Integral von $$$\tan^{2}{\left(u \right)}$$$

Bestimme $$$\int \tan^{2}{\left(u \right)}\, du$$$.

Sei $$$v=\tan{\left(u \right)}$$$.

Dann gelten $$$u=\operatorname{atan}{\left(v \right)}$$$ und $$$du=\left(\operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{dv}{v^{2} + 1}$$$ (die Schritte sind » zu sehen).

Daher,

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

Forme den Bruch um und zerlege ihn:

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dv = c v$$$ mit $$$c=1$$$ an:

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

Das Integral von $$$\frac{1}{v^{2} + 1}$$$ ist $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

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

Zur Erinnerung: $$$v=\tan{\left(u \right)}$$$:

$$- \operatorname{atan}{\left({\color{red}{v}} \right)} + {\color{red}{v}} = - \operatorname{atan}{\left({\color{red}{\tan{\left(u \right)}}} \right)} + {\color{red}{\tan{\left(u \right)}}}$$

Daher,

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

Vereinfachen:

$$\int{\tan^{2}{\left(u \right)} d u} = - u + \tan{\left(u \right)}$$

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \tan^{2}{\left(u \right)}\, du = \left(- u + \tan{\left(u \right)}\right) + C$$$A