$$$\sin{\left(3 x \right)} \cos{\left(x \right)}$$$ 的积分

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

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 $$$\alpha=3 x$$$ 和 $$$\beta=x$$$ 重写被积函数:

$${\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(4 x \right)}}{2}\right)d x}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \sin{\left(2 x \right)} + \sin{\left(4 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\left(\frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(4 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(2 x \right)} + \sin{\left(4 x \right)}\right)d x}}{2}\right)}}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(\sin{\left(2 x \right)} + \sin{\left(4 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(2 x \right)} d x} + \int{\sin{\left(4 x \right)} d x}\right)}}}{2}$$

设$$$u=2 x$$$。

则$$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{2}$$$。

所以，

$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}$$

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

$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{2}$$

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

$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$

回忆一下 $$$u=2 x$$$:

$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{4} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

设$$$u=4 x$$$。

则$$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{4}$$$。

积分变为

$$- \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\sin{\left(4 x \right)} d x}}}}{2} = - \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{2}$$

对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$- \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{2} = - \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{4}\right)}}}{2}$$

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

$$- \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = - \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

回忆一下 $$$u=4 x$$$:

$$- \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = - \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{\left(4 x\right)}} \right)}}{8}$$

因此，

$$\int{\sin{\left(3 x \right)} \cos{\left(x \right)} d x} = - \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left(4 x \right)}}{8}$$

加上积分常数：

$$\int{\sin{\left(3 x \right)} \cos{\left(x \right)} d x} = - \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left(4 x \right)}}{8}+C$$

$$$\int \sin{\left(3 x \right)} \cos{\left(x \right)}\, dx = \left(- \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left(4 x \right)}}{8}\right) + C$$$A