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

Let $I = \int \sqrt{1-4x^2} dx = \int \sqrt{(1)^2 - (2x)^2} dx$.
Substitute $2x = t$,then $2 dx = dt$,which implies $dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \frac{1}{2} \int \sqrt{1^2 - t^2} dt$.
Using the standard formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C$:
$I = \frac{1}{2} \left[ \frac{t}{2} \sqrt{1 - t^2} + \frac{1^2}{2} \sin^{-1} \left( \frac{t}{1} \right) \right] + C$.
Simplifying the expression:
$I = \frac{t}{4} \sqrt{1 - t^2} + \frac{1}{4} \sin^{-1} (t) + C$.
Substituting $t = 2x$ back into the expression:
$I = \frac{2x}{4} \sqrt{1 - (2x)^2} + \frac{1}{4} \sin^{-1} (2x) + C$.
$I = \frac{x}{2} \sqrt{1 - 4x^2} + \frac{1}{4} \sin^{-1} (2x) + C$,where $C$ is the constant of integration.