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

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

使用公式 $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ 将被积函数改写，其中 $$$\alpha=x$$$ 和 $$$\beta=2 x$$$:

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \cos{\left(x \right)} - \cos{\left(3 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

逐项积分：

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

设$$$u=3 x$$$。

则$$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{3}$$$。

因此，

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

对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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

$$\frac{\int{\cos{\left(x \right)} d x}}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{6} = \frac{\int{\cos{\left(x \right)} d x}}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{6}$$

回忆一下 $$$u=3 x$$$:

$$\frac{\int{\cos{\left(x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{6} = \frac{\int{\cos{\left(x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{6}$$

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

$$- \frac{\sin{\left(3 x \right)}}{6} + \frac{{\color{red}{\int{\cos{\left(x \right)} d x}}}}{2} = - \frac{\sin{\left(3 x \right)}}{6} + \frac{{\color{red}{\sin{\left(x \right)}}}}{2}$$

因此，

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} d x} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{6}$$

加上积分常数：

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} d x} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{6}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)}\, dx = \left(\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{6}\right) + C$$$A