Integral von $$$\frac{1}{11} - \tan^{2}{\left(x \right)}$$$

Bestimme $$$\int \left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)\, dx$$$.

Gliedweise integrieren:

$${\color{red}{\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{11} d x} - \int{\tan^{2}{\left(x \right)} d x}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=\frac{1}{11}$$$ an:

$$- \int{\tan^{2}{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{11} d x}}} = - \int{\tan^{2}{\left(x \right)} d x} + {\color{red}{\left(\frac{x}{11}\right)}}$$

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).

Also,

$$\frac{x}{11} - {\color{red}{\int{\tan^{2}{\left(x \right)} d x}}} = \frac{x}{11} - {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}}$$

Forme den Bruch um und zerlege ihn:

$$\frac{x}{11} - {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}} = \frac{x}{11} - {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}}$$

Gliedweise integrieren:

$$\frac{x}{11} - {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}} = \frac{x}{11} - {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$

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

$$\frac{x}{11} + \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{1 d u}}} = \frac{x}{11} + \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{u}}$$

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

$$- u + \frac{x}{11} + {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = - u + \frac{x}{11} + {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

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

$$\frac{x}{11} + \operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}} = \frac{x}{11} + \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} - {\color{red}{\tan{\left(x \right)}}}$$

Daher,

$$\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x} = \frac{x}{11} - \tan{\left(x \right)} + \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$

Vereinfachen:

$$\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x} = \frac{12 x}{11} - \tan{\left(x \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)d x} = \frac{12 x}{11} - \tan{\left(x \right)}+C$$

$$$\int \left(\frac{1}{11} - \tan^{2}{\left(x \right)}\right)\, dx = \left(\frac{12 x}{11} - \tan{\left(x \right)}\right) + C$$$A