Answer
$\frac{{{x^3}}}{2}\sin 2x + \frac{3}{4}{x^2}\cos 2x - \frac{3}{4}x\sin 2x - \frac{3}{8}\cos 2x + C$
Work Step by Step
$$\eqalign{
& \int {{x^3}\cos 2x} dx \cr
& {\text{Integrate by tabulation method }} \cr
& {\text{First}}{\text{, let }}u = {x^3}{\text{ and }}dv = v'dx = \cos 2xdx \cr
& \cr
& {\text{*Differentiating }}u{\text{ until obtain 0 as derivative}} \cr
& \left( 1 \right){\text{ }}\frac{d}{{dx}}\left[ {{x^3}} \right] = 3{x^2} \cr
& \left( 2 \right){\text{ }}\frac{d}{{dx}}\left[ {3{x^2}} \right] = 6x \cr
& \left( 3 \right){\text{ }}\frac{d}{{dx}}\left[ {6x} \right] = 6 \cr
& \left( 4 \right){\text{ }}\frac{d}{{dx}}\left[ 6 \right] = 0 \cr
& \cr
& {\text{*Integrating }}dv = \cos 2xdx{\text{ four times}} \cr
& \left( 1 \right){\text{ }}\int {\cos 2xdx} = \frac{1}{2}\sin 2x \cr
& \left( 2 \right){\text{ }}\int {\left( {\frac{1}{2}\sin 2x} \right)dx} = - \frac{1}{4}\cos 2x \cr
& \left( 3 \right){\text{ }}\int {\left( { - \frac{1}{4}\cos 2x} \right)dx} = - \frac{1}{8}\sin 2x \cr
& \left( 4 \right){\text{ }}\int {\left( { - \frac{1}{8}\sin 2x} \right)dx} = \frac{1}{{16}}\cos 2x \cr
& \cr
& {\text{Now we will create a table with }}u{\text{ and its derivatives}} \cr
& {\text{and }}v'dx{\text{ and its derivatives}}{\text{, and a column alternating}} \cr
& {\text{the signs }} + - + - ,{\text{ }}\left( {{\text{ the image bellow}}} \right) \cr
& {\text{We can obtain the integral solution by adding the signed }} \cr
& {\text{products of the diagonal entries}} \cr
& \int {{x^3}\cos 2x} dx = {x^3}\left( {\frac{1}{2}\sin 2x} \right) - \left( {3{x^2}} \right)\left( { - \frac{1}{4}\cos 2x} \right) \cr
& + \left( {6x} \right)\left( { - \frac{1}{8}\sin 2x} \right) - 6\left( {\frac{1}{{16}}\cos 2x} \right) + C \cr
& = \frac{{{x^3}}}{2}\sin 2x + \frac{3}{4}{x^2}\cos 2x - \frac{3}{4}x\sin 2x - \frac{3}{8}\cos 2x + C \cr} $$
