Integral von $$$2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}$$$

Bestimme $$$\int 2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=2$$$ und $$$f{\left(x \right)} = \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}$$$ an:

$${\color{red}{\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x}}} = {\color{red}{\left(2 \int{\cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x}\right)}}$$

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 $$$\alpha=x$$$:

$$2 {\color{red}{\int{\cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x}}} = 2 {\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \cot^{6}{\left(x \right)} \csc^{2}{\left(x \right)} d x}}}$$

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 $$$\csc^{2}{\left(x \right)} dx = - du$$$.

Somit,

$$2 {\color{red}{\int{\left(\cot^{2}{\left(x \right)} + 1\right) \cot^{6}{\left(x \right)} \csc^{2}{\left(x \right)} d x}}} = 2 {\color{red}{\int{\left(- u^{6} \left(u^{2} + 1\right)\right)d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = u^{6} \left(u^{2} + 1\right)$$$ an:

$$2 {\color{red}{\int{\left(- u^{6} \left(u^{2} + 1\right)\right)d u}}} = 2 {\color{red}{\left(- \int{u^{6} \left(u^{2} + 1\right) d u}\right)}}$$

Expand the expression:

$$- 2 {\color{red}{\int{u^{6} \left(u^{2} + 1\right) d u}}} = - 2 {\color{red}{\int{\left(u^{8} + u^{6}\right)d u}}}$$

Gliedweise integrieren:

$$- 2 {\color{red}{\int{\left(u^{8} + u^{6}\right)d u}}} = - 2 {\color{red}{\left(\int{u^{6} d u} + \int{u^{8} d u}\right)}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=6$$$ an:

$$- 2 \int{u^{8} d u} - 2 {\color{red}{\int{u^{6} d u}}}=- 2 \int{u^{8} d u} - 2 {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=- 2 \int{u^{8} d u} - 2 {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=8$$$ an:

$$- \frac{2 u^{7}}{7} - 2 {\color{red}{\int{u^{8} d u}}}=- \frac{2 u^{7}}{7} - 2 {\color{red}{\frac{u^{1 + 8}}{1 + 8}}}=- \frac{2 u^{7}}{7} - 2 {\color{red}{\left(\frac{u^{9}}{9}\right)}}$$

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

$$- \frac{2 {\color{red}{u}}^{7}}{7} - \frac{2 {\color{red}{u}}^{9}}{9} = - \frac{2 {\color{red}{\cot{\left(x \right)}}}^{7}}{7} - \frac{2 {\color{red}{\cot{\left(x \right)}}}^{9}}{9}$$

Daher,

$$\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x} = - \frac{2 \cot^{9}{\left(x \right)}}{9} - \frac{2 \cot^{7}{\left(x \right)}}{7}$$

Vereinfachen:

$$\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x} = \frac{2 \left(- 7 \cot^{2}{\left(x \right)} - 9\right) \cot^{7}{\left(x \right)}}{63}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)} d x} = \frac{2 \left(- 7 \cot^{2}{\left(x \right)} - 9\right) \cot^{7}{\left(x \right)}}{63}+C$$

$$$\int 2 \cot^{6}{\left(x \right)} \csc^{4}{\left(x \right)}\, dx = \frac{2 \left(- 7 \cot^{2}{\left(x \right)} - 9\right) \cot^{7}{\left(x \right)}}{63} + C$$$A