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

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

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

Entonces $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\sec^{2}{\left(x \right)} dx = du$$$.

La integral puede reescribirse como

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

La integral de la función exponencial es $$$\int{e^{u} d u} = e^{u}$$$:

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

Recordemos que $$$u=\tan{\left(x \right)}$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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