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

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

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,

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

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

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

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

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

Integral becomes

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

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

Apply the power reducing formula $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ with $$$\alpha= u $$$:

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$:

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

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{4} = \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{4}$$

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

$$\frac{\int{\cos{\left(2 u \right)} d u}}{4} + \frac{{\color{red}{\int{1 d u}}}}{4} = \frac{\int{\cos{\left(2 u \right)} d u}}{4} + \frac{{\color{red}{u}}}{4}$$

Let $$$v=2 u$$$.

Then $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (steps can be seen »), and we have that $$$du = \frac{dv}{2}$$$.

So,

$$\frac{u}{4} + \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{4} = \frac{u}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{u}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{4} = \frac{u}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{4}$$

The integral of the cosine is $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{u}{4} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{8} = \frac{u}{4} + \frac{{\color{red}{\sin{\left(v \right)}}}}{8}$$

Recall that $$$v=2 u$$$:

$$\frac{u}{4} + \frac{\sin{\left({\color{red}{v}} \right)}}{8} = \frac{u}{4} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{8}$$

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

$$\frac{\sin{\left(2 {\color{red}{u}} \right)}}{8} + \frac{{\color{red}{u}}}{4} = \frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(2 x \right)}}} \right)}}{8} + \frac{{\color{red}{\operatorname{asin}{\left(2 x \right)}}}}{4}$$

Therefore,

$$\int{\sqrt{1 - 4 x^{2}} d x} = \frac{\sin{\left(2 \operatorname{asin}{\left(2 x \right)} \right)}}{8} + \frac{\operatorname{asin}{\left(2 x \right)}}{4}$$

Using the formulas $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, simplify the expression:

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

Add the constant of integration:

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

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