$$$\sin{\left(5 x \right)}$$$ 的积分

求$$$\int \sin{\left(5 x \right)}\, dx$$$。

设$$$u=5 x$$$。

则$$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{5}$$$。

所以，

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

对 $$$c=\frac{1}{5}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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

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

回忆一下 $$$u=5 x$$$:

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

因此，

$$\int{\sin{\left(5 x \right)} d x} = - \frac{\cos{\left(5 x \right)}}{5}$$

加上积分常数：

$$\int{\sin{\left(5 x \right)} d x} = - \frac{\cos{\left(5 x \right)}}{5}+C$$

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