$\int \cos \left(t\right)\cdot \sin ^5\left(t\right) \,\dd t=$ ${{\sin ^6\left(t\right)}\over{6}} + C$

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