by Step
$$\eqalign{
& \int {8{x^4}\cos 2x} dx \cr
& {\text{Integrate by tabulation, see image below}} \cr
& = 8{x^4}\left( {\frac{1}{2}\sin 2x} \right) + 32{x^3}\left( {\frac{1}{4}\cos 2x} \right) + 96{x^2}\left( { - \frac{1}{8}\sin 2x} \right) \cr
& - 192x\left( { - \frac{1}{{16}}\cos 2x} \right) + 192\left( {\frac{1}{{32}}\sin 2x} \right) + C \cr
& = 4{x^4}\sin 2x + 8{x^3}\cos 2x - 12{x^2}\sin 2x + 12x\cos 2x + 6\sin 2x + C \cr
& {\text{Factoring}} \cr
& \left( {4{x^4} - 12{x^2} + 6} \right)\sin 2x + \left( {8{x^3} - 12x} \right)\cos 2x + C \cr} $$
