Integral von $$$n \tan{\left(x \right)}$$$ nach $$$x$$$

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=n$$$ und $$$f{\left(x \right)} = \tan{\left(x \right)}$$$ an:

$${\color{red}{\int{n \tan{\left(x \right)} d x}}} = {\color{red}{n \int{\tan{\left(x \right)} d x}}}$$

Schreibe die Tangente als $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$ um:

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

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

Dann $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\sin{\left(x \right)} dx = - du$$$.

Daher,

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

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

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

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

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

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

$$- n \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - n \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)}$$

Daher,

$$\int{n \tan{\left(x \right)} d x} = - n \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{n \tan{\left(x \right)} d x} = - n \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}+C$$

$$$\int n \tan{\left(x \right)}\, dx = - n \ln\left(\left|{\cos{\left(x \right)}}\right|\right) + C$$$A