#\int \cos(6\cdot y)^3\cdot \sin(6\cdot y) \,\dd y=# #-{{\cos(6\cdot y)^4}\over{24}} + C#

---

We apply the substitution method with #g(y)=-{{y^3}\over{6}}# and #h(y)=\cos(6\cdot y)#, because in that case #g(h(y)) \cdot h'(y)=\cos(6\cdot y)^3\cdot \sin(6\cdot y)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos(6\cdot y)^3\cdot \sin(6\cdot y) \,\dd y&=& \displaystyle \int -{{\cos(6\cdot y)^3}\over{6}} \cdot -6\cdot \sin(6\cdot y) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-6\cdot \sin(6\cdot y)} \\ &=& \displaystyle \int \left(-{{\cos(6\cdot y)^3}\over{6}} \right) \, \dd(\cos(6\cdot y)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int -{{u^3}\over{6}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(6\cdot y)=u} \\ &=& \displaystyle -{{u^4}\over{24}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos(6\cdot y)^4}\over{24}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(6\cdot y)}
\end{array}\]