AP Calculus AB and BC

Integral of 1/sqrt(1-x^2): Answer, Proof, Mistakes

Q: What is the integral of 1/sqrt(1-x^2)?

It is \arcsin x + C, that is, \int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C. The result holds on the open interval -1 < x < 1, where the integrand is defined. You can confirm it instantly by differentiating: \frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}.

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

For a > 0 it is \arcsin\left(\frac{x}{a}\right) + C. Factor a^2 out of the root to get a\sqrt{1 - (x/a)^2}, then substitute u = \frac{x}{a}; the a from dx = a\,du cancels the a from the root, so no constant survives out front. For example, \int \frac{1}{\sqrt{25-x^2}}\,dx = \arcsin\left(\frac{x}{5}\right) + C.

Q: Why is it arcsin and not arccos?

Both work, because \frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}} means -\arccos x is also an antiderivative. Since \arcsin x + \arccos x = \frac{\pi}{2}, the two answers differ only by a constant, which the + C absorbs. Write \arcsin x + C so your answer matches the AP scoring guidelines and the formula sheet convention.

The integral of 1/sqrt(1 - x^2) is arcsin x + C, valid on the open interval -1 < x < 1. It works because the derivative of arcsin x is exactly 1/sqrt(1 - x^2), so antidifferentiation just runs that rule backwards. The general version is the integral of 1/sqrt(a^2 - x^2) dx = arcsin(x/a) + C for a > 0.

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

Why the antiderivative is arcsin x

This integral is not solved by a technique. It is recognized. The whole justification is that some function already has $\frac{1}{\sqrt{1-x^2}}$ as its derivative, and that function is $\arcsin x$ .

\frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\sqrt{1-x^2}}

Antidifferentiation reverses that statement, and the constant of integration covers every other function with the same derivative.

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

If you want the derivative rule itself rather than taking it on faith, it comes from implicit differentiation. Write $y = \arcsin x$ , which says the same thing as $\sin y = x$ with $y$ restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

\sin y = x \quad \Longrightarrow \quad \cos y \cdot \frac{dy}{dx} = 1

Solve for $\frac{dy}{dx}$ and rewrite $\cos y$ using $\sin^2 y + \cos^2 y = 1$ . Cosine is never negative on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , so the positive root is the right one.

\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}

The domain is open, not closed

$\arcsin x$ is defined on $[-1, 1]$ , but the integrand $\frac{1}{\sqrt{1-x^2}}$ needs $1 - x^2 > 0$ , so the antiderivative statement is valid only on $-1 < x < 1$ . At $x = \pm 1$ the integrand blows up, which is why a definite integral reaching an endpoint is improper.

The general 1/sqrt(a^2 - x^2) form

AP problems almost never hand you a bare 1 under the root. The pattern to memorize is the version with a constant.

\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin\left(\frac{x}{a}\right) + C, \qquad a > 0

To see where it comes from, factor $a^2$ out of the root. Since $a > 0$ , $\sqrt{a^2} = a$ .

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

Now substitute $u = \frac{x}{a}$ , so $du = \frac{1}{a}\,dx$ and $dx = a\,du$ . The $a$ in the numerator cancels the $a$ from the root, leaving the basic form.

\int \frac{a\,du}{a\sqrt{1-u^2}} = \int \frac{1}{\sqrt{1-u^2}}\,du = \arcsin u + C = \arcsin\left(\frac{x}{a}\right) + C

Example 1. Evaluate $\int \frac{1}{\sqrt{9-x^2}}\,dx$ . Here $a^2 = 9$ , so $a = 3$ .

\int \frac{1}{\sqrt{9-x^2}}\,dx = \arcsin\left(\frac{x}{3}\right) + C

Example 2. Evaluate $\int \frac{1}{\sqrt{1-4x^2}}\,dx$ . The inside is $(2x)^2$ , so let $u = 2x$ and $du = 2\,dx$ . That introduces a compensating factor of $\frac{1}{2}$ .

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

Check every answer by differentiating

Differentiating $\arcsin\left(\frac{x}{3}\right)$ gives $\frac{1/3}{\sqrt{1 - x^2/9}} = \frac{1}{\sqrt{9-x^2}}$ , which is the integrand you started with. That five second check catches nearly every dropped or duplicated constant.

Where this integral shows up on the AP exam

This antiderivative belongs to Unit 6, Integration and Accumulation of Change, which carries 17 to 20 percent of the AB exam and 17 to 20 percent of the BC exam. Its home topic is 6.14, Selecting Techniques for Antidifferentiation, where the skill being tested is recognition rather than manipulation.

