$$$\sin{\left(14 x \right)} \cos{\left(9 x \right)}$$$ 的積分

求$$$\int \sin{\left(14 x \right)} \cos{\left(9 x \right)}\, dx$$$。

使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$，令 $$$\alpha=14 x$$$ 與 $$$\beta=9 x$$$，將被積函數改寫:

$${\color{red}{\int{\sin{\left(14 x \right)} \cos{\left(9 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(5 x \right)}}{2} + \frac{\sin{\left(23 x \right)}}{2}\right)d x}}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \sin{\left(5 x \right)} + \sin{\left(23 x \right)}$$$：

$${\color{red}{\int{\left(\frac{\sin{\left(5 x \right)}}{2} + \frac{\sin{\left(23 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(5 x \right)} + \sin{\left(23 x \right)}\right)d x}}{2}\right)}}$$

逐項積分：

$$\frac{{\color{red}{\int{\left(\sin{\left(5 x \right)} + \sin{\left(23 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(5 x \right)} d x} + \int{\sin{\left(23 x \right)} d x}\right)}}}{2}$$

令 $$$u=5 x$$$。

則 $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{5}$$$。

該積分可改寫為

$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{2}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{5}$$$ 與 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$：

$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{2} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{2}$$

正弦函數的積分為 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$：

$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{10} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{10}$$

回顧一下 $$$u=5 x$$$：

$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{10} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{10}$$

令 $$$u=23 x$$$。

則 $$$du=\left(23 x\right)^{\prime }dx = 23 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{23}$$$。

因此，

$$- \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\sin{\left(23 x \right)} d x}}}}{2} = - \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{23} d u}}}}{2}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{23}$$$ 與 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$：

$$- \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{23} d u}}}}{2} = - \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{23}\right)}}}{2}$$

正弦函數的積分為 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$：

$$- \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{46} = - \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{46}$$

回顧一下 $$$u=23 x$$$：

$$- \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left({\color{red}{u}} \right)}}{46} = - \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left({\color{red}{\left(23 x\right)}} \right)}}{46}$$

因此，

$$\int{\sin{\left(14 x \right)} \cos{\left(9 x \right)} d x} = - \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left(23 x \right)}}{46}$$

加上積分常數：

$$\int{\sin{\left(14 x \right)} \cos{\left(9 x \right)} d x} = - \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left(23 x \right)}}{46}+C$$

$$$\int \sin{\left(14 x \right)} \cos{\left(9 x \right)}\, dx = \left(- \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left(23 x \right)}}{46}\right) + C$$$A