Integral of $$$\frac{4}{\sqrt{16 - x^{2}}}$$$

Find $$$\int \frac{4}{\sqrt{16 - x^{2}}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{1}{\sqrt{16 - x^{2}}}$$$:

$${\color{red}{\int{\frac{4}{\sqrt{16 - x^{2}}} d x}}} = {\color{red}{\left(4 \int{\frac{1}{\sqrt{16 - x^{2}}} d x}\right)}}$$

Let $$$x=4 \sin{\left(u \right)}$$$.

Then $$$dx=\left(4 \sin{\left(u \right)}\right)^{\prime }du = 4 \cos{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asin}{\left(\frac{x}{4} \right)}$$$.

Integrand becomes

$$$\frac{1}{\sqrt{16 - x^{2}}} = \frac{1}{\sqrt{16 - 16 \sin^{2}{\left( u \right)}}}$$$

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

$$$\frac{1}{\sqrt{16 - 16 \sin^{2}{\left( u \right)}}}=\frac{1}{4 \sqrt{1 - \sin^{2}{\left( u \right)}}}=\frac{1}{4 \sqrt{\cos^{2}{\left( u \right)}}}$$$

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

$$$\frac{1}{4 \sqrt{\cos^{2}{\left( u \right)}}} = \frac{1}{4 \cos{\left( u \right)}}$$$

Thus,

$$4 {\color{red}{\int{\frac{1}{\sqrt{16 - x^{2}}} d x}}} = 4 {\color{red}{\int{1 d u}}}$$

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

$$4 {\color{red}{\int{1 d u}}} = 4 {\color{red}{u}}$$

Recall that $$$u=\operatorname{asin}{\left(\frac{x}{4} \right)}$$$:

$$4 {\color{red}{u}} = 4 {\color{red}{\operatorname{asin}{\left(\frac{x}{4} \right)}}}$$

Therefore,

$$\int{\frac{4}{\sqrt{16 - x^{2}}} d x} = 4 \operatorname{asin}{\left(\frac{x}{4} \right)}$$

Add the constant of integration:

$$\int{\frac{4}{\sqrt{16 - x^{2}}} d x} = 4 \operatorname{asin}{\left(\frac{x}{4} \right)}+C$$

$$$\int \frac{4}{\sqrt{16 - x^{2}}}\, dx = 4 \operatorname{asin}{\left(\frac{x}{4} \right)} + C$$$A