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

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We apply the substitution method with #g(x)=x^2-1# and #h(x)=\cos(x)#, because in that case #g(h(x)) \cdot h'(x)=\sin(x)^3# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \sin(x)^3 \,\dd x&=& \displaystyle \int \left(\cos(x)^2-1\right) \cdot -\sin(x) \, \dd x \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(x)) \cdot h'(x) \, \dd x \text{ with } h'(x)=-\sin(x)} \\ &&\phantom{xxx}\blue{\text{used the trigonometric rule }\sin^2(x)=1-\cos^2(x)} \\ &=& \displaystyle \int \left(\cos(x)^2-1 \right) \, \dd(\cos(x)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(x)=\dd (h(x))} \\ &=& \displaystyle \int u^2-1 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos(x)=u} \\ &=& \displaystyle {{u^3}\over{3}}-u +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos(x)^3}\over{3}}-\cos(x) +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos(x)}
\end{array}\]