Integral von $$$\sin{\left(x \right)} \sec^{2}{\left(\cos{\left(x \right)} \right)}$$$

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

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 $$$\sin{\left(x \right)} dx = - du$$$.

Somit,

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

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

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

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

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

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

$$- \tan{\left({\color{red}{u}} \right)} = - \tan{\left({\color{red}{\cos{\left(x \right)}}} \right)}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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