Integral von $$$\tan^{6}{\left(x \right)}$$$

Bestimme $$$\int \tan^{6}{\left(x \right)}\, dx$$$.

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

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

Das Integral wird zu

$${\color{red}{\int{\tan^{6}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{u^{6}}{u^{2} + 1} d u}}}$$

Da der Grad des Zählers mindestens so groß ist wie der des Nenners, führen Sie eine Polynomdivision durch (die Schritte sind » zu sehen):

$${\color{red}{\int{\frac{u^{6}}{u^{2} + 1} d u}}} = {\color{red}{\int{\left(u^{4} - u^{2} + 1 - \frac{1}{u^{2} + 1}\right)d u}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(u^{4} - u^{2} + 1 - \frac{1}{u^{2} + 1}\right)d u}}} = {\color{red}{\left(\int{1 d u} - \int{u^{2} d u} + \int{u^{4} d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$

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

$$- \int{u^{2} d u} + \int{u^{4} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{1 d u}}} = - \int{u^{2} d u} + \int{u^{4} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{u}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=4$$$ an:

$$u - \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{u^{4} d u}}}=u - \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=u - \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

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

$$\frac{u^{5}}{5} + u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{u^{2} d u}}}=\frac{u^{5}}{5} + u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=\frac{u^{5}}{5} + u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$\frac{u^{5}}{5} - \frac{u^{3}}{3} + u - {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = \frac{u^{5}}{5} - \frac{u^{3}}{3} + u - {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

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

$$- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}} - \frac{{\color{red}{u}}^{3}}{3} + \frac{{\color{red}{u}}^{5}}{5} = - \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} + {\color{red}{\tan{\left(x \right)}}} - \frac{{\color{red}{\tan{\left(x \right)}}}^{3}}{3} + \frac{{\color{red}{\tan{\left(x \right)}}}^{5}}{5}$$

Daher,

$$\int{\tan^{6}{\left(x \right)} d x} = \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)} - \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$

Vereinfachen:

$$\int{\tan^{6}{\left(x \right)} d x} = - x + \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\tan^{6}{\left(x \right)} d x} = - x + \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+C$$

$$$\int \tan^{6}{\left(x \right)}\, dx = \left(- x + \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}\right) + C$$$A