Задача №1515

\[ \int{x^2}\cos{wx}dx =\left[\begin{aligned} & u=x^2;\;du=2xdx.\\ & dv=\cos{wx}dx;\;v=\frac{\sin{wx}}{w}. \end{aligned}\right] =\frac{x^2\sin{wx}}{w}-\frac{2}{w}\cdot\int{x}\sin{wx}dx=\\ =\left[\begin{aligned} & u=x;\;du=dx.\\ & dv=\sin{wx}dx;\;v=-\frac{\cos{wx}}{w}. \end{aligned}\right] =\frac{x^2\sin{wx}}{w}-\frac{2}{w}\cdot\left(-\frac{x\cos{wx}}{w}+\frac{1}{w}\cdot\int\cos{wx}dx\right)=\\ =\frac{x^2\sin{wx}}{w}+\frac{2x\cos{wx}}{w^2}-\frac{2\sin{wx}}{w^3}+C \]