Answer
$\frac{{{{\cos }^9}y}}{9} - \frac{{{{\cos }^7}y}}{7} + C$
Work Step by Step
$$\eqalign{
&\text{Let }I= \int {{{\cos }^6}y{{\sin }^3}y} dy \cr
& {\text{Split }}{\sin ^3}y{\text{ as si}}{{\text{n}}^2}y\sin y \cr
& I= \int {{{\cos }^6}y{{\sin }^2}y} \sin ydy \cr
& {\text{Rewrite }}{\sin ^2}y{\text{ using }}{\sin ^2}y + {\cos ^2}y = 1 \cr
& I = \int {{{\cos }^6}y\left( {1 - {{\cos }^2}y} \right)} \sin ydy \cr
& {\text{Multiply}} \cr
& I= \int {\left( {{{\cos }^6}y - {{\cos }^8}y} \right)} \sin ydy \cr
& \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \cos y,{\text{ }}du = - \sin ydy,{\text{ }} - du = \sin ydy,{\text{ then}} \cr
& I= \int {\left( {{u^6} - {u^8}} \right)} \left( { - 1} \right)du \cr
& = \int {\left( {{u^8} - {u^6}} \right)} du \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& I= \frac{{{u^9}}}{9} - \frac{{{u^7}}}{7} + C \cr
& {\text{Write in terms of }}y,{\text{ substitute }}\cos y{\text{ for }}u \cr
& I= \frac{{{{\cos }^9}y}}{9} - \frac{{{{\cos }^7}y}}{7} + C \cr} $$
