$$$18 \pi^{2} \tan{\left(18 x \right)}$$$ 的积分

求$$$\int 18 \pi^{2} \tan{\left(18 x \right)}\, dx$$$。

对 $$$c=18 \pi^{2}$$$ 和 $$$f{\left(x \right)} = \tan{\left(18 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{18 \pi^{2} \tan{\left(18 x \right)} d x}}} = {\color{red}{\left(18 \pi^{2} \int{\tan{\left(18 x \right)} d x}\right)}}$$

设$$$u=18 x$$$。

则$$$du=\left(18 x\right)^{\prime }dx = 18 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{18}$$$。

该积分可以改写为

$$18 \pi^{2} {\color{red}{\int{\tan{\left(18 x \right)} d x}}} = 18 \pi^{2} {\color{red}{\int{\frac{\tan{\left(u \right)}}{18} d u}}}$$

对 $$$c=\frac{1}{18}$$$ 和 $$$f{\left(u \right)} = \tan{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$18 \pi^{2} {\color{red}{\int{\frac{\tan{\left(u \right)}}{18} d u}}} = 18 \pi^{2} {\color{red}{\left(\frac{\int{\tan{\left(u \right)} d u}}{18}\right)}}$$

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

$$\pi^{2} {\color{red}{\int{\tan{\left(u \right)} d u}}} = \pi^{2} {\color{red}{\int{\frac{\sin{\left(u \right)}}{\cos{\left(u \right)}} d u}}}$$

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$$$\sin{\left(u \right)} du = - dv$$$。

因此，

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

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

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

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

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

回忆一下 $$$v=\cos{\left(u \right)}$$$:

$$- \pi^{2} \ln{\left(\left|{{\color{red}{v}}}\right| \right)} = - \pi^{2} \ln{\left(\left|{{\color{red}{\cos{\left(u \right)}}}}\right| \right)}$$

回忆一下 $$$u=18 x$$$:

$$- \pi^{2} \ln{\left(\left|{\cos{\left({\color{red}{u}} \right)}}\right| \right)} = - \pi^{2} \ln{\left(\left|{\cos{\left({\color{red}{\left(18 x\right)}} \right)}}\right| \right)}$$

因此，

$$\int{18 \pi^{2} \tan{\left(18 x \right)} d x} = - \pi^{2} \ln{\left(\left|{\cos{\left(18 x \right)}}\right| \right)}$$

加上积分常数：

$$\int{18 \pi^{2} \tan{\left(18 x \right)} d x} = - \pi^{2} \ln{\left(\left|{\cos{\left(18 x \right)}}\right| \right)}+C$$

$$$\int 18 \pi^{2} \tan{\left(18 x \right)}\, dx = - \pi^{2} \ln\left(\left|{\cos{\left(18 x \right)}}\right|\right) + C$$$A