#\int \cos(y)^6\cdot \sin(y) \,\dd y=# #-{{\cos(y)^7}\over{7}} + C#

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We apply the substitution method with #g(y)=-y^6# and #h(y)=\cos(y)#, because in that case #g(h(y)) \cdot h'(y)=\cos(y)^6\cdot \sin(y)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos(y)^6\cdot \sin(y) \,\dd y&=& \displaystyle \int -\cos(y)^6 \cdot -\sin(y) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-\sin(y)} \\ &=& \displaystyle \int \left(-\cos(y)^6 \right) \, \dd(\cos(y)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int -u^6 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(y)=u} \\ &=& \displaystyle -{{u^7}\over{7}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos(y)^7}\over{7}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(y)}
\end{array}\]