Integral of 1/√(1−x²): Formula, Steps and Examples

The integral of \dfrac{1}{\sqrt{1-x^2}} is \arcsin x + C. It is the reverse of the derivative of arcsin x, and it works on the interval -1 < x < 1 where the square root is real and nonzero.

For any constant a > 0:

Graph of 1/√(1−x²)

How to integrate 1/√(1−x²) step by step

The square root of 1 - x^2 looks like the Pythagorean identity 1 - \sin^2\theta = \cos^2\theta, so substitute x = \sin\theta.

1. Let x = \sin\theta with -\frac{\pi}{2} < \theta < \frac{\pi}{2}. Then dx = \cos\theta\,d\theta.
2. Simplify the root. On this interval \cos\theta > 0, so \sqrt{1 - \sin^2\theta} = \cos\theta.
3. Integrate: \int \frac{\cos\theta\,d\theta}{\cos\theta} = \int 1\,d\theta = \theta + C
4. Go back to x: \theta = \arcsin x.

For \frac{1}{\sqrt{a^2-x^2}}, use x = a\sin\theta. Then dx = a\cos\theta\,d\theta and \sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta. The two factors of a cancel, which is why the answer has no \frac{1}{a} in front. This is the first case of trig substitution.

Check by differentiating

The step uses a\sqrt{1 - \frac{x^2}{a^2}} = \sqrt{a^2 - x^2}, which holds because a > 0.

Definite integrals

The last two are improper integrals, since the integrand is unbounded at x = \pm 1. They still converge because the blow-up is only like \frac{1}{\sqrt{1-x}}.

Worked examples

Example 1. \displaystyle\int \frac{dx}{\sqrt{9 - x^2}} = \arcsin\frac{x}{3} + C, using a = 3.

Example 2: a coefficient on x². \displaystyle\int \frac{dx}{\sqrt{1 - 4x^2}}. Let u = 2x, du = 2\,dx:

Example 3: complete the square. \displaystyle\int \frac{dx}{\sqrt{4x - x^2}}. Write 4x - x^2 = 4 - (x - 2)^2, then use a = 2 and u = x - 2:

Example 4: a definite integral. \displaystyle\int_0^{3} \frac{dx}{\sqrt{36 - x^2}} = \arcsin\frac{3}{6} - \arcsin 0 = \frac{\pi}{6}.

Common mistakes

1. Adding a \frac{1}{a}. That belongs to the arctan formula \int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\frac{x}{a}. The arcsin formula has none. Compare with the integral of 1/(1+x²).
2. Reaching for arcsin when there is an x on top. \int \frac{x}{\sqrt{1-x^2}}\,dx is a plain u-substitution with answer -\sqrt{1-x^2} + C.
3. Confusing \frac{1}{\sqrt{1-x^2}} with \sqrt{1-x^2}. The square root on its own integrates to something longer; see the integral of √(1−x²).
4. Using the formula outside \lvert x\rvert < a. For \lvert x\rvert > a the root is not real. Integrals like \int \frac{dx}{\sqrt{x^2-1}} give \ln\lvert x + \sqrt{x^2-1}\rvert instead.

To integrate the inverse sine function itself, see the integral of arcsin x. The arcsine page covers its domain and graph.