## Calculus 8th Edition

$$\frac{-t^2}{ \beta} \cos \beta t+ \frac{2}{ \beta} \left(\frac{t}{ \beta} \sin \beta t + \frac{1}{ \beta^2}\cos \beta t\right)+c$$
Given $$\int t^2\sin \beta t \ dt$$ Let \begin{align*} u&= t^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\sin \beta t dt\\ u&=2tdt\ \ \ \ \ \ \ \ \ \ \ \ \ dv= \frac{-1}{ \beta} \cos \beta t \end{align*} Then using integration by parts \begin{align*} \int t^2\sin \beta t \ dt &=uv-\int vdu\\ &=\frac{-t^2}{ \beta} \cos \beta t+ \frac{2}{ \beta} \int t \cos \beta t \ dt \end{align*} To evaluate $\displaystyle \int t \cos \beta t \ dt$ Let \begin{align*} u&= t \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\cos \beta t dt\\ u&= dt\ \ \ \ \ \ \ \ \ \ \ \ \ dv= \frac{1}{ \beta} \sin \beta t \end{align*} then \begin{align*} \int t \cos \beta t \ dt &=uv-\int vdu\\ &=\frac{t}{ \beta} \sin \beta t- \frac{1}{ \beta} \int \sin \beta t \ dt\\ &= \frac{t}{ \beta} \sin \beta t + \frac{1}{ \beta^2}\cos \beta t \end{align*} Hence \begin{align*} \int t^2\sin \beta t \ dt &=\frac{-t^2}{ \beta} \cos \beta t+ \frac{2}{ \beta} \int t \cos \beta t \ dt \\ &=\frac{-t^2}{ \beta} \cos \beta t+ \frac{2}{ \beta} \left(\frac{t}{ \beta} \sin \beta t + \frac{1}{ \beta^2}\cos \beta t\right)+c \end{align*}