#\int \cos(8\cdot t)^5\cdot \sin(8\cdot t) \,\dd t=# #-{{\cos(8\cdot t)^6}\over{48}} + C#

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We apply the substitution method with #g(t)=-{{t^5}\over{8}}# and #h(t)=\cos(8\cdot t)#, because in that case #g(h(t)) \cdot h'(t)=\cos(8\cdot t)^5\cdot \sin(8\cdot t)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos(8\cdot t)^5\cdot \sin(8\cdot t) \,\dd t&=& \displaystyle \int -{{\cos(8\cdot t)^5}\over{8}} \cdot -8\cdot \sin(8\cdot t) \, \dd t \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(t)) \cdot h'(t) \, \dd t \text{ with } h'(t)=-8\cdot \sin(8\cdot t)} \\ &=& \displaystyle \int \left(-{{\cos(8\cdot t)^5}\over{8}} \right) \, \dd(\cos(8\cdot t)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(t)=\dd (h(t))} \\ &=& \displaystyle \int -{{u^5}\over{8}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(8\cdot t)=u} \\ &=& \displaystyle -{{u^6}\over{48}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos(8\cdot t)^6}\over{48}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(8\cdot t)}
\end{array}\]