$$$\left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)}$$$ 的积分

求$$$\int \left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)}\, dx$$$。

Expand the expression:

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

逐项积分：

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

对 $$$c=3$$$ 和 $$$f{\left(x \right)} = \sin^{3}{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

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$$$\alpha=x$$$:

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

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，

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

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

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

逐项积分：

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

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

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

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

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

回忆一下 $$$u=\cos{\left(x \right)}$$$:

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

对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \sin{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

正弦函数的积分为 $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

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

因此，

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

加上积分常数：

$$\int{\left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)} d x} = - \cos^{3}{\left(x \right)} + \cos{\left(x \right)}+C$$

$$$\int \left(2 - 3 \sin^{2}{\left(x \right)}\right) \sin{\left(x \right)}\, dx = \left(- \cos^{3}{\left(x \right)} + \cos{\left(x \right)}\right) + C$$$A