Topic 3.4, Differentiating Inverse Trigonometric Functions, supplies the derivative rule this integral reverses. Learn the two together and each one checks the other.
Topic 6.9, Integrating Using Substitution, handles the $\frac{1}{\sqrt{a^2-x^2}}$ and $\frac{1}{\sqrt{1-4x^2}}$ variants through $u = \frac{x}{a}$ or $u = 2x$ .
Topic 6.14 is where it actually gets tested: a multiple choice item shows an integrand and asks which antiderivative matches, with $\arctan$ and logarithm answers sitting next to it as distractors.
Topic 8.1 and the area topics can wrap it in a definite integral, most often $\int_0^{1/2} \frac{1}{\sqrt{1-x^2}}\,dx = \frac{\pi}{6}$ or a similar exact value.

A definite version evaluates through the Fundamental Theorem exactly as usual, provided the interval stays inside $(-1, 1)$ .

\int_{0}^{1/2} \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin\left(\frac{1}{2}\right) - \arcsin(0) = \frac{\pi}{6}

BC only: the endpoints make it improper

$\int_0^1 \frac{1}{\sqrt{1-x^2}}\,dx$ has an infinite integrand at $x = 1$ , so on the BC exam it must be written as a limit. It converges: $\lim_{b \to 1^-} \arcsin b = \frac{\pi}{2}$ . AB students will not be asked to evaluate it.

Common mistakes with this integral

Confusing it with the $\arctan$ integral. A square root in the denominator means $\arcsin$ ; no square root means $\arctan$ . Compare $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$ with $\int \frac{1}{1+x^2}\,dx = \arctan x + C$ .
Adding a $\frac{1}{a}$ that does not belong. The arcsin form has no outside constant: $\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin\left(\frac{x}{a}\right) + C$ . The $\frac{1}{a}$ belongs to the arctan form, $\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$ .
Reaching for a power rule. Rewriting the integrand as $(1-x^2)^{-1/2}$ and trying to raise the exponent fails, because the power rule for integrals requires the base to be the variable itself, not a composite like $1 - x^2$ .
Attempting $u = 1 - x^2$ . That substitution gives $du = -2x\,dx$ , and there is no $x$ in the numerator to absorb, so it goes nowhere. Substitute on the inside of the square instead, $u = \frac{x}{a}$ .
Forgetting the compensating constant when the inside is not just $\frac{x}{a}$ . For $\int \frac{1}{\sqrt{1-4x^2}}\,dx$ the answer is $\frac{1}{2}\arcsin(2x) + C$ ; dropping the $\frac{1}{2}$ makes the derivative twice too large.
Integrating straight through $x = \pm 1$ . The integrand is undefined at the endpoints, so $\int_{-2}^{2} \frac{1}{\sqrt{1-x^2}}\,dx$ is not a legal definite integral at all, and the answer is not $\arcsin 2 - \arcsin(-2)$ .
Losing the $+ C$ . On a free response question asking for the general antiderivative or solving a differential equation, the missing constant costs a point on its own.

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(1-x^2)?

It is $\arcsin x + C$ , that is, $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$ . The result holds on the open interval $-1 < x < 1$ , where the integrand is defined. You can confirm it instantly by differentiating: $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$ .

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

For $a > 0$ it is $\arcsin\left(\frac{x}{a}\right) + C$ . Factor $a^2$ out of the root to get $a\sqrt{1 - (x/a)^2}$ , then substitute $u = \frac{x}{a}$ ; the $a$ from $dx = a\,du$ cancels the $a$ from the root, so no constant survives out front. For example, $\int \frac{1}{\sqrt{25-x^2}}\,dx = \arcsin\left(\frac{x}{5}\right) + C$ .

Why is it arcsin and not arccos?

Both work, because $\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}$ means $-\arccos x$ is also an antiderivative. Since $\arcsin x + \arccos x = \frac{\pi}{2}$ , the two answers differ only by a constant, which the $+ C$ absorbs. Write $\arcsin x + C$ so your answer matches the AP scoring guidelines and the formula sheet convention.

How is this different from the integral giving arctan?

Look for the square root. A square root in the denominator, as in $\frac{1}{\sqrt{1-x^2}}$ , points to $\arcsin$ ; no square root, as in $\frac{1}{1+x^2}$ , points to $\arctan$ . The signs also differ: arcsin comes from $1 - x^2$ (a difference), arctan from $1 + x^2$ (a sum).

Can I evaluate this integral from 0 to 1?

Yes, but only as an improper integral, because the integrand is undefined at $x = 1$ . Write it as $\lim_{b \to 1^-} \int_0^b \frac{1}{\sqrt{1-x^2}}\,dx = \lim_{b \to 1^-} \arcsin b = \frac{\pi}{2}$ . Improper integrals are BC only; on AB the interval will stay strictly inside $(-1, 1)$ .