Integral of 1/sqrt(1-4x^2): Arcsine With a Half

Substitute for the inner coefficient

The template is $\int\frac{du}{\sqrt{1-u^{2}}} = \arcsin u + C$. Here $4x^{2} = \left(2x\right)^{2}$, so the inner function is $u = 2x$, and $du = 2\,dx$ means $dx = \frac{du}{2}$.

Differentiating confirms where the coefficient goes: $\frac{d}{dx}\left[\frac{1}{2}\arcsin 2x\right] = \frac{1}{2}\cdot\frac{2}{\sqrt{1-4x^{2}}} = \frac{1}{\sqrt{1-4x^{2}}}$. The $\frac{1}{2}$ exists to cancel the chain rule factor of $2$.

The same answer by factoring the radical

Some courses learn the form $\int\frac{du}{\sqrt{a^{2}-u^{2}}} = \arcsin\left(\frac{u}{a}\right) + C$ instead. To reach it, pull the $4$ out of the radical: $\sqrt{1-4x^{2}} = 2\sqrt{\frac{1}{4}-x^{2}}$, so $a = \frac{1}{2}$ and $u = x$.

Both routes need $1-4x^{2} > 0$, so the antiderivative lives on $-\frac{1}{2} < x < \frac{1}{2}$. A definite integral reaching $\pm\frac{1}{2}$ is improper, and one written across those points has no value at all.

The mistakes students make

The arcsine template is easy to recall and easy to misapply, because the inner coefficient is the only thing that changes.

• Answering $\arcsin\left(2x\right)$. Differentiating it returns $\frac{2}{\sqrt{1-4x^{2}}}$, so the answer is double what it should be.
• Multiplying by $2$ rather than dividing, giving $2\arcsin\left(2x\right)$. Solving $du = 2\,dx$ for $dx$ puts the $2$ underneath.
• Treating the square root as an arctangent setup and writing $\frac{1}{2}\arctan\left(2x\right)$. A constant numerator over $\sqrt{a^{2}-u^{2}}$ points to arcsine; a constant numerator over a bare quadratic points to arctangent. Check the numerator first, though: if it is a multiple of the derivative of what sits under the radical, the problem is a plain substitution and the answer is a square root, as in $\int\frac{x\,dx}{\sqrt{1-4x^{2}}} = -\frac{\sqrt{1-4x^{2}}}{4} + C$.