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