#\int \cos ^8\left(y\right)\cdot \sin \left(y\right) \,\dd y=# #-{{\cos ^9\left(y\right)}\over{9}} + C#

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We apply the substitution method with #g(y)=-y^8# and #h(y)=\cos \left(y\right)#, because in that case #g(h(y)) \cdot h'(y)=\cos ^8\left(y\right)\cdot \sin \left(y\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos ^8\left(y\right)\cdot \sin \left(y\right) \,\dd y&=& \displaystyle \int -\cos ^8\left(y\right) \cdot -\sin \left(y\right) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-\sin \left(y\right)} \\ &=& \displaystyle \int \left(-\cos ^8\left(y\right) \right) \, \dd(\cos \left(y\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int -u^8 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos \left(y\right)=u} \\ &=& \displaystyle -{{u^9}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos ^9\left(y\right)}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos \left(y\right)}
\end{array}\]