Integral von $$$\frac{1}{4 \cos{\left(x \right)} + 5}$$$

Bestimme $$$\int \frac{1}{4 \cos{\left(x \right)} + 5}\, dx$$$.

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.:

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

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).

Daher,

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

Vereinfachen:

$${\color{red}{\int{\frac{2}{\left(u^{2} + 1\right) \left(\frac{4 \left(1 - u^{2}\right)}{u^{2} + 1} + 5\right)} d u}}} = {\color{red}{\int{\frac{2}{u^{2} + 9} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=2$$$ und $$$f{\left(u \right)} = \frac{1}{u^{2} + 9}$$$ an:

$${\color{red}{\int{\frac{2}{u^{2} + 9} d u}}} = {\color{red}{\left(2 \int{\frac{1}{u^{2} + 9} d u}\right)}}$$

Sei $$$v=\frac{u}{3}$$$.

Dann $$$dv=\left(\frac{u}{3}\right)^{\prime }du = \frac{du}{3}$$$ (die Schritte sind » zu sehen), und es gilt $$$du = 3 dv$$$.

Also,

$$2 {\color{red}{\int{\frac{1}{u^{2} + 9} d u}}} = 2 {\color{red}{\int{\frac{1}{3 \left(v^{2} + 1\right)} d v}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(v \right)} = \frac{1}{v^{2} + 1}$$$ an:

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

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

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

Zur Erinnerung: $$$v=\frac{u}{3}$$$:

$$\frac{2 \operatorname{atan}{\left({\color{red}{v}} \right)}}{3} = \frac{2 \operatorname{atan}{\left({\color{red}{\left(\frac{u}{3}\right)}} \right)}}{3}$$

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

$$\frac{2 \operatorname{atan}{\left(\frac{{\color{red}{u}}}{3} \right)}}{3} = \frac{2 \operatorname{atan}{\left(\frac{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}{3} \right)}}{3}$$

Daher,

$$\int{\frac{1}{4 \cos{\left(x \right)} + 5} d x} = \frac{2 \operatorname{atan}{\left(\frac{\tan{\left(\frac{x}{2} \right)}}{3} \right)}}{3}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{4 \cos{\left(x \right)} + 5} d x} = \frac{2 \operatorname{atan}{\left(\frac{\tan{\left(\frac{x}{2} \right)}}{3} \right)}}{3}+C$$

$$$\int \frac{1}{4 \cos{\left(x \right)} + 5}\, dx = \frac{2 \operatorname{atan}{\left(\frac{\tan{\left(\frac{x}{2} \right)}}{3} \right)}}{3} + C$$$A