#\int \cos ^5\left(2\cdot y\right)\cdot \sin \left(2\cdot y\right) \,\dd y=# #-{{\cos ^6\left(2\cdot y\right)}\over{12}} + C#

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We apply the substitution method with #g(y)=-{{y^5}\over{2}}# and #h(y)=\cos \left(2\cdot y\right)#, because in that case #g(h(y)) \cdot h'(y)=\cos ^5\left(2\cdot y\right)\cdot \sin \left(2\cdot y\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos ^5\left(2\cdot y\right)\cdot \sin \left(2\cdot y\right) \,\dd y&=& \displaystyle \int -{{\cos ^5\left(2\cdot y\right)}\over{2}} \cdot -2\cdot \sin \left(2\cdot y\right) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-2\cdot \sin \left(2\cdot y\right)} \\ &=& \displaystyle \int \left(-{{\cos ^5\left(2\cdot y\right)}\over{2}} \right) \, \dd(\cos \left(2\cdot y\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int -{{u^5}\over{2}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos \left(2\cdot y\right)=u} \\ &=& \displaystyle -{{u^6}\over{12}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos ^6\left(2\cdot y\right)}\over{12}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos \left(2\cdot y\right)}
\end{array}\]