Answer
$$\frac{{{{\sin }^3}t}}{3} - \frac{{{{\sin }^5}t}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}t{{\cos }^3}t} dt \cr
& {\text{split off }}{\cos ^3}t \cr
& = \int {{{\sin }^2}t{{\cos }^2}t\cos t} dt \cr
& {\text{identity }}{\sin ^2}t + {\cos ^2}t = 1 \cr
& = \int {{{\sin }^2}t\left( {1 - {{\sin }^2}t} \right)\cos t} dt \cr
& = \int {\left( {{{\sin }^2}t - {{\sin }^4}t} \right)\cos t} dt \cr
& {\text{substitute }}u = \sin t,{\text{ }}du = \cos tdt \cr
& = \int {\left( {{u^2} - {u^4}} \right)du} \cr
& {\text{find the antiderivatives by the power rule}} \cr
& = \frac{{{u^3}}}{3} - \frac{{{u^5}}}{5} + C \cr
& {\text{write in terms of }}t,{\text{ replace }}u = \sin t \cr
& = \frac{{{{\sin }^3}t}}{3} - \frac{{{{\sin }^5}t}}{5} + C \cr} $$
