sinx三次方的原函数

$\int \sin^3x \, dx = \int \sin^2x \cdot \sin x \, dx = \int (1-\cos^2x) \cdot \sin x \, dx$

令 $u = \cos x, \, du = -\sin x \, dx$，则

$\begin{aligned} \int (1-\cos^2x) \cdot \sin x \, dx &= \int (1-u^2) \cdot (-du) \\ &= u - \frac{u^3}{3} + C \\ &= \cos x - \frac{\cos^3x}{3} + C \end{aligned}$

因此，$\int \sin^3x \, dx = \cos x - \frac{\cos^3x}{3} + C$。