Integraal van $$$\frac{1}{34 \cosh{\left(x \right)}}$$$

Bepaal $$$\int \frac{1}{34 \cosh{\left(x \right)}}\, dx$$$.

Herschrijf de hyperbolische functie in termen van de exponentiële functie:

$${\color{red}{\int{\frac{1}{34 \cosh{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{34 \left(\frac{e^{x}}{2} + \frac{e^{- x}}{2}\right)} d x}}}$$

Vereenvoudig de integraand:

$${\color{red}{\int{\frac{1}{34 \left(\frac{e^{x}}{2} + \frac{e^{- x}}{2}\right)} d x}}} = {\color{red}{\int{\frac{1}{17 \left(e^{x} + e^{- x}\right)} d x}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{17}$$$ en $$$f{\left(x \right)} = \frac{1}{e^{x} + e^{- x}}$$$:

$${\color{red}{\int{\frac{1}{17 \left(e^{x} + e^{- x}\right)} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{e^{x} + e^{- x}} d x}}{17}\right)}}$$

Simplify:

$$\frac{{\color{red}{\int{\frac{1}{e^{x} + e^{- x}} d x}}}}{17} = \frac{{\color{red}{\int{\frac{e^{x}}{e^{2 x} + 1} d x}}}}{17}$$

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

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

Dus,

$$\frac{{\color{red}{\int{\frac{e^{x}}{e^{2 x} + 1} d x}}}}{17} = \frac{{\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{17}$$

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

$$\frac{{\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{17} = \frac{{\color{red}{\operatorname{atan}{\left(u \right)}}}}{17}$$

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

$$\frac{\operatorname{atan}{\left({\color{red}{u}} \right)}}{17} = \frac{\operatorname{atan}{\left({\color{red}{e^{x}}} \right)}}{17}$$

Dus,

$$\int{\frac{1}{34 \cosh{\left(x \right)}} d x} = \frac{\operatorname{atan}{\left(e^{x} \right)}}{17}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{34 \cosh{\left(x \right)}} d x} = \frac{\operatorname{atan}{\left(e^{x} \right)}}{17}+C$$

$$$\int \frac{1}{34 \cosh{\left(x \right)}}\, dx = \frac{\operatorname{atan}{\left(e^{x} \right)}}{17} + C$$$A