$\begin{align}  & \int{{{x}^{3}}\sin x}dx \\ & \text{Using integration by part} \\ & \int{udv}=uv-\int{vdu} \\ & \text{Let }u={{x}^{3}},\text{ }du=3{{x}^{2}}dx \\ & dv=\sin xdx,\text{ }v=-\cos x \\ & \int{{{x}^{3}}\sin xdx}={{x}^{3}}(-\cos x)-\int{-\cos x(3{{x}^{2}}dx)} \\ & =-{{x}^{3}}\cos x+3\int{{{x}^{2}}\cos xdx} \\ & \text{Using integrating for the }\int{{{x}^{2}}\cos xdx} \\ & \int{{{x}^{3}}\sin xdx}=-{{x}^{3}}\cos x+3\left[ {{x}^{2}}\int{\cos xdx-\int{2x\left( \int{\cos xdx} \right)}} \right] \\ & \int{{{x}^{3}}\sin xdx}=-{{x}^{3}}\cos x+3\left[ {{x}^{2}}\sin x-2\int{x\sin xdx} \right] \\ & \text{    }=-{{x}^{3}}\cos x+3{{x}^{2}}\sin x-6\int{x\sin xdx} \\ & \text{   }=-{{x}^{3}}\cos x+3{{x}^{2}}\sin x-6\left[ x\int{\sin dx-\int{\left( \tfrac{d}{dx}(x)\int{\sin xdx} \right)dx}} \right] \\ & \text{   }=-{{x}^{3}}\cos x+3{{x}^{2}}\sin x-6\left[ -x\cos x-\int{\cos xdx} \right] \\ & \text{  }=-{{x}^{3}}\cos x+3{{x}^{2}}\sin x+6x\cos x+6\sin x+C \\\end{align}$