Question

Integrate the function:
$\sqrt{1-4x^2}$

Answer 1

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

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

$=2\int \sqrt{{\left(\frac{1}{2}\right)}^2-x^2}.dx$

Using $\int \sqrt{a^2-x^2}.dx$

$=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$

$=2[\frac{x}{2}\sqrt{{\left(\frac{1}{2}\right)}^2-x^2}+{\left(\frac{1}{2}\right)}^2.\frac{1}{2}{\sin }^{-1}\frac{x}{\frac{1}{2}}+C_1]$

Where $C_1$ is constant of integration

$=2[\frac{x}{2}\sqrt{\left(\frac{1}{4}\right)-x^2}+\frac{1}{4}.\frac{1}{2}{\sin }^{-1}2x+C_1]$

$=x\sqrt{\frac{1}{4}-x^2}+\frac{1}{4}{\sin }^{-1}2x+2C_1$

$=x\sqrt{\frac{1-4x^2}{4}}+\frac{1}{4}{\sin }^{-1}2x+C$

$=\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}{\sin }^{-1}2x+C$

Where $C=2C_1$