Integral von $$$\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}}$$$

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

Drücke den Integranden mithilfe von Sinus und/oder Kosinus aus.:

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

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

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

Somit,

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

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

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

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

$$- {\color{red}{u}}^{-1} = - {\color{red}{\sin{\left(x \right)}}}^{-1}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x} = - \frac{1}{\sin{\left(x \right)}}+C$$

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