AP Calculus AB and BC

Integral of 1/sqrt(4-x^2): Answer and Steps

Q: How do you know to use arcsine instead of arctangent?

The square root over a difference of squares, \sqrt{a^2 - x^2}, is the arcsine signature. A sum a^2 + x^2 with no root would give arctangent.

The integral of 1/sqrt(4-x^2) is arcsin(x/2) + C. This matches the standard arcsine form with a = 2, since 4 = 2^2. Differentiating arcsin(x/2) simplifies back to 1/sqrt(4-x^2), which confirms the answer.

\int \frac{1}{\sqrt{4-x^2}}\,dx = \arcsin\frac{x}{2} + C

Matching the arcsine form

A constant over the square root of a difference of squares integrates to an arcsine. The standard form has $\sqrt{a^2 - x^2}$ in the denominator.

\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + C

Here $4 = 2^2$ , so $a = 2$ . Substitute into the formula.

\int \frac{1}{\sqrt{4 - x^2}}\,dx = \int \frac{1}{\sqrt{2^2 - x^2}}\,dx = \arcsin\frac{x}{2} + C

No leading fraction

Unlike the arctangent form, the arcsine form has no $\frac{1}{a}$ out front. That factor is absorbed by the square root.

Verifying by differentiating

Differentiate the result. The chain rule contributes an inner factor of $\frac{1}{2}$ from the argument $\frac{x}{2}$ .

\frac{d}{dx}\left(\arcsin\frac{x}{2}\right) = \frac{1}{\sqrt{1 - \left(\frac{x}{2}\right)^2}}\cdot\frac{1}{2} = \frac{1}{2\sqrt{1 - \frac{x^2}{4}}}

Pulling the leading $2$ inside the root as a factor of $4$ clears the inner fraction and returns the integrand.

\frac{1}{2\sqrt{1 - \frac{x^2}{4}}} = \frac{1}{\sqrt{4\left(1 - \frac{x^2}{4}\right)}} = \frac{1}{\sqrt{4 - x^2}}

Common mistakes

Adding a $\frac{1}{a}$ factor. The arcsine form has none, so the answer is $\arcsin\frac{x}{2}$ , not $\frac{1}{2}\arcsin\frac{x}{2}$ .
Using $a = 4$ instead of $a = 2$ . The constant under the root is $a^2 = 4$ , so $a = 2$ .
Confusing this with the arctangent form. A square root over a difference gives arcsine; a plain sum $4 + x^2$ gives arctangent.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of 1/sqrt(4-x^2) from 0 to 1?

$\int_0^1 \frac{1}{\sqrt{4-x^2}}\,dx = \left[\arcsin\frac{x}{2}\right]_0^1 = \arcsin\frac{1}{2} = \frac{\pi}{6}$ .

How do you know to use arcsine instead of arctangent?

The square root over a difference of squares, $\sqrt{a^2 - x^2}$ , is the arcsine signature. A sum $a^2 + x^2$ with no root would give arctangent.

What is the general integral of 1/sqrt(a^2-x^2)?

$\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + C$ for $a > 0$ and $|x| < a$ . Here $a = 2$ .