#\int -\cos(4\cdot y)^9\cdot \sin(4\cdot y) \,\dd y=# #{{\cos(4\cdot y)^{10}}\over{40}} + C#

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We passen de substitutiemethode toe met #g(y)={{y^9}\over{4}}# en #h(y)=\cos(4\cdot y)#, want dan geldt #g(h(y)) \cdot h'(y)=-\cos(4\cdot y)^9\cdot \sin(4\cdot y)#. Dit gaat als volgt:
\[\begin{array}{rcl}\displaystyle \int -\cos(4\cdot y)^9\cdot \sin(4\cdot y) \,\dd y&=& \displaystyle \int {{\cos(4\cdot y)^9}\over{4}} \cdot -4\cdot \sin(4\cdot y) \, \dd y \\&&\phantom{xxx}\blue{\text{stap 2: herschreven in de vorm }\int g(h(y)) \cdot h'(y) \, \dd y \text{ met } h'(y)=-4\cdot \sin(4\cdot y)} \\ &=& \displaystyle \int {{\cos(4\cdot y)^9}\over{4}} \, \dd(\cos(4\cdot y)) \\ &&\phantom{xxx}\blue{\text{stap 3: herschreven met het gebruik van }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int {{u^9}\over{4}} \, \dd u \\ &&\phantom{xxx}\blue{\text{stap 4: gesubstitueerd }\cos(4\cdot y)=u} \\ &=& \displaystyle {{u^{10}}\over{40}} +C \\ &&\phantom{xxx}\blue{\text{stap 5: geprimitvieerd}} \\ &=& \displaystyle {{\cos(4\cdot y)^{10}}\over{40}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: gesubstitueerd }u=\cos(4\cdot y)}
\end{array}\]