## Thomas' Calculus 13th Edition

$$\frac{{{{\sin }^3}\theta }}{3} - \frac{{2{{\sin }^5}\theta }}{5} + \frac{{{{\sin }^7}\theta }}{7} + C$$
\eqalign{ & \int {{{\sin }^2}\theta {{\cos }^5}\theta } d\theta \cr & = \int {{{\sin }^2}\theta {{\cos }^{4 + 1}}\theta } d\theta \cr & {\text{use }}{a^{m + n}} = {a^m}{a^n} \cr & = \int {{{\sin }^2}\theta {{\cos }^4}\theta \cos \theta } d\theta \cr & {\text{Write co}}{{\text{s}}^4}\theta {\text{ as }}{\left( {{{\cos }^2}\theta } \right)^2} \cr & = \int {{{\sin }^2}\theta {{\left( {{{\cos }^2}\theta } \right)}^2}\cos \theta } d\theta \cr & {\text{Use the identity co}}{{\text{s}}^2}\theta = 1 - {\sin ^2}\theta \cr & = \int {{{\sin }^2}\theta {{\left( {1 - {{\sin }^2}\theta } \right)}^2}\cos \theta } d\theta \cr & \cr & {\text{Use the substitution method}}{\text{, }}u = \sin \theta ,\,\,\,du = \cos \theta d\theta \cr & \int {{{\sin }^2}\theta {{\left( {1 - {{\sin }^2}\theta } \right)}^2}\cos \theta } d\theta = \int {{u^2}{{\left( {1 - {u^2}} \right)}^2}} du \cr & {\text{expand and multiply}} \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{integrating}} \cr & = \frac{{{u^3}}}{3} - \frac{{2{u^5}}}{5} + \frac{{{u^7}}}{7} + C \cr & \cr & {\text{replacing }}u = \sin \theta \cr & = \frac{{{{\sin }^3}\theta }}{3} - \frac{{2{{\sin }^5}\theta }}{5} + \frac{{{{\sin }^7}\theta }}{7} + C \cr}