$$$n \tan{\left(x \right)}$$$ 关于$$$x$$$的积分

求$$$\int n \tan{\left(x \right)}\, dx$$$。

对 $$$c=n$$$ 和 $$$f{\left(x \right)} = \tan{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{n \tan{\left(x \right)} d x}}} = {\color{red}{n \int{\tan{\left(x \right)} d x}}}$$

将正切表示为 $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

$$n {\color{red}{\int{\tan{\left(x \right)} d x}}} = n {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

设$$$u=\cos{\left(x \right)}$$$。

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

因此，

$$n {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = n {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

$$n {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = n {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- n {\color{red}{\int{\frac{1}{u} d u}}} = - n {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回忆一下 $$$u=\cos{\left(x \right)}$$$:

$$- n \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - n \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)}$$

因此，

$$\int{n \tan{\left(x \right)} d x} = - n \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}$$

加上积分常数：

$$\int{n \tan{\left(x \right)} d x} = - n \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}+C$$

$$$\int n \tan{\left(x \right)}\, dx = - n \ln\left(\left|{\cos{\left(x \right)}}\right|\right) + C$$$A