Integraal van $$$\frac{1}{\cosh{\left(x \right)}}$$$

Bepaal $$$\int \frac{1}{\cosh{\left(x \right)}}\, dx$$$.

Herschrijf de hyperbolische functie in termen van de exponentiële functie:

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

Vereenvoudig de integraand:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=2$$$ en $$$f{\left(x \right)} = \frac{1}{e^{x} + e^{- x}}$$$:

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

Simplify:

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

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

Dan $$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$e^{x} dx = du$$$.

De integraal kan worden herschreven als

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

De integraal van $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

We herinneren eraan dat $$$u=e^{x}$$$:

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

Dus,

$$\int{\frac{1}{\cosh{\left(x \right)}} d x} = 2 \operatorname{atan}{\left(e^{x} \right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{\cosh{\left(x \right)}} d x} = 2 \operatorname{atan}{\left(e^{x} \right)}+C$$

$$$\int \frac{1}{\cosh{\left(x \right)}}\, dx = 2 \operatorname{atan}{\left(e^{x} \right)} + C$$$A