Integral of arcsin x: Formula, Steps and Examples

The integral of \arcsin x is x\arcsin x + \sqrt{1-x^2} + C. Like \ln x and \arctan x, it has no direct reverse rule, so you integrate it by parts using a factor of 1.

Graph of arcsin x with the area from 0 to 1

How to integrate arcsin x step by step

Use integration by parts, \int u\,dv = uv - \int v\,du.

1. Choose the parts. u = \arcsin x and dv = dx, so du = \frac{1}{\sqrt{1-x^2}}\,dx and v = x.
2. Apply the formula. \int \arcsin x\,dx = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,dx
3. Solve the leftover integral with w = 1 - x^2, dw = -2x\,dx: \int\frac{x}{\sqrt{1-x^2}}\,dx = -\frac{1}{2}\int w^{-1/2}\,dw = -\sqrt{w} + C = -\sqrt{1-x^2} + C
4. Combine. Subtracting a negative gives a plus: \int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C

Check by differentiating

The two fractions cancel, leaving only \arcsin x. This uses the derivative of arcsin x, \frac{1}{\sqrt{1-x^2}}.

Definite integrals

From 0 to 1:

The graph gives the same answer without any antiderivative. The region under y = \arcsin x plus the region to its left (between the curve and the y-axis) fill a rectangle of area \frac{\pi}{2}. The left region is the area under x = \sin y for 0 \le y \le \frac{\pi}{2}, which is 1.

From 0 to 1/2:

From −1 to 1: \arcsin x is odd, so the integral is 0.

Worked examples

Example 1: arccos x. Same parts setup, but du = -\frac{1}{\sqrt{1-x^2}}\,dx:

As a check, adding the arcsin and arccos results gives x\left(\arcsin x + \arccos x\right) = \frac{\pi}{2}x, which is the integral of the constant \frac{\pi}{2}.

Example 2: a scaled input. \int\arcsin\frac{x}{2}\,dx. Let u = \frac{x}{2}, dx = 2\,du:

Here 2\sqrt{1 - \frac{x^2}{4}} = \sqrt{4 - x^2}.

Example 3: a substitution first. \int \frac{\arcsin x}{\sqrt{1-x^2}}\,dx is not a parts problem. With u = \arcsin x, du = \frac{dx}{\sqrt{1-x^2}}, so the answer is \frac{1}{2}(\arcsin x)^2 + C.

Common mistakes

1. A sign error in the leftover integral. \int\frac{x}{\sqrt{1-x^2}}\,dx is -\sqrt{1-x^2}, so the final answer has +\sqrt{1-x^2}. Differentiating catches this at once.
2. Reading \sin^{-1}x as \frac{1}{\sin x}. The reciprocal is \csc x, a different function; see the integral of csc x.
3. Using parts when a substitution is there. If \frac{1}{\sqrt{1-x^2}} already sits next to \arcsin x, substitute instead.
4. Ignoring the domain. \arcsin x only exists for -1 \le x \le 1, so a definite integral must stay inside that interval.

To get arcsin as an answer, see the integral of 1/√(1−x²). The arcsine page has more on the function itself, and the integral of arctan x uses the same method.