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

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

设$$$u=\ln{\left(x \right)}$$$。

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

积分变为

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

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

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

设$$$v=\cos{\left(u \right)}$$$。

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

所以，

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

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

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

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

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

回忆一下 $$$v=\cos{\left(u \right)}$$$:

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

回忆一下 $$$u=\ln{\left(x \right)}$$$:

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

因此，

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

加上积分常数：

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

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