#\int \cos(3\cdot y)^4\cdot \sin(3\cdot y) \,\dd y=# #-{{\cos(3\cdot y)^5}\over{15}} + C#

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We apply the substitution method with #g(y)=-{{y^4}\over{3}}# and #h(y)=\cos(3\cdot y)#, because in that case #g(h(y)) \cdot h'(y)=\cos(3\cdot y)^4\cdot \sin(3\cdot y)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos(3\cdot y)^4\cdot \sin(3\cdot y) \,\dd y&=& \displaystyle \int -{{\cos(3\cdot y)^4}\over{3}} \cdot -3\cdot \sin(3\cdot y) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-3\cdot \sin(3\cdot y)} \\ &=& \displaystyle \int \left(-{{\cos(3\cdot y)^4}\over{3}} \right) \, \dd(\cos(3\cdot y)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int -{{u^4}\over{3}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(3\cdot y)=u} \\ &=& \displaystyle -{{u^5}\over{15}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos(3\cdot y)^5}\over{15}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(3\cdot y)}
\end{array}\]