$$$e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)}$$$ 的积分

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

设$$$u=\sec^{2}{\left(x \right)}$$$。

则$$$du=\left(\sec^{2}{\left(x \right)}\right)^{\prime }dx = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$$$ (步骤见»)，并有$$$\tan{\left(x \right)} \sec^{2}{\left(x \right)} dx = \frac{du}{2}$$$。

该积分可以改写为

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{e^{u}}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

该积分（指数积分）没有闭式表达式：

$$\frac{{\color{red}{\int{\frac{e^{u}}{u} d u}}}}{2} = \frac{{\color{red}{\operatorname{Ei}{\left(u \right)}}}}{2}$$

回忆一下 $$$u=\sec^{2}{\left(x \right)}$$$:

$$\frac{\operatorname{Ei}{\left({\color{red}{u}} \right)}}{2} = \frac{\operatorname{Ei}{\left({\color{red}{\sec^{2}{\left(x \right)}}} \right)}}{2}$$

因此，

$$\int{e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)} d x} = \frac{\operatorname{Ei}{\left(\sec^{2}{\left(x \right)} \right)}}{2}$$

加上积分常数：

$$\int{e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)} d x} = \frac{\operatorname{Ei}{\left(\sec^{2}{\left(x \right)} \right)}}{2}+C$$

$$$\int e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)}\, dx = \frac{\operatorname{Ei}{\left(\sec^{2}{\left(x \right)} \right)}}{2} + C$$$A