Integral de $$$\operatorname{sech}{\left(x \right)}$$$

Halla $$$\int \operatorname{sech}{\left(x \right)}\, dx$$$.

Reescribe la secante hiperbólica usando el exponente $$$\operatorname{sech}\left(x\right)=\frac{2}{e^{\left(x\right)}+e^{-\left(x\right)}}$$$:

$${\color{red}{\int{\operatorname{sech}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{2}{e^{x} + e^{- x}} d x}}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=2$$$ y $$$f{\left(x \right)} = \frac{1}{e^{x} + e^{- x}}$$$:

$${\color{red}{\int{\frac{2}{e^{x} + e^{- x}} d x}}} = {\color{red}{\left(2 \int{\frac{1}{e^{x} + e^{- x}} d x}\right)}}$$

Simplify:

$$2 {\color{red}{\int{\frac{1}{e^{x} + e^{- x}} d x}}} = 2 {\color{red}{\int{\frac{e^{x}}{e^{2 x} + 1} d x}}}$$

Sea $$$u=e^{x}$$$.

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

La integral puede reescribirse como

$$2 {\color{red}{\int{\frac{e^{x}}{e^{2 x} + 1} d x}}} = 2 {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}$$

La integral de $$$\frac{1}{u^{2} + 1}$$$ es $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$2 {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = 2 {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

Recordemos que $$$u=e^{x}$$$:

$$2 \operatorname{atan}{\left({\color{red}{u}} \right)} = 2 \operatorname{atan}{\left({\color{red}{e^{x}}} \right)}$$

Por lo tanto,

$$\int{\operatorname{sech}{\left(x \right)} d x} = 2 \operatorname{atan}{\left(e^{x} \right)}$$

Añade la constante de integración:

$$\int{\operatorname{sech}{\left(x \right)} d x} = 2 \operatorname{atan}{\left(e^{x} \right)}+C$$

$$$\int \operatorname{sech}{\left(x \right)}\, dx = 2 \operatorname{atan}{\left(e^{x} \right)} + C$$$A