Integral von $$$\frac{1}{\sin{\left(x \right)}}$$$

Bestimme $$$\int \frac{1}{\sin{\left(x \right)}}\, dx$$$.

Schreibe den Sinus mithilfe der Doppelwinkel-Formel um $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

$${\color{red}{\int{\frac{1}{\sin{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}}$$

Multipliziere Zähler und Nenner mit $$$\sec^2\left(\frac{x}{2} \right)$$$:

$${\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}}$$

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

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

Das Integral wird zu

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

Das Integral von $$$\frac{1}{u}$$$ ist $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}\right| \right)}$$

Daher,

$$\int{\frac{1}{\sin{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sin{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}+C$$

$$$\int \frac{1}{\sin{\left(x \right)}}\, dx = \ln\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right|\right) + C$$$A