$$$6 \sin{\left(3 x \right)}$$$ 的積分

求$$$\int 6 \sin{\left(3 x \right)}\, dx$$$。

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=6$$$ 與 $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$：

$${\color{red}{\int{6 \sin{\left(3 x \right)} d x}}} = {\color{red}{\left(6 \int{\sin{\left(3 x \right)} d x}\right)}}$$

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

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

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$：

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

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

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

回顧一下 $$$u=3 x$$$：

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

因此，

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

加上積分常數：

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

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