Answer
$ - \frac{{{t^2}}}{\beta }\cos \beta t + \frac{{2t}}{{{\beta ^2}}}\sin \beta t + \frac{2}{{{\beta ^3}}}\cos \beta t + C$
Work Step by Step
$$\eqalign{
& \int {{t^2}\sin \beta t} dt \cr
& {\text{Integrating by parts}}{\text{, }} \cr
& {\text{Let }}u = {t^2},{\text{ }}du = 2tdt \cr
& {\text{ }}dv = \sin \beta tdt,{\text{ }}v = - \frac{1}{\beta }\cos \beta t \cr
& {\text{Using integration by parts formula}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {{t^2}\sin \beta t} dt = \left( {{t^2}} \right)\left( { - \frac{1}{\beta }\cos \beta t} \right) - \int {\left( { - \frac{1}{\beta }\cos \beta t} \right)\left( {2t} \right)dt} \cr
& \int {{t^2}\sin \beta t} dt = - \frac{{{t^2}}}{\beta }\cos \beta t + \frac{2}{\beta }\int {t\cos \beta tdt} \cr
& \cr
& {\text{Integrate by parts }}\int {t\cos \beta tdt} \cr
& {\text{Let }}u = t,{\text{ }}du = dt \cr
& {\text{ }}dv = \cos \beta tdt,{\text{ }}v = \frac{1}{\beta }\sin \beta t \cr
& \int {{t^2}\sin \beta t} dt = - \frac{{{t^2}}}{\beta }\cos \beta t + \frac{2}{\beta }\left[ {\frac{t}{\beta }\sin \beta t - \int {\frac{1}{\beta }\sin \beta tdt} } \right] \cr
& \int {{t^2}\sin \beta t} dt = - \frac{{{t^2}}}{\beta }\cos \beta t + \frac{{2t}}{{{\beta ^2}}}\sin \beta t - \frac{2}{\beta }\int {\frac{1}{\beta }\sin \beta tdt} \cr
& {\text{Integrating}} \cr
& \int {{t^2}\sin \beta t} dt = - \frac{{{t^2}}}{\beta }\cos \beta t + \frac{{2t}}{{{\beta ^2}}}\sin \beta t - \frac{2}{{{\beta ^2}}}\left( { - \frac{1}{\beta }\cos \beta t} \right) + C \cr
& {\text{Simplifying}} \cr
& \int {{t^2}\sin \beta t} dt = - \frac{{{t^2}}}{\beta }\cos \beta t + \frac{{2t}}{{{\beta ^2}}}\sin \beta t + \frac{2}{{{\beta ^3}}}\cos \beta t + C \cr} $$
