#\int -\cos(3\cdot x)^5\cdot \sin(3\cdot x) \,\dd x=# #{{\cos(3\cdot x)^6}\over{18}} + C#

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We apply the substitution method with #g(x)={{x^5}\over{3}}# and #h(x)=\cos(3\cdot x)#, because in that case #g(h(x)) \cdot h'(x)=-\cos(3\cdot x)^5\cdot \sin(3\cdot x)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int -\cos(3\cdot x)^5\cdot \sin(3\cdot x) \,\dd x&=& \displaystyle \int {{\cos(3\cdot x)^5}\over{3}} \cdot -3\cdot \sin(3\cdot x) \, \dd x \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(x)) \cdot h'(x) \, \dd x \text{ with } h'(x)=-3\cdot \sin(3\cdot x)} \\ &=& \displaystyle \int \left({{\cos(3\cdot x)^5}\over{3}} \right) \, \dd(\cos(3\cdot x)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(x)=\dd (h(x))} \\ &=& \displaystyle \int {{u^5}\over{3}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(3\cdot x)=u} \\ &=& \displaystyle {{u^6}\over{18}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos(3\cdot x)^6}\over{18}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(3\cdot x)}
\end{array}\]