Answer
$$\frac{-t^2}{ \beta} \cos \beta t+ \frac{2}{ \beta} \left(\frac{t}{ \beta} \sin \beta t + \frac{1}{ \beta^2}\cos \beta t\right)+c$$
Work Step by Step
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*}
