Integral of $$$\frac{1}{\sqrt{9 - x^{2}}}$$$

Find $$$\int \frac{1}{\sqrt{9 - x^{2}}}\, dx$$$.

Let $$$x=3 \sin{\left(u \right)}$$$.

Then $$$dx=\left(3 \sin{\left(u \right)}\right)^{\prime }du = 3 \cos{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asin}{\left(\frac{x}{3} \right)}$$$.

Thus,

$$$\frac{1}{\sqrt{9 - x^{2}}} = \frac{1}{\sqrt{9 - 9 \sin^{2}{\left( u \right)}}}$$$

Use the identity $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{9 - 9 \sin^{2}{\left( u \right)}}}=\frac{1}{3 \sqrt{1 - \sin^{2}{\left( u \right)}}}=\frac{1}{3 \sqrt{\cos^{2}{\left( u \right)}}}$$$

Assuming that $$$\cos{\left( u \right)} \ge 0$$$, we obtain the following:

$$$\frac{1}{3 \sqrt{\cos^{2}{\left( u \right)}}} = \frac{1}{3 \cos{\left( u \right)}}$$$

Integral can be rewritten as

$${\color{red}{\int{\frac{1}{\sqrt{9 - x^{2}}} d x}}} = {\color{red}{\int{1 d u}}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$

Recall that $$$u=\operatorname{asin}{\left(\frac{x}{3} \right)}$$$:

$${\color{red}{u}} = {\color{red}{\operatorname{asin}{\left(\frac{x}{3} \right)}}}$$

Therefore,

$$\int{\frac{1}{\sqrt{9 - x^{2}}} d x} = \operatorname{asin}{\left(\frac{x}{3} \right)}$$

Add the constant of integration:

$$\int{\frac{1}{\sqrt{9 - x^{2}}} d x} = \operatorname{asin}{\left(\frac{x}{3} \right)}+C$$

$$$\int \frac{1}{\sqrt{9 - x^{2}}}\, dx = \operatorname{asin}{\left(\frac{x}{3} \right)} + C$$$A