$$$\sin{\left(8 x \right)} \cos{\left(2 x \right)}$$$ 的积分

求$$$\int \sin{\left(8 x \right)} \cos{\left(2 x \right)}\, dx$$$。

使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ 并结合 $$$\alpha=8 x$$$ 和 $$$\beta=2 x$$$ 重写被积函数:

$${\color{red}{\int{\sin{\left(8 x \right)} \cos{\left(2 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(6 x \right)}}{2} + \frac{\sin{\left(10 x \right)}}{2}\right)d x}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \sin{\left(6 x \right)} + \sin{\left(10 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\left(\frac{\sin{\left(6 x \right)}}{2} + \frac{\sin{\left(10 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(6 x \right)} + \sin{\left(10 x \right)}\right)d x}}{2}\right)}}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(\sin{\left(6 x \right)} + \sin{\left(10 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(6 x \right)} d x} + \int{\sin{\left(10 x \right)} d x}\right)}}}{2}$$

设$$$u=6 x$$$。

则$$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{6}$$$。

因此，

$$\frac{\int{\sin{\left(10 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(6 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(10 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}}{2}$$

对 $$$c=\frac{1}{6}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{\int{\sin{\left(10 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}}{2} = \frac{\int{\sin{\left(10 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{6}\right)}}}{2}$$

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

$$\frac{\int{\sin{\left(10 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{12} = \frac{\int{\sin{\left(10 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{12}$$

回忆一下 $$$u=6 x$$$:

$$\frac{\int{\sin{\left(10 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{12} = \frac{\int{\sin{\left(10 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(6 x\right)}} \right)}}{12}$$

设$$$u=10 x$$$。

则$$$du=\left(10 x\right)^{\prime }dx = 10 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{10}$$$。

积分变为

$$- \frac{\cos{\left(6 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(10 x \right)} d x}}}}{2} = - \frac{\cos{\left(6 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{10} d u}}}}{2}$$

对 $$$c=\frac{1}{10}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$- \frac{\cos{\left(6 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{10} d u}}}}{2} = - \frac{\cos{\left(6 x \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{10}\right)}}}{2}$$

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

$$- \frac{\cos{\left(6 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{20} = - \frac{\cos{\left(6 x \right)}}{12} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{20}$$

回忆一下 $$$u=10 x$$$:

$$- \frac{\cos{\left(6 x \right)}}{12} - \frac{\cos{\left({\color{red}{u}} \right)}}{20} = - \frac{\cos{\left(6 x \right)}}{12} - \frac{\cos{\left({\color{red}{\left(10 x\right)}} \right)}}{20}$$

因此，

$$\int{\sin{\left(8 x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(6 x \right)}}{12} - \frac{\cos{\left(10 x \right)}}{20}$$

加上积分常数：

$$\int{\sin{\left(8 x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(6 x \right)}}{12} - \frac{\cos{\left(10 x \right)}}{20}+C$$

$$$\int \sin{\left(8 x \right)} \cos{\left(2 x \right)}\, dx = \left(- \frac{\cos{\left(6 x \right)}}{12} - \frac{\cos{\left(10 x \right)}}{20}\right) + C$$$A