Integral von $$$\sec^{2}{\left(x y \right)}$$$ nach $$$x$$$

Bestimme $$$\int \sec^{2}{\left(x y \right)}\, dx$$$.

Sei $$$u=x y$$$.

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

Somit,

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

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

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

Das Integral von $$$\sec^{2}{\left(u \right)}$$$ ist $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$:

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

Zur Erinnerung: $$$u=x y$$$:

$$\frac{\tan{\left({\color{red}{u}} \right)}}{y} = \frac{\tan{\left({\color{red}{x y}} \right)}}{y}$$

Daher,

$$\int{\sec^{2}{\left(x y \right)} d x} = \frac{\tan{\left(x y \right)}}{y}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\sec^{2}{\left(x y \right)} d x} = \frac{\tan{\left(x y \right)}}{y}+C$$

$$$\int \sec^{2}{\left(x y \right)}\, dx = \frac{\tan{\left(x y \right)}}{y} + C$$$A