# Arcsine arcsin(x): Graph, Domain, Range, Derivative and Integral

> The arcsine function arcsin(x), the inverse of sine: domain [−1, 1], range [−π/2, π/2], graph, derivative 1/√(1−x²), integral and examples.

Source: https://practicecalculus.com/functions/arcsine
Updated: 2026-11-03

$$ f(x) = \arcsin x = \sin^{-1} x $$

## At a glance

- **Domain:** $[-1, 1]$
- **Range:** $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
- **Derivative:** $\frac{1}{\sqrt{1 - x^2}}$
- **Integral:** $x\arcsin x + \sqrt{1 - x^2} + C$
- **Meaning:** The angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $x$
- **Symmetry:** Odd: $\arcsin(-x) = -\arcsin x$

## Key takeaways

- $\arcsin x$ is the angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $x$, defined only for $-1 \le x \le 1$.
- Its derivative is $\frac{1}{\sqrt{1-x^2}}$, so $\int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C$.
- $\sin^{-1}x$ means arcsine, not $\frac{1}{\sin x}$.

The **arcsine function** undoes the sine function: $\arcsin x$ is the angle whose sine is $x$. Because [sine](/functions/sine-function) repeats, it is not one-to-one, so arcsine picks the single answer in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. In calculus it matters mostly as the antiderivative of $\frac{1}{\sqrt{1-x^2}}$.

## Definition of arcsin x

$$y = \arcsin x \quad\Longleftrightarrow\quad \sin y = x \text{ and } -\frac{\pi}{2} \le y \le \frac{\pi}{2}$$

So $\arcsin\frac{1}{2} = \frac{\pi}{6}$, because $\sin\frac{\pi}{6} = \frac{1}{2}$ and $\frac{\pi}{6}$ is in the allowed range. The notation $\sin^{-1}x$ means the same thing.

## Graph of y = arcsin x

```graph
title: 'Graph of y = arcsin x on [−1, 1]'
caption: 'Arcsine rises from $-\frac{\pi}{2}$ at $x = -1$ to $\frac{\pi}{2}$ at $x = 1$. The tangent lines at both endpoints are vertical.'
x: [-1.8, 1.8]
y: [-1.9, 1.9]
functions:
  - expr: arcsin(x)
    label: y = arcsin x
    domain: [-1, 1]
points:
  - [-1, -1.5708, '(−1, −π/2)']
  - [0, 0, '(0, 0)']
  - [1, 1.5708, '(1, π/2)']
```

The graph is the piece of $y = \sin x$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, reflected across the line $y = x$.

## Key values of arcsine

| $x$ | $-1$ | $-\frac{\sqrt{3}}{2}$ | $-\frac{1}{2}$ | $0$ | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | $1$ |
| --- | --- | --- | --- | --- | --- | --- | --- | --- |
| $\arcsin x$ | $-\frac{\pi}{2}$ | $-\frac{\pi}{3}$ | $-\frac{\pi}{6}$ | $0$ | $\frac{\pi}{6}$ | $\frac{\pi}{4}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ |

## Properties of arcsine

- **Domain $[-1, 1]$, range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.** Inputs outside $[-1, 1]$ are undefined, since no angle has sine $2$.
- **Increasing.** Arcsine is strictly increasing on its whole domain.
- **Odd.** $\arcsin(-x) = -\arcsin x$.
- **Cancellation.** $\sin(\arcsin x) = x$ for every $x$ in $[-1, 1]$, but $\arcsin(\sin x) = x$ only for $x$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
- **Relation to arccosine.** $\arcsin x + \arccos x = \frac{\pi}{2}$. See [arccosine](/functions/arccosine).

## Derivative of arcsin x

$$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1$$

**Why.** Let $y = \arcsin x$, so $\sin y = x$. [Implicit differentiation](/rules/implicit-differentiation) gives $\cos y \cdot y' = 1$, so $y' = \frac{1}{\cos y}$. Since $y$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\cos y \ge 0$, and $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$. This is a special case of the [inverse function rule](/rules/inverse-function-rule).

At $x = \pm 1$ the denominator is zero, which is why the graph has vertical tangents at its endpoints.

## Integrals involving arcsine

The derivative formula, read backward, is the integral you will use most:

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

The integral of arcsine itself comes from [integration by parts](/blog/integration-by-parts) with $u = \arcsin x$, $dv = dx$:

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

## Worked examples

**Example 1: chain rule.** Differentiate $f(x) = \arcsin(3x)$:

$$f'(x) = \frac{3}{\sqrt{1 - 9x^2}}$$

**Example 2: a definite integral.** $\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}} = \arcsin\frac{1}{2} - \arcsin 0 = \frac{\pi}{6}$.

**Example 3: using $a$.** $\int\frac{dx}{\sqrt{9 - x^2}} = \arcsin\frac{x}{3} + C$.

**Example 4: composition.** Simplify $\cos(\arcsin x)$. If $\theta = \arcsin x$ then $\sin\theta = x$ and $\cos\theta \ge 0$, so $\cos(\arcsin x) = \sqrt{1 - x^2}$.

## Common mistakes

1. **Reading $\sin^{-1}x$ as $\frac{1}{\sin x}$.** The reciprocal is $\csc x$, covered on [reciprocal trig functions](/functions/reciprocal-trig-functions).
2. **Giving an angle outside the range.** $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$, not $\frac{7\pi}{6}$ or $\frac{11\pi}{6}$.
3. **Plugging in $|x| > 1$.** $\arcsin 2$ does not exist as a real number.
4. **Forgetting the chain rule factor.** $\frac{d}{dx}\arcsin(x^2) = \frac{2x}{\sqrt{1 - x^4}}$, not $\frac{1}{\sqrt{1-x^4}}$.

## How arcsine connects to other topics

Arcsine is the backbone of the substitution $x = a\sin\theta$ in [trig substitution](/blog/trig-substitution), used for integrals with $\sqrt{a^2 - x^2}$. Its siblings are [arccosine](/functions/arccosine) and [arctangent](/functions/arctangent), and all three derivatives are derived the same way in the guide to [implicit differentiation](/blog/implicit-differentiation).

## Frequently asked questions

### What is the derivative of arcsin x?

The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$ for $-1 < x < 1$. With the chain rule, $\frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1-u^2}}$.

### What is the domain and range of arcsin x?

The domain is $[-1, 1]$, because sine only produces values in that interval. The range is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the interval on which sine is one-to-one.

### Is sin inverse the same as 1/sin x?

No. $\sin^{-1}x$ is the inverse function arcsine. The reciprocal $\frac{1}{\sin x}$ is the cosecant, $\csc x$.

### What is the integral of arcsin x?

Using integration by parts with $u = \arcsin x$ and $dv = dx$ gives $\int\arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C$.

### Why is arcsin(sin x) not always equal to x?

Arcsine always returns an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. For example $\arcsin(\sin\pi) = \arcsin 0 = 0$, not $\pi$.