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

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

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

Då gäller $$$x=\operatorname{atan}{\left(u \right)}$$$ och $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (stegen kan ses »).

Integralen kan omskrivas som

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

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

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

Integralen av $$$\frac{1}{u^{2} + 1}$$$ är $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

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

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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