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

> The integral of 1/√(1−x²) is arcsin x + C. See the sine substitution, the 1/√(a²−x²) version, a graph, definite integrals and examples.

Source: https://practicecalculus.com/integrals/1-over-sqrt-1-minus-x-squared
Updated: 2026-11-06

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

## At a glance

- **Antiderivative:** $\arcsin x + C$
- **General form:** $\displaystyle\int \frac{dx}{\sqrt{a^2-x^2}} = \arcsin\frac{x}{a} + C$
- **Valid for:** $-1 < x < 1$ (or $\lvert x\rvert < a$)
- **Method:** Substitution $x = \sin\theta$
- **From 0 to 1:** $\dfrac{\pi}{2}$ (an improper integral that converges)

## Key takeaways

- The integral of $\frac{1}{\sqrt{1-x^2}}$ is $\arcsin x + C$, valid for $-1 < x < 1$.
- For $a > 0$, $\int \frac{dx}{\sqrt{a^2-x^2}} = \arcsin\frac{x}{a} + C$. Unlike the arctan formula, there is no $\frac{1}{a}$ in front.
- $-\arccos x + C$ is also a correct answer, because $\arcsin x + \arccos x = \frac{\pi}{2}$, a constant.
- If there is an $x$ on top, as in $\frac{x}{\sqrt{1-x^2}}$, use $u = 1 - x^2$ instead and get $-\sqrt{1-x^2} + C$.

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

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

For any constant $a > 0$:

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

## Graph of 1/√(1−x²)

```graph
title: 'y = 1/√(1−x²) with the area from 0 to 1/2 shaded'
caption: 'The shaded area is $\arcsin\frac{1}{2} - \arcsin 0 = \frac{\pi}{6} \approx 0.524$. The curve shoots up at $x = \pm 1$, but the total area from $-1$ to $1$ is still finite: $\pi$.'
x: [-1.5, 1.5]
y: [-0.5, 5]
functions:
  - expr: 1/sqrt(1-x^2)
    label: y = 1/√(1−x²)
    domain: [-0.98, 0.98]
shade:
  expr: 1/sqrt(1-x^2)
  from: 0
  to: 0.5
vertical:
  - x: -1
    label: x = −1
  - x: 1
    label: x = 1
points:
  - [0, 1, '(0, 1)']
```

## 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](/blog/trig-substitution).

## Check by differentiating

$$\frac{d}{dx}\arcsin\frac{x}{a} = \frac{1}{\sqrt{1 - \frac{x^2}{a^2}}}\cdot\frac{1}{a} = \frac{1}{\sqrt{a^2 - x^2}}$$

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

## Definite integrals

| Integral | Value |
| --- | --- |
| $\int_0^{1/2} \frac{dx}{\sqrt{1-x^2}}$ | $\frac{\pi}{6}$ |
| $\int_0^{\sqrt{2}/2} \frac{dx}{\sqrt{1-x^2}}$ | $\frac{\pi}{4}$ |
| $\int_0^{1} \frac{dx}{\sqrt{1-x^2}}$ | $\frac{\pi}{2}$ |
| $\int_{-1}^{1} \frac{dx}{\sqrt{1-x^2}}$ | $\pi$ |

The last two are [improper integrals](/blog/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 $u = 2x$, $du = 2\,dx$:

$$\frac{1}{2}\int \frac{du}{\sqrt{1-u^2}} = \frac{1}{2}\arcsin(2x) + C$$

**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 $u = x - 2$:

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

**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)](/integrals/1-over-1-plus-x-squared).
2. **Reaching for arcsin when there is an $x$ on top.** $\int \frac{x}{\sqrt{1-x^2}}\,dx$ is a plain [u-substitution](/rules/u-substitution-rule) 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)](/integrals/sqrt-1-minus-x-squared).
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](/integrals/arcsin-x). The [arcsine](/functions/arcsine) page covers its domain and graph.

## Frequently asked questions

### What is the integral of 1/√(1−x²)?

It is $\arcsin x + C$, also written $\sin^{-1} x + C$.

### What is the integral of 1/√(a²−x²)?

It is $\arcsin\left(\frac{x}{a}\right) + C$ for $a > 0$. For example, $\int \frac{dx}{\sqrt{4-x^2}} = \arcsin\frac{x}{2} + C$.

### Is the integral of 1/√(1−x²) arcsin x or −arccos x?

Both are correct. They differ by the constant $\frac{\pi}{2}$, and an antiderivative is only defined up to a constant. Most books write $\arcsin x$.

### What is the integral of x/√(1−x²)?

It is $-\sqrt{1-x^2} + C$. Let $u = 1 - x^2$, $du = -2x\,dx$, so the integral becomes $-\frac{1}{2}\int u^{-1/2}\,du = -\sqrt{u} + C$.