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

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We apply the substitution method with #g(t)=-t^8# and #h(t)=\cos(t)#, because in that case #g(h(t)) \cdot h'(t)=\cos(t)^8\cdot \sin(t)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos(t)^8\cdot \sin(t) \,\dd t&=& \displaystyle \int -\cos(t)^8 \cdot -\sin(t) \, \dd t \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(t)) \cdot h'(t) \, \dd t \text{ with } h'(t)=-\sin(t)} \\ &=& \displaystyle \int \left(-\cos(t)^8 \right) \, \dd(\cos(t)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(t)=\dd (h(t))} \\ &=& \displaystyle \int -u^8 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(t)=u} \\ &=& \displaystyle -{{u^9}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos(t)^9}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(t)}
\end{array}\]