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Let $I=\int \sqrt{1-4 x^{2}} d x=\int \sqrt{(1)^{2}-(2 x)^{2}} d x$

Let $2 x=t \Rightarrow 2 d x=d t$

$\therefore I=\frac{1}{2} \int \sqrt{(1)^{2}-(t)^{2}} d t$

It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+\mathrm{C}$

$\begin{aligned} \Rightarrow I &=\frac{1}{2}\left[\frac{t}{2} \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1} t\right]+\mathrm{C} \\ &=\frac{t}{4} \sqrt{1-t^{2}}+\frac{1}{4} \sin ^{-1} t+\mathrm{C} \\ &=\frac{2 x}{4} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+\mathrm{C} \\ &=\frac{x}{2} \sqrt{1-4 x^{2}}+\frac{1}{4} \sin ^{-1} 2 x+\mathrm{C} \end{aligned}$