Answer
$$\frac{{{{\sin }^3}\theta }}{3} - \frac{{2{{\sin }^5}\theta }}{5} + \frac{{{{\sin }^7}\theta }}{7} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^2}\theta {{\cos }^5}\theta } d\theta \cr
& {\text{split off }}\cos \theta \cr
& = \int {{{\sin }^2}\theta {{\cos }^4}\theta \cos \theta } d\theta \cr
& {\text{Pythagorean identity}} \cr
& = \int {{{\sin }^2}\theta {{\left( {1 - {{\sin }^2}\theta } \right)}^2}\cos \theta } d\theta \cr
& u = \sin \theta ,{\text{ }}du = \cos \theta d\theta \cr
& = \int {{u^2}{{\left( {1 - {u^2}} \right)}^2}du} \cr
& = \int {{u^2}\left( {1 - 2{u^2} + {u^4}} \right)du} \cr
& = \int {\left( {{u^2} - 2{u^4} + {u^6}} \right)du} \cr
& {\text{evaluate the integral}} \cr
& = \frac{{{u^3}}}{3} - \frac{{2{u^5}}}{5} + \frac{{{u^7}}}{7} + C \cr
& {\text{replace }}u = \sin \theta \cr
& = \frac{{{{\sin }^3}\theta }}{3} - \frac{{2{{\sin }^5}\theta }}{5} + \frac{{{{\sin }^7}\theta }}{7} + C \cr} $$
