Answer
$$ - 2{x^4}\cos 2x + 4{x^3}\sin 2x + 6{x^2}\cos 2x - 6x\sin 2x - 3\cos 2x + C$$
Work Step by Step
$$\eqalign{
& \int {4{x^4}\sin 2x} dx \cr
& {\text{Evaluate using tabular integration by parts}} \cr
& \frac{d}{{dx}}{\text{ }}\int {dx} \cr
& 4{x^4}{\text{ + }}\sin 2x \cr
& 16{x^3}{\text{ }} - {\text{ }} - \frac{1}{2}\cos 2x \cr
& 48{x^2}{\text{ + }} - \frac{1}{4}\sin 2x \cr
& 96x{\text{ }} - {\text{ }}\frac{1}{8}\cos 2x \cr
& 96{\text{ + }}\frac{1}{{16}}\sin 2x \cr
& {\text{0 }} - {\text{ }} - \frac{1}{{32}}\cos 2x \cr
& \cr
& {\text{Then,}} \cr
& = \left( {4{x^4}{\text{ }}} \right)\left( + \right)\left( { - \frac{1}{2}\cos 2x} \right) + \left( {16{x^3}} \right)\left( - \right)\left( { - \frac{1}{4}\sin 2x} \right) \cr
& + \left( {48{x^2}} \right)\left( + \right)\left( {\frac{1}{8}\cos 2x} \right) + 96x\left( - \right)\left( {\frac{1}{{16}}\sin 2x} \right) + 96\left( + \right)\left( { - \frac{1}{{32}}\cos 2x} \right) \cr
& + C \cr
& {\text{Simplifying}} \cr
& = - 2{x^4}\cos 2x + 4{x^3}\sin 2x + 6{x^2}\cos 2x - 6x\sin 2x - 3\cos 2x + C \cr} $$
