Integral von $$$\frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}$$$

Bestimme $$$\int \frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx$$$.

Multiplizieren Sie Zähler und Nenner mit $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ und wandeln Sie $$$\frac{\cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$ in $$$\frac{1}{\tan^{2}{\left(x \right)}}$$$ um:

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

Ziehen Sie zwei Kosinusfaktoren heraus und schreiben Sie diese mithilfe der Formel $$$\frac{1}{\cos^{2}{\left(x \right)}}=\sec^{2}{\left(x \right)}$$$ in Abhängigkeit von der Sekans um.:

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

Schreiben Sie den Kosinus in Abhängigkeit vom Tangens mithilfe der Formel $$$\cos^{2}{\left(x \right)}=\frac{1}{\tan^{2}{\left(x \right)} + 1}$$$ um.:

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

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

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

Das Integral wird zu

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

Expand the expression:

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

Gliedweise integrieren:

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

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

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

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

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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