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

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

设$$$u=\tan{\left(x \right)}$$$。

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

因此，

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

指数函数的积分为 $$$\int{e^{u} d u} = e^{u}$$$：

$${\color{red}{\int{e^{u} d u}}} = {\color{red}{e^{u}}}$$

回忆一下 $$$u=\tan{\left(x \right)}$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{\tan{\left(x \right)}}}}$$

因此，

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

加上积分常数：

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

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