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

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

Sei $$$u=\sec^{2}{\left(x \right)}$$$.

Dann $$$du=\left(\sec^{2}{\left(x \right)}\right)^{\prime }dx = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\tan{\left(x \right)} \sec^{2}{\left(x \right)} dx = \frac{du}{2}$$$.

Das Integral wird zu

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

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

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

Dieses Integral (Exponentialintegral) besitzt keine geschlossene Form:

$$\frac{{\color{red}{\int{\frac{e^{u}}{u} d u}}}}{2} = \frac{{\color{red}{\operatorname{Ei}{\left(u \right)}}}}{2}$$

Zur Erinnerung: $$$u=\sec^{2}{\left(x \right)}$$$:

$$\frac{\operatorname{Ei}{\left({\color{red}{u}} \right)}}{2} = \frac{\operatorname{Ei}{\left({\color{red}{\sec^{2}{\left(x \right)}}} \right)}}{2}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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