Integral of $$$e^{\tan{\left(x \right)}} \sec^{2}{\left(x \right)}$$$

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

Let $$$u=\tan{\left(x \right)}$$$.

Then $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (steps can be seen »), and we have that $$$\sec^{2}{\left(x \right)} dx = du$$$.

Thus,

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

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

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

Recall that $$$u=\tan{\left(x \right)}$$$:

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

Therefore,

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

Add the constant of integration:

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

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