Integral von $$$\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}$$$

Bestimme $$$\int \frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}\, dx$$$.

Multiplizieren Sie Zähler und Nenner mit $$$\frac{1}{\cosh^{2}{\left(x \right)}}$$$ und wandeln Sie $$$\frac{\cosh^{2}{\left(x \right)}}{\sinh^{2}{\left(x \right)}}$$$ in $$$\frac{1}{\tanh^{2}{\left(x \right)}}$$$ um:

$${\color{red}{\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cosh^{4}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}}$$

Erhalte zwei hyperbolische Kosinusse und schreibe die anderen hyperbolischen Kosinusse mithilfe der Formel $$$\cosh^{2}{\left(x \right)}=\frac{1}{1 - \tanh^{2}{\left(x \right)}}$$$ in Abhängigkeit vom hyperbolischen Tangens um.:

$${\color{red}{\int{\frac{1}{\cosh^{4}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1 - \tanh^{2}{\left(x \right)}}{\cosh^{2}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}}$$

Sei $$$u=\tanh{\left(x \right)}$$$.

Dann $$$du=\left(\tanh{\left(x \right)}\right)^{\prime }dx = \operatorname{sech}^{2}{\left(x \right)} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\operatorname{sech}^{2}{\left(x \right)} dx = du$$$.

Also,

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

Expand the expression:

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

$$\int{\frac{1}{u^{2}} d u} - {\color{red}{\int{1 d u}}} = \int{\frac{1}{u^{2}} d u} - {\color{red}{u}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

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

Zur Erinnerung: $$$u=\tanh{\left(x \right)}$$$:

$$- {\color{red}{u}}^{-1} - {\color{red}{u}} = - {\color{red}{\tanh{\left(x \right)}}}^{-1} - {\color{red}{\tanh{\left(x \right)}}}$$

Daher,

$$\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x} = - \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x} = - \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}+C$$

$$$\int \frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}\, dx = \left(- \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}\right) + C$$$A