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

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