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

---

We apply the substitution method with #g(x)=-x^9# and #h(x)=\cos(x)#, because in that case #g(h(x)) \cdot h'(x)=\cos(x)^9\cdot \sin(x)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos(x)^9\cdot \sin(x) \,\dd x&=& \displaystyle \int -\cos(x)^9 \cdot -\sin(x) \, \dd x \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(x)) \cdot h'(x) \, \dd x \text{ with } h'(x)=-\sin(x)} \\ &=& \displaystyle \int \left(-\cos(x)^9 \right) \, \dd(\cos(x)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(x)=\dd (h(x))} \\ &=& \displaystyle \int -u^9 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(x)=u} \\ &=& \displaystyle -{{u^{10}}\over{10}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos(x)^{10}}\over{10}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(x)}
\end{array}\]