Answer
$$ - \frac{1}{4}\sqrt {9 - 4{t^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{tdt}}{{\sqrt {9 - 4{t^2}} }}} \cr
& {\text{Integrate by the substitution method}} \cr
& \,\,\,u = 9 - 4{t^2},\,\,\,\,du = - 8tdt,\,\,\,\,\,tdt = - \frac{1}{8}du \cr
& \int {\frac{{tdt}}{{\sqrt {9 - 4{t^2}} }}} = \int {\frac{{\left( { - 1/8} \right)du}}{{\sqrt u }}} \cr
& = - \frac{1}{8}\int {{u^{ - 1/2}}} du \cr
& {\text{integrate}} \cr
& = - \frac{1}{8}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = - \frac{1}{4}\sqrt u + C \cr
& {\text{Write in terms of }}x,{\text{ replace }}9 - 4{t^2}{\text{ for }}u \cr
& = - \frac{1}{4}\sqrt {9 - 4{t^2}} + C \cr} $$
