Reduce the power by applying the identity,
[tex]\sin^2x+\cos^2x=1[/tex]
[tex]\implies\displaystyle\int\sin^3x\,\mathrm dx=\int\sin x(1-\cos^2x)\,\mathrm dx[/tex]
Let [tex]u=\cos x\implies\mathrm du=-\sin x\,\mathrm dx[/tex]:
[tex]\implies\displaystyle\int\sin^3x\,\mathrm dx=-\int(1-u^2)\,\mathrm du[/tex]
[tex]=\dfrac{u^3}3-u+C=\boxed{\dfrac{\cos^3x}3-\cos x+C}[/tex]