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

---

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