Integralen av $$$\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}$$$

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

Multiplicera täljaren och nämnaren med $$$\frac{1}{\cosh^{2}{\left(x \right)}}$$$ och omvandla $$$\frac{\cosh^{2}{\left(x \right)}}{\sinh^{2}{\left(x \right)}}$$$ till $$$\frac{1}{\tanh^{2}{\left(x \right)}}$$$:

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

Ta två hyperboliska cosinus och skriv om övriga hyperboliska cosinus i termer av den hyperboliska tangenten med hjälp av formeln $$$\cosh^{2}{\left(x \right)}=\frac{1}{1 - \tanh^{2}{\left(x \right)}}$$$:

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

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

Då $$$du=\left(\tanh{\left(x \right)}\right)^{\prime }dx = \operatorname{sech}^{2}{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\operatorname{sech}^{2}{\left(x \right)} dx = du$$$.

Alltså,

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

Expand the expression:

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

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=1$$$:

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

Kom ihåg att $$$u=\tanh{\left(x \right)}$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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