#\int -\cos(5\cdot y)^6\cdot \sin(5\cdot y) \,\dd y=# #{{\cos(5\cdot y)^7}\over{35}} + C#

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We apply the substitution method with #g(y)={{y^6}\over{5}}# and #h(y)=\cos(5\cdot y)#, because in that case #g(h(y)) \cdot h'(y)=-\cos(5\cdot y)^6\cdot \sin(5\cdot y)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int -\cos(5\cdot y)^6\cdot \sin(5\cdot y) \,\dd y&=& \displaystyle \int {{\cos(5\cdot y)^6}\over{5}} \cdot -5\cdot \sin(5\cdot y) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-5\cdot \sin(5\cdot y)} \\ &=& \displaystyle \int \left({{\cos(5\cdot y)^6}\over{5}} \right) \, \dd(\cos(5\cdot y)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int {{u^6}\over{5}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(5\cdot y)=u} \\ &=& \displaystyle {{u^7}\over{35}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos(5\cdot y)^7}\over{35}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(5\cdot y)}
\end{array}\]