Primeiro tomamos \(u=t^2\) e \(dv=\operatorname{sen}(\beta t)\,dt\). Então

\[ du=2t\,dt, \qquad v=-\frac{1}{\beta}\cos(\beta t). \]

Logo,

\[ \int t^2\operatorname{sen}(\beta t)\,dt = -\frac{t^2}{\beta}\cos(\beta t) + \frac{2}{\beta}\int t\cos(\beta t)\,dt. \]

Agora aplicamos integração por partes de novo em \(\int t\cos(\beta t)\,dt\):

\[ \int t\cos(\beta t)\,dt = \frac{t}{\beta}\operatorname{sen}(\beta t) + \frac{1}{\beta^2}\cos(\beta t). \]

Substituindo, chegamos a

\[ \colorbox{green!20}{\int t^2\operatorname{sen}(\beta t)\,dt = -\frac{t^2}{\beta}\cos(\beta t) + \frac{2t}{\beta^2}\operatorname{sen}(\beta t) + \frac{2}{\beta^3}\cos(\beta t)+C}. \]