Integral of $$$\frac{1}{\sqrt{10 - x^{2}}}$$$
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Find $$$\int \frac{1}{\sqrt{10 - x^{2}}}\, dx$$$.
Solution
Let $$$x=\sqrt{10} \sin{\left(u \right)}$$$.
Then $$$dx=\left(\sqrt{10} \sin{\left(u \right)}\right)^{\prime }du = \sqrt{10} \cos{\left(u \right)} du$$$ (steps can be seen »).
Also, it follows that $$$u=\operatorname{asin}{\left(\frac{\sqrt{10} x}{10} \right)}$$$.
So,
$$$\frac{1}{\sqrt{10 - x^{2}}} = \frac{1}{\sqrt{10 - 10 \sin^{2}{\left( u \right)}}}$$$
Use the identity $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:
$$$\frac{1}{\sqrt{10 - 10 \sin^{2}{\left( u \right)}}}=\frac{\sqrt{10}}{10 \sqrt{1 - \sin^{2}{\left( u \right)}}}=\frac{\sqrt{10}}{10 \sqrt{\cos^{2}{\left( u \right)}}}$$$
Assuming that $$$\cos{\left( u \right)} \ge 0$$$, we obtain the following:
$$$\frac{\sqrt{10}}{10 \sqrt{\cos^{2}{\left( u \right)}}} = \frac{\sqrt{10}}{10 \cos{\left( u \right)}}$$$
Integral can be rewritten as
$${\color{red}{\int{\frac{1}{\sqrt{10 - x^{2}}} d x}}} = {\color{red}{\int{1 d u}}}$$
Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:
$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$
Recall that $$$u=\operatorname{asin}{\left(\frac{\sqrt{10} x}{10} \right)}$$$:
$${\color{red}{u}} = {\color{red}{\operatorname{asin}{\left(\frac{\sqrt{10} x}{10} \right)}}}$$
Therefore,
$$\int{\frac{1}{\sqrt{10 - x^{2}}} d x} = \operatorname{asin}{\left(\frac{\sqrt{10} x}{10} \right)}$$
Add the constant of integration:
$$\int{\frac{1}{\sqrt{10 - x^{2}}} d x} = \operatorname{asin}{\left(\frac{\sqrt{10} x}{10} \right)}+C$$
Answer
$$$\int \frac{1}{\sqrt{10 - x^{2}}}\, dx = \operatorname{asin}{\left(\frac{\sqrt{10} x}{10} \right)} + C$$$A