Integral von $$$\sec^{4}{\left(\theta \right)}$$$

Bestimme $$$\int \sec^{4}{\left(\theta \right)}\, d\theta$$$.

Ziehen Sie zwei Sekansfaktoren heraus und drücken Sie alles andere in Abhängigkeit vom Tangens aus, mithilfe der Formel $$$\sec^2\left( \alpha \right)=\tan^2\left( \alpha \right) + 1$$$ mit $$$\alpha=\theta$$$:

$${\color{red}{\int{\sec^{4}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\left(\tan^{2}{\left(\theta \right)} + 1\right) \sec^{2}{\left(\theta \right)} d \theta}}}$$

Sei $$$u=\tan{\left(\theta \right)}$$$.

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

Also,

$${\color{red}{\int{\left(\tan^{2}{\left(\theta \right)} + 1\right) \sec^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\left(u^{2} + 1\right)d u}}}$$

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

$$\int{u^{2} d u} + {\color{red}{\int{1 d u}}} = \int{u^{2} d u} + {\color{red}{u}}$$

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

$$u + {\color{red}{\int{u^{2} d u}}}=u + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=u + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Zur Erinnerung: $$$u=\tan{\left(\theta \right)}$$$:

$${\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = {\color{red}{\tan{\left(\theta \right)}}} + \frac{{\color{red}{\tan{\left(\theta \right)}}}^{3}}{3}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \sec^{4}{\left(\theta \right)}\, d\theta = \left(\frac{\tan^{3}{\left(\theta \right)}}{3} + \tan{\left(\theta \right)}\right) + C$$$A