${\displaystyle {}{\begin{aligned}\int _{0}^{x}\sin ^{3}t\,dt&=\int _{0}^{x}\sin t\cdot \sin ^{2}t\,dt\\&=\int _{0}^{x}\sin t\cdot {\left(1-\cos ^{2}t\right)}\,dt\\&=\int _{0}^{x}\sin t\,dt-\int _{0}^{x}(\sin t\cos t)\cos t\,dt\\&=\int _{0}^{x}\sin t\,dt-{\left({\frac {\sin ^{2}t}{2}}\cos t\right)}|_{0}^{x}-{\frac {1}{2}}{\left(\int _{0}^{x}\sin ^{3}t\,dt\right)}.\end{aligned}}}$

Durch Multiplikation mit ${\displaystyle {}2}$ und Umstellen erhält man

${\displaystyle {}{\begin{aligned}3\int _{0}^{x}\sin ^{3}t\,dt&=2\int _{0}^{x}\sin t\,dt-\sin ^{2}x\cos x\\&=-2\cos x-\sin ^{2}x\cos x.\end{aligned}}}$

Also ist

${\displaystyle -{\frac {2}{3}}\cos x-{\frac {1}{3}}\sin ^{2}x\cos x}$

${\displaystyle {}\sin ^{3}x}$