Integral of $$$\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}$$$
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Find $$$\int \frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=8$$$ and $$$f{\left(x \right)} = \frac{e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}$$$:
$${\color{red}{\int{\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x}}} = {\color{red}{\left(8 \int{\frac{e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x}\right)}}$$
Let $$$u=\operatorname{acos}{\left(x \right)}$$$.
Then $$$du=\left(\operatorname{acos}{\left(x \right)}\right)^{\prime }dx = - \frac{1}{\sqrt{1 - x^{2}}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{\sqrt{1 - x^{2}}} = - du$$$.
Thus,
$$8 {\color{red}{\int{\frac{e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x}}} = 8 {\color{red}{\int{\left(- e^{u}\right)d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = e^{u}$$$:
$$8 {\color{red}{\int{\left(- e^{u}\right)d u}}} = 8 {\color{red}{\left(- \int{e^{u} d u}\right)}}$$
The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:
$$- 8 {\color{red}{\int{e^{u} d u}}} = - 8 {\color{red}{e^{u}}}$$
Recall that $$$u=\operatorname{acos}{\left(x \right)}$$$:
$$- 8 e^{{\color{red}{u}}} = - 8 e^{{\color{red}{\operatorname{acos}{\left(x \right)}}}}$$
Therefore,
$$\int{\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x} = - 8 e^{\operatorname{acos}{\left(x \right)}}$$
Add the constant of integration:
$$\int{\frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}} d x} = - 8 e^{\operatorname{acos}{\left(x \right)}}+C$$
Answer
$$$\int \frac{8 e^{\operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}\, dx = - 8 e^{\operatorname{acos}{\left(x \right)}} + C$$$A