Задача №1410

\[ \int{x^3\sin{x}}dx =\left[\begin{aligned}& u=x^3;\,du=3x^2dx.\\& dv=\sin{x}dx;\;v=-\cos{x}.\end{aligned}\right] =-x^3\cos{x}+3\int{x^2\cos{x}}dx=\\ =\left[\begin{aligned}& u=x^2;\,du=2xdx.\\& dv=\cos{x}dx;\;v=\sin{x}.\end{aligned}\right] =-x^3\cos{x}+3\cdot\left(x^2\sin{x}-2\int{x\sin{x}}dx\right)=\\ =-x^3\cos{x}+3x^2\sin{x}-6\int{x\sin{x}}dx =\left[\begin{aligned}& u=x;\,du=dx.\\& dv=\sin{x}dx;\;v=-\cos{x}.\end{aligned}\right]=\\ =-x^3\cos{x}+3x^2\sin{x}-6\cdot\left(-x\cos{x}+\int{\cos{x}}dx\right) =-x^3\cos{x}+3x^2\sin{x}+6x\cos{x}-6\sin{x}+C \]