Integral von $$$\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)}$$$
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Ihre Eingabe
Bestimme $$$\int \sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)}\, dx$$$.
Lösung
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,
$${\color{red}{\int{\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \sqrt{u}\right)d u}}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = \sqrt{u}$$$ an:
$${\color{red}{\int{\left(- \sqrt{u}\right)d u}}} = {\color{red}{\left(- \int{\sqrt{u} d u}\right)}}$$
Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=\frac{1}{2}$$$ an:
$$- {\color{red}{\int{\sqrt{u} d u}}}=- {\color{red}{\int{u^{\frac{1}{2}} d u}}}=- {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$
Zur Erinnerung: $$$u=\cot{\left(x \right)}$$$:
$$- \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} = - \frac{2 {\color{red}{\cot{\left(x \right)}}}^{\frac{3}{2}}}{3}$$
Daher,
$$\int{\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)} d x} = - \frac{2 \cot^{\frac{3}{2}}{\left(x \right)}}{3}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)} d x} = - \frac{2 \cot^{\frac{3}{2}}{\left(x \right)}}{3}+C$$
Antwort
$$$\int \sqrt{\cot{\left(x \right)}} \csc^{2}{\left(x \right)}\, dx = - \frac{2 \cot^{\frac{3}{2}}{\left(x \right)}}{3} + C$$$A