Integral von $$$\sec^{5}{\left(x \right)}$$$

Bestimme $$$\int \sec^{5}{\left(x \right)}\, dx$$$.

Für das Integral $$$\int{\sec^{5}{\left(x \right)} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=\sec^{3}{\left(x \right)}$$$ und $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Dann gilt $$$\operatorname{du}=\left(\sec^{3}{\left(x \right)}\right)^{\prime }dx=3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (Rechenschritte siehe »).

Somit,

$$\int{\sec^{5}{\left(x \right)} d x}=\sec^{3}{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$

Konstante ausklammern:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$

Wenden Sie die Formel $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$ an:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}$$

Ausmultiplizieren:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}$$

Das Integral einer Differenz ist die Differenz der Integrale:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 \int{\sec^{5}{\left(x \right)} d x}$$

Somit erhalten wir die folgende einfache lineare Gleichung für das Integral:

$${\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 {\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}$$

Löst man es, erhält man, dass

$$\int{\sec^{5}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \int{\sec^{3}{\left(x \right)} d x}}{4}$$

Für das Integral $$$\int{\sec^{3}{\left(x \right)} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=\sec{\left(x \right)}$$$ und $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Dann gilt $$$\operatorname{du}=\left(\sec{\left(x \right)}\right)^{\prime }dx=\tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (Rechenschritte siehe »).

Das Integral lässt sich umschreiben als

$$\int{\sec^{3}{\left(x \right)} d x}=\sec{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot \tan{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}$$

Wenden Sie die Formel $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$ an:

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

Ausmultiplizieren:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}$$

Das Integral einer Differenz ist die Differenz der Integrale:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - \int{\sec^{3}{\left(x \right)} d x}$$

Somit erhalten wir die folgende einfache lineare Gleichung für das Integral:

$${\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}$$

Löst man es, erhält man, dass

$$\int{\sec^{3}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}$$

Daher,

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}}{4} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\left(\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}\right)}}}{4}$$

Schreibe die Sekante als $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$ um:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\sec{\left(x \right)} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8}$$

Schreibe den Kosinus mithilfe der Formel $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ in Abhängigkeit vom Sinus um und schreibe anschließend den Sinus mithilfe der Doppelwinkel-Formel $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$ um.:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$

Multipliziere Zähler und Nenner mit $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$

Sei $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.

Dann $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

Das Integral lässt sich umschreiben als

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

Das Integral von $$$\frac{1}{u}$$$ ist $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Zur Erinnerung: $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} = \frac{3 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$

Daher,

$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}+C$$

$$$\int \sec^{5}{\left(x \right)}\, dx = \left(\frac{3 \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}\right) + C$$$A