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

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=8$$$ y $$$f{\left(x \right)} = \frac{e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}$$$:

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

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

Entonces $$$du=\left(\operatorname{acos}{\left(x \right)}\right)^{\prime }dx = - \frac{1}{\sqrt{1 - x^{2}}} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{dx}{\sqrt{1 - x^{2}}} = - du$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = e^{u}$$$:

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

La integral de la función exponencial es $$$\int{e^{u} d u} = e^{u}$$$:

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

Recordemos que $$$u=\operatorname{acos}{\left(x \right)}$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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