$$$\frac{1}{\tan{\left(x \right)}}$$$ 的积分

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

设$$$u=\tan{\left(x \right)}$$$。

则 $$$x=\operatorname{atan}{\left(u \right)}$$$ 且 $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$（步骤见»）。

因此，

$${\color{red}{\int{\frac{1}{\tan{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{u \left(u^{2} + 1\right)} d u}}}$$

设$$$v=u^{2} + 1$$$。

则$$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (步骤见»)，并有$$$u du = \frac{dv}{2}$$$。

所以，

$${\color{red}{\int{\frac{1}{u \left(u^{2} + 1\right)} d u}}} = {\color{red}{\int{\frac{1}{2 v \left(v - 1\right)} d v}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(v \right)} = \frac{1}{v \left(v - 1\right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$${\color{red}{\int{\frac{1}{2 v \left(v - 1\right)} d v}}} = {\color{red}{\left(\frac{\int{\frac{1}{v \left(v - 1\right)} d v}}{2}\right)}}$$

进行部分分式分解（步骤可见»）:

$$\frac{{\color{red}{\int{\frac{1}{v \left(v - 1\right)} d v}}}}{2} = \frac{{\color{red}{\int{\left(\frac{1}{v - 1} - \frac{1}{v}\right)d v}}}}{2}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(\frac{1}{v - 1} - \frac{1}{v}\right)d v}}}}{2} = \frac{{\color{red}{\left(- \int{\frac{1}{v} d v} + \int{\frac{1}{v - 1} d v}\right)}}}{2}$$

设$$$w=v - 1$$$。

则$$$dw=\left(v - 1\right)^{\prime }dv = 1 dv$$$ (步骤见»)，并有$$$dv = dw$$$。

所以，

$$- \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\int{\frac{1}{v - 1} d v}}}}{2} = - \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2}$$

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

$$- \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2} = - \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{2}$$

回忆一下 $$$w=v - 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{2} - \frac{\int{\frac{1}{v} d v}}{2} = \frac{\ln{\left(\left|{{\color{red}{\left(v - 1\right)}}}\right| \right)}}{2} - \frac{\int{\frac{1}{v} d v}}{2}$$

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

$$\frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

回忆一下 $$$v=u^{2} + 1$$$:

$$\frac{\ln{\left(\left|{-1 + {\color{red}{v}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{-1 + {\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{2}$$

回忆一下 $$$u=\tan{\left(x \right)}$$$:

$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{2} + \frac{\ln{\left({\color{red}{u}}^{2} \right)}}{2} = - \frac{\ln{\left(1 + {\color{red}{\tan{\left(x \right)}}}^{2} \right)}}{2} + \frac{\ln{\left({\color{red}{\tan{\left(x \right)}}}^{2} \right)}}{2}$$

因此，

$$\int{\frac{1}{\tan{\left(x \right)}} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \frac{\ln{\left(\tan^{2}{\left(x \right)} \right)}}{2}$$

化简：

$$\int{\frac{1}{\tan{\left(x \right)}} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(x \right)} \right)}$$

加上积分常数：

$$\int{\frac{1}{\tan{\left(x \right)}} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(x \right)} \right)}+C$$

$$$\int \frac{1}{\tan{\left(x \right)}}\, dx = \left(- \frac{\ln\left(\tan^{2}{\left(x \right)} + 1\right)}{2} + \ln\left(\tan{\left(x \right)}\right)\right) + C$$$A