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