Answer
$$ - {x^3}\cos x + 3{x^2}\sin x + 6x\cos x - 6\sin x + C$$
Work Step by Step
$$\eqalign{ & \int {{x^3}\sin } dx \cr & {\text{Using the reduction formula }} \cr & \int {{x^n}\sin axdx} = - \frac{{{x^n}\cos ax}}{a} + \frac{n}{a}\int {{x^{n - 1}}\cos axdx} ,\,\,\,\,\,{\text{for }}a \ne 0 \cr & {\text{then,}} \cr & \int {{x^3}\sin } dx = - \frac{{{x^3}\cos x}}{1} + \frac{3}{1}\int {{x^{3 - 1}}\cos xdx} \cr & \int {{x^3}\sin } dx = - {x^3}\cos x + 3\int {{x^2}\cos xdx} \cr & {\text{Using the reduction formula }} \cr & \int {{x^n}\cos axdx} = \frac{{{x^n}\sin ax}}{a} - \frac{n}{a}\int {{x^{n - 1}}\sin axdx} ,\,\,\,\,\,{\text{for }}a \ne 0 \cr & \int {{x^3}\sin } dx = - {x^3}\cos x + 3\left( {{x^2}\sin x - 2\int {{x^{2 - 1}}\sin xdx} } \right) \cr & \int {{x^3}\sin } dx = - {x^3}\cos x + 3{x^2}\sin x - 6\int {x\sin xdx} \cr & {\text{Using the reduction formula }} \cr & \int {{x^n}\sin axdx} = - \frac{{{x^n}\cos ax}}{a} + \frac{n}{a}\int {{x^{n - 1}}\cos axdx} ,\,\,\,\,\,{\text{for }}a \ne 0 \cr & \int {{x^3}\sin } dx = - {x^3}\cos x + 3{x^2}\sin x - 6\left( { - x\cos x + \int {{x^{1 - 1}}\cos xdx} } \right) \cr & \int {{x^3}\sin } dx = - {x^3}\cos x + 3{x^2}\sin x + 6x\cos x - 6\int {\cos xdx} \cr & \int {{x^3}\sin } dx = - {x^3}\cos x + 3{x^2}\sin x + 6x\cos x - 6\sin x + C \cr} $$
