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

> The integral of √(1−x²) is ½(x√(1−x²) + arcsin x) + C. Learn the trig substitution, the link to the area of a circle and worked examples.

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

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

## At a glance

- **Antiderivative:** $\frac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right) + C$
- **General form:** $\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C$
- **Method:** Trig substitution $x = \sin\theta$, then $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$
- **From −1 to 1:** $\dfrac{\pi}{2}$, the area of a half unit circle
- **From 0 to 1:** $\dfrac{\pi}{4}$, a quarter circle

## Key takeaways

- The integral of $\sqrt{1-x^2}$ is $\frac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right) + C$.
- The graph of $y = \sqrt{1-x^2}$ is the top half of the unit circle, so $\int_{-1}^{1}\sqrt{1-x^2}\,dx = \frac{\pi}{2}$.
- The method is trig substitution: $x = \sin\theta$ turns the root into $\cos\theta$ and the integral into $\int\cos^2\theta\,d\theta$.
- For a radius $a$, $\int_{-a}^{a}\sqrt{a^2-x^2}\,dx = \frac{\pi a^2}{2}$, half the area of a circle.

The **integral of $\sqrt{1-x^2}$ is $\dfrac{1}{2}\left(x\sqrt{1-x^2} + \arcsin x\right) + C$**. It is the classic first example of trig substitution, and it is how calculus proves the area of a circle.

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

For a circle of radius $a$:

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

## Graph: half of the unit circle

```graph
title: 'y = √(1−x²) is the upper half of the unit circle'
caption: 'The shaded area from $-1$ to $1$ is half of a circle of radius 1, so it equals $\frac{\pi}{2} \approx 1.571$.'
x: [-1.6, 1.6]
y: [-0.6, 1.4]
functions:
  - expr: sqrt(1-x^2)
    label: y = √(1−x²)
    domain: [-1, 1]
shade:
  expr: sqrt(1-x^2)
  from: -1
  to: 1
points:
  - [0, 1, '(0, 1)']
  - [-1, 0, '(−1, 0)']
  - [1, 0, '(1, 0)']
```

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

1. **Substitute.** Let $x = \sin\theta$ with $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$. Then $dx = \cos\theta\,d\theta$ and $\sqrt{1-\sin^2\theta} = \cos\theta$, which is not negative on this interval.
2. **Rewrite.**
$$\int \sqrt{1-x^2}\,dx = \int \cos\theta\cdot\cos\theta\,d\theta = \int \cos^2\theta\,d\theta$$
3. **Use the half-angle identity** $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$, as on the [integral of cos²x](/integrals/cos-squared-x) page:
$$\int \cos^2\theta\,d\theta = \frac{\theta}{2} + \frac{\sin 2\theta}{4} + C = \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2} + C$$
4. **Go back to $x$.** $\theta = \arcsin x$, $\sin\theta = x$ and $\cos\theta = \sqrt{1-x^2}$:
$$\frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt{1-x^2} + C$$

The general version uses $x = a\sin\theta$ in the same way. The [trig substitution guide](/blog/trig-substitution) covers the other two root shapes, $\sqrt{a^2+x^2}$ and $\sqrt{x^2-a^2}$.

## Check by differentiating

Differentiate each piece. By the [product rule](/rules/product-rule):

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

and $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$. Adding them:

$$\sqrt{1-x^2} + \frac{1 - x^2}{\sqrt{1-x^2}} = \sqrt{1-x^2} + \sqrt{1-x^2} = 2\sqrt{1-x^2}$$

Half of that is $\sqrt{1-x^2}$, as it should be.

## Definite integrals and the area of a circle

Over the full interval, the root term vanishes at both ends:

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

Doubling it to include the lower half gives the area of the unit circle, $\pi$. With radius $a$, the same steps give $\int_{-a}^{a}\sqrt{a^2-x^2}\,dx = \frac{\pi a^2}{2}$, so a full circle has area $\pi a^2$.

## Worked examples

**Example 1: part of a circle.** Find $\int_0^{1/2}\sqrt{1-x^2}\,dx$.

$$\frac{1}{2}\left(\frac{1}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{8} + \frac{\pi}{12} \approx 0.478$$

The picture: a $30^\circ$ wedge of area $\frac{\pi}{12}$ plus a triangle of area $\frac{\sqrt{3}}{8}$.

**Example 2: radius 3.** $\int_{-3}^{3}\sqrt{9-x^2}\,dx = \frac{9\pi}{2}$, half of a circle of radius 3. No antiderivative work is needed once you see the shape.

**Example 3: a general antiderivative.** With $a = 2$:

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

**Example 4: shift the circle.** $\int_0^2\sqrt{2x - x^2}\,dx$. Complete the square: $2x - x^2 = 1 - (x-1)^2$. This is the unit half circle centred at $x = 1$, so the answer is $\frac{\pi}{2}$.

## Common mistakes

1. **Trying $u = 1 - x^2$.** You would need an $x\,dx$ to match $du$, and there is none.
2. **Forgetting to convert back.** An answer in $\theta$ is not finished. Use a right triangle with opposite side $x$ and hypotenuse $1$ to read off $\cos\theta = \sqrt{1-x^2}$.
3. **Dropping the $\frac{a^2}{2}$.** In the general formula the arcsin term carries $\frac{a^2}{2}$, not $\frac{1}{2}$.
4. **Confusing it with $\frac{1}{\sqrt{1-x^2}}$.** That one integrates straight to $\arcsin x$; see the [integral of 1/√(1−x²)](/integrals/1-over-sqrt-1-minus-x-squared).

## Frequently asked questions

### What is the integral of √(1−x²)?

It is $\frac{1}{2}x\sqrt{1-x^2} + \frac{1}{2}\arcsin x + C$, valid for $-1 \le x \le 1$.

### What is the integral of √(a²−x²)?

It is $\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C$. For $a = 2$ this gives $\frac{x}{2}\sqrt{4-x^2} + 2\arcsin\frac{x}{2} + C$.

### Why is the integral of √(1−x²) from −1 to 1 equal to π/2?

Because $y = \sqrt{1-x^2}$ traces the upper half of the circle $x^2 + y^2 = 1$. The area under it is half of $\pi r^2$ with $r = 1$, which is $\frac{\pi}{2}$.

### Can I integrate √(1−x²) with u-substitution?

No. There is no $x$ outside the root to supply $du$. A plain $u = 1 - x^2$ fails, which is why trig substitution is needed.