Integral of $$$\frac{1}{\sqrt{1 - 4 x^{2}}}$$$

Find $$$\int \frac{1}{\sqrt{1 - 4 x^{2}}}\, dx$$$.

Let $$$x=\frac{\sin{\left(u \right)}}{2}$$$.

Then $$$dx=\left(\frac{\sin{\left(u \right)}}{2}\right)^{\prime }du = \frac{\cos{\left(u \right)}}{2} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asin}{\left(2 x \right)}$$$.

Therefore,

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

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

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

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

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

Thus,

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=\frac{1}{2}$$$:

$${\color{red}{\int{\frac{1}{2} d u}}} = {\color{red}{\left(\frac{u}{2}\right)}}$$

Recall that $$$u=\operatorname{asin}{\left(2 x \right)}$$$:

$$\frac{{\color{red}{u}}}{2} = \frac{{\color{red}{\operatorname{asin}{\left(2 x \right)}}}}{2}$$

Therefore,

$$\int{\frac{1}{\sqrt{1 - 4 x^{2}}} d x} = \frac{\operatorname{asin}{\left(2 x \right)}}{2}$$

Add the constant of integration:

$$\int{\frac{1}{\sqrt{1 - 4 x^{2}}} d x} = \frac{\operatorname{asin}{\left(2 x \right)}}{2}+C$$

$$$\int \frac{1}{\sqrt{1 - 4 x^{2}}}\, dx = \frac{\operatorname{asin}{\left(2 x \right)}}{2} + C$$$A