$$$1 - \tan{\left(x \right)}$$$ 的积分

求$$$\int \left(1 - \tan{\left(x \right)}\right)\, dx$$$。

逐项积分：

$${\color{red}{\int{\left(1 - \tan{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{\tan{\left(x \right)} d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

$$- \int{\tan{\left(x \right)} d x} + {\color{red}{\int{1 d x}}} = - \int{\tan{\left(x \right)} d x} + {\color{red}{x}}$$

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

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

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$$$\sin{\left(x \right)} dx = - du$$$。

该积分可以改写为

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

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

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

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

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

回忆一下 $$$u=\cos{\left(x \right)}$$$:

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

因此，

$$\int{\left(1 - \tan{\left(x \right)}\right)d x} = x + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}$$

加上积分常数：

$$\int{\left(1 - \tan{\left(x \right)}\right)d x} = x + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}+C$$

$$$\int \left(1 - \tan{\left(x \right)}\right)\, dx = \left(x + \ln\left(\left|{\cos{\left(x \right)}}\right|\right)\right) + C$$$A