$$$- 2 \sin{\left(x \right)} \cos{\left(2 x \right)}$$$ 的積分

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

使用公式 $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$，以 $$$\alpha=x$$$ 和 $$$\beta=2 x$$$ 將 $$$\sin\left(x \right)\cos\left(2 x \right)$$$ 改寫:

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

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 $$$f{\left(x \right)} = 2 \sin{\left(x \right)} - 2 \sin{\left(3 x \right)}$$$：

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

逐項積分：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=2$$$ 與 $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$：

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

令 $$$u=3 x$$$。

則 $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{3}$$$。

該積分可改寫為

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{3}$$$ 與 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$：

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

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

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

回顧一下 $$$u=3 x$$$：

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

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

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

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

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

因此，

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

加上積分常數：

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

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