Integral von $$$\csc^{4}{\left(x \right)}$$$

Bestimme $$$\int \csc^{4}{\left(x \right)}\, dx$$$.

Klammere zwei Kosekans-Faktoren aus und schreibe alles andere in Abhängigkeit vom Kotangens, mithilfe der Formel $$$\csc^2\left( \alpha \right)=\cot^2\left( \alpha \right)+1$$$ mit $$$\alpha=x$$$:

$${\color{red}{\int{\csc^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \csc^{2}{\left(x \right)} d x}}}$$

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

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

Das Integral wird zu

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

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

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

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: $$$u=\cot{\left(x \right)}$$$:

$$- {\color{red}{u}} - \frac{{\color{red}{u}}^{3}}{3} = - {\color{red}{\cot{\left(x \right)}}} - \frac{{\color{red}{\cot{\left(x \right)}}}^{3}}{3}$$

Daher,

$$\int{\csc^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\csc^{4}{\left(x \right)} d x} = - \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}+C$$

$$$\int \csc^{4}{\left(x \right)}\, dx = \left(- \frac{\cot^{3}{\left(x \right)}}{3} - \cot{\left(x \right)}\right) + C$$$A