\(I=\int \sin ^{3} x\, {\rm d}x =\int \sin x\left(1-\cos ^{2} x\right)\, {\rm d}x\)
Đặt\(t=\cos x\Leftrightarrow {\rm d}t=-\sin x{\rm d}x\)
Khi đó: \(I=-\int \left(1-t^{2} \right){\rm d}t=\int \left(t^{2} -1\right){\rm d}t =\frac{t^{3} }{3} -t+C=\frac{\cos ^{3} x}{3} -\cos x+C.\)