Question

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

Answer 1

Let $I=\int \sqrt{1-4x^2}dx=\int \sqrt{(1{)}^2-(2x{)}^2}dx$
Let $2x=t\Rightarrow 2dx=dt$
$\therefore I=\frac{1}{2}\int \sqrt{(1{)}^2-(t{)}^2}dt$
We know that, $\int \sqrt{(a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$
$\Rightarrow I=\frac{1}{2}[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t]+C$
$=\frac{t}{4}\sqrt{1-t^2}+\frac{1}{4}{\sin }^{-1}t+C$
$=\frac{2x}{4}\sqrt{1-4x^2}+\frac{1}{4}{\sin }^{-1}2x+C$
$=\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}{\sin }^{-1}2x+C$