Integral von $$$\frac{1}{3 \cos{\left(x \right)} - 5}$$$

Bestimme $$$\int \frac{1}{3 \cos{\left(x \right)} - 5}\, dx$$$.

Schreibe den Integranden mithilfe der Formel $$$\cos{\left(x \right)}=\frac{1 - \tan^{2}{\left(\frac{x}{2} \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} + 1}$$$ um.:

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

Sei $$$u=\tan{\left(\frac{x}{2} \right)}$$$.

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

Das Integral lässt sich umschreiben als

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

Vereinfachen:

$${\color{red}{\int{\frac{2}{\left(u^{2} + 1\right) \left(\frac{3 \left(1 - u^{2}\right)}{u^{2} + 1} - 5\right)} d u}}} = {\color{red}{\int{\left(- \frac{1}{4 u^{2} + 1}\right)d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = \frac{1}{4 u^{2} + 1}$$$ an:

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

Sei $$$v=2 u$$$.

Dann $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (die Schritte sind » zu sehen), und es gilt $$$du = \frac{dv}{2}$$$.

Also,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(v \right)} = \frac{1}{v^{2} + 1}$$$ an:

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

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

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

Zur Erinnerung: $$$v=2 u$$$:

$$- \frac{\operatorname{atan}{\left({\color{red}{v}} \right)}}{2} = - \frac{\operatorname{atan}{\left({\color{red}{\left(2 u\right)}} \right)}}{2}$$

Zur Erinnerung: $$$u=\tan{\left(\frac{x}{2} \right)}$$$:

$$- \frac{\operatorname{atan}{\left(2 {\color{red}{u}} \right)}}{2} = - \frac{\operatorname{atan}{\left(2 {\color{red}{\tan{\left(\frac{x}{2} \right)}}} \right)}}{2}$$

Daher,

$$\int{\frac{1}{3 \cos{\left(x \right)} - 5} d x} = - \frac{\operatorname{atan}{\left(2 \tan{\left(\frac{x}{2} \right)} \right)}}{2}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{3 \cos{\left(x \right)} - 5} d x} = - \frac{\operatorname{atan}{\left(2 \tan{\left(\frac{x}{2} \right)} \right)}}{2}+C$$

$$$\int \frac{1}{3 \cos{\left(x \right)} - 5}\, dx = - \frac{\operatorname{atan}{\left(2 \tan{\left(\frac{x}{2} \right)} \right)}}{2} + C$$$A