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

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

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

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

Das Integral von $$$\tan{\left(x \right)} \sec{\left(x \right)}$$$ ist $$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} = \sec{\left(x \right)}$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

$$\int{n \tan{\left(x \right)} \sec{\left(x \right)} d x} = n \sec{\left(x \right)}+C$$

$$$\int n \tan{\left(x \right)} \sec{\left(x \right)}\, dx = n \sec{\left(x \right)} + C$$$A