#\int \cos(t)\cdot \sin(t)^4 \,\dd t=# #{{\sin(t)^5}\over{5}} + C#

---

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