Integral von $$$\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}$$$

Bestimme $$$\int \frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=8$$$ und $$$f{\left(x \right)} = \frac{e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}$$$ an:

$${\color{red}{\int{\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x}}} = {\color{red}{\left(8 \int{\frac{e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x}\right)}}$$

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

Dann $$$du=\left(\operatorname{acos}{\left(x \right)}\right)^{\prime }dx = - \frac{1}{\sqrt{1 - x^{2}}} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\frac{dx}{\sqrt{1 - x^{2}}} = - du$$$.

Das Integral lässt sich umschreiben als

$$8 {\color{red}{\int{\frac{e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x}}} = 8 {\color{red}{\int{\left(- e^{u}\right)d u}}}$$

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

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

Das Integral der Exponentialfunktion lautet $$$\int{e^{u} d u} = e^{u}$$$:

$$- 8 {\color{red}{\int{e^{u} d u}}} = - 8 {\color{red}{e^{u}}}$$

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

$$- 8 e^{{\color{red}{u}}} = - 8 e^{{\color{red}{\operatorname{acos}{\left(x \right)}}}}$$

Daher,

$$\int{\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x} = - 8 e^{\operatorname{acos}{\left(x \right)}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x} = - 8 e^{\operatorname{acos}{\left(x \right)}}+C$$

$$$\int \frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}\, dx = - 8 e^{\operatorname{acos}{\left(x \right)}} + C$$$A