#\int \sin(t)^3 \,\dd t=# #{{\cos(t)^3}\over{3}}-\cos(t) + C#

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We apply the substitution method with #g(t)=t^2-1# and #h(t)=\cos(t)#, because in that case #g(h(t)) \cdot h'(t)=\sin(t)^3# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \sin(t)^3 \,\dd t&=& \displaystyle \int \left(\cos(t)^2-1\right) \cdot -\sin(t) \, \dd t \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(t)) \cdot h'(t) \, \dd t \text{ with } h'(t)=-\sin(t)} \\ &&\phantom{xxx}\blue{\text{used the trigonometric rule }\sin^2(t)=1-\cos^2(t)} \\ &=& \displaystyle \int \left(\cos(t)^2-1 \right) \, \dd(\cos(t)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(t)=\dd (h(t))} \\ &=& \displaystyle \int u^2-1 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(t)=u} \\ &=& \displaystyle {{u^3}\over{3}}-u +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos(t)^3}\over{3}}-\cos(t) +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(t)}
\end{array}\]