#\int -\cos(7\cdot y)^8\cdot \sin(7\cdot y) \,\dd y=# #{{\cos(7\cdot y)^9}\over{63}} + C#

---

Aplicamos el método de sustitución con #g(y)={{y^8}\over{7}}# y #h(y)=\cos(7\cdot y)#, porque en ese caso se aplica #g(h(y)) \cdot h'(y)=-\cos(7\cdot y)^8\cdot \sin(7\cdot y)#. Esto se realiza de la siguiente manera:
\[\begin{array}{rcl}\displaystyle \int -\cos(7\cdot y)^8\cdot \sin(7\cdot y) \,\dd y&=& \displaystyle \int {{\cos(7\cdot y)^8}\over{7}} \cdot -7\cdot \sin(7\cdot y) \, \dd y \\&&\phantom{xxx}\blue{\text{paso 2: volver a escribir en la forma }\int g(h(y)) \cdot h'(y) \, \dd y \\
\text{ con } h'(y)=-7\cdot \sin(7\cdot y)} \\ &=& \displaystyle \int \left({{\cos(7\cdot y)^8}\over{7}} \right) \, \dd(\cos(7\cdot y)) \\ &&\phantom{xxx}\blue{\text{paso 3: volver a escribir usando }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int {{u^8}\over{7}} \, \dd u \\ &&\phantom{xxx}\blue{\text{paso 4: sustituir }\cos(7\cdot y)=u} \\ &=& \displaystyle {{u^9}\over{63}} +C \\ &&\phantom{xxx}\blue{\text{paso 5: encontrar la antiderivada}} \\ &=& \displaystyle {{\cos(7\cdot y)^9}\over{63}} +C \\ &&\phantom{xxx}\blue{\text{paso 6: sustituir }u=\cos(7\cdot y)}
\end{array}\]