Đặt $\begin{cases}u=x^{2} \\ dv=sin3x .dx \end{cases} =>\begin{cases}du=2x.dx \\ v=-\frac{1}{3}cos3x \end{cases}$
$\int x^{2}sin3xdx=-x^{2}.\frac{1}{3}cos3x+ \frac{2}{3}\int\limits xcos3xdx $
Đặt $\begin{cases}u_{1}=x \\ dv_{1}=cos3x.dx \end{cases} => \begin{cases}u_{1}=dx \\ v_{1}=\frac{1}{3}sin3x \end{cases}$
$\int\limits x^{2}sin3xdx= -x^{2}.\frac{1}{3}cos3x+\frac{2}{9}.x.sin3x -\frac{2}{9}\int sin3xdx$
$=-x^{2}.\frac{1}{3}cos3x+\frac{2}{9}.x.sin3x+\frac{2}{9}cos3x$ + C