#\int -\cos(7\cdot t)^5\cdot \sin(7\cdot t) \,\dd t=# #{{\cos(7\cdot t)^6}\over{42}} + C#

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We apply the substitution method with #g(t)={{t^5}\over{7}}# and #h(t)=\cos(7\cdot t)#, because in that case #g(h(t)) \cdot h'(t)=-\cos(7\cdot t)^5\cdot \sin(7\cdot t)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int -\cos(7\cdot t)^5\cdot \sin(7\cdot t) \,\dd t&=& \displaystyle \int {{\cos(7\cdot t)^5}\over{7}} \cdot -7\cdot \sin(7\cdot t) \, \dd t \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(t)) \cdot h'(t) \, \dd t \text{ with } h'(t)=-7\cdot \sin(7\cdot t)} \\ &=& \displaystyle \int \left({{\cos(7\cdot t)^5}\over{7}} \right) \, \dd(\cos(7\cdot t)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(t)=\dd (h(t))} \\ &=& \displaystyle \int {{u^5}\over{7}} \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(7\cdot t)=u} \\ &=& \displaystyle {{u^6}\over{42}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos(7\cdot t)^6}\over{42}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(7\cdot t)}
\end{array}\]