Answer
$$ - \frac{1}{3}\cos \left( {2{t^{3/2}}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\sqrt t } \sin \left( {2{t^{3/2}}} \right)dt \cr
& = \int {{t^{1/2}}} \sin \left( {2{t^{3/2}}} \right)dt \cr
& {\text{use the substitution method }} \cr
& u = 2{t^{3/2}},\,\,\,\,du = 3{t^{1/2}}dt,\,\,\,\,\,dt = \frac{{du}}{{3{t^{1/2}}}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {{t^{1/2}}} \sin \left( {2{t^{3/2}}} \right)dt = \int {{t^{1/2}}} \sin u\left( {\frac{{du}}{{3{t^{1/2}}}}} \right) \cr
& = \frac{1}{3}\int {\sin u} du \cr
& {\text{integrate}} \cr
& = \frac{1}{3}\left( { - \cos u} \right) + C \cr
& = - \frac{1}{3}\cos u + C \cr
& {\text{write in terms of }}t;{\text{ replace }}u = 2{t^{3/2}} \cr
& = - \frac{1}{3}\cos \left( {2{t^{3/2}}} \right) + C \cr} $$
