$$$\sin^{3}{\left(x \right)}$$$ 的积分

求$$$\int \sin^{3}{\left(x \right)}\, dx$$$。

提出一个正弦，并将其余部分用余弦表示，使用公式$$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$，其中$$$\alpha=x$$$:

$${\color{red}{\int{\sin^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x}}}$$

设$$$u=\cos{\left(x \right)}$$$。

则$$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (步骤见»)，并有$$$\sin{\left(x \right)} dx = - du$$$。

因此，

$${\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x}}} = {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$

对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = 1 - u^{2}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\left(u^{2} - 1\right)d u}}} = {\color{red}{\left(- \int{\left(1 - u^{2}\right)d u}\right)}}$$

逐项积分：

$$- {\color{red}{\int{\left(1 - u^{2}\right)d u}}} = - {\color{red}{\left(\int{1 d u} - \int{u^{2} d u}\right)}}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

$$\int{u^{2} d u} - {\color{red}{\int{1 d u}}} = \int{u^{2} d u} - {\color{red}{u}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=2$$$：

$$- u + {\color{red}{\int{u^{2} d u}}}=- u + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- u + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回忆一下 $$$u=\cos{\left(x \right)}$$$:

$$- {\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = - {\color{red}{\cos{\left(x \right)}}} + \frac{{\color{red}{\cos{\left(x \right)}}}^{3}}{3}$$

因此，

$$\int{\sin^{3}{\left(x \right)} d x} = \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$$

加上积分常数：

$$\int{\sin^{3}{\left(x \right)} d x} = \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+C$$

$$$\int \sin^{3}{\left(x \right)}\, dx = \left(\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) + C$$$A