# Integral of arcsin x: Formula, Steps and Examples

> The integral of arcsin x is x arcsin x + √(1−x²) + C. Learn the integration by parts steps, the arccos version, a graph and worked examples.

Source: https://practicecalculus.com/integrals/arcsin-x
Updated: 2026-11-06

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

## At a glance

- **Antiderivative:** $x\arcsin x + \sqrt{1-x^2} + C$
- **Domain:** $-1 \le x \le 1$
- **Method:** Integration by parts with $u = \arcsin x$, $dv = dx$
- **Arccos version:** $\int \arccos x\,dx = x\arccos x - \sqrt{1-x^2} + C$
- **From 0 to 1:** $\frac{\pi}{2} - 1 \approx 0.571$

## Key takeaways

- The integral of $\arcsin x$ is $x\arcsin x + \sqrt{1-x^2} + C$.
- Use integration by parts with $u = \arcsin x$ and $dv = dx$, then a $u$-substitution for $\int \frac{x}{\sqrt{1-x^2}}\,dx$.
- The integral of $\arccos x$ has the same shape with a minus sign: $x\arccos x - \sqrt{1-x^2} + C$.
- The value $\int_0^1 \arcsin x\,dx = \frac{\pi}{2} - 1$ can also be found from the graph by subtracting areas.

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.

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

## Graph of arcsin x with the area from 0 to 1

```graph
title: 'y = arcsin x with the area from 0 to 1 shaded'
caption: 'The shaded area is $\frac{\pi}{2} - 1 \approx 0.571$. It is the $1 \times \frac{\pi}{2}$ rectangle minus the area between the curve and the $y$-axis, which is $\int_0^{\pi/2}\sin y\,dy = 1$.'
x: [-1.5, 1.5]
y: [-2, 2]
functions:
  - expr: arcsin(x)
    label: y = arcsin x
    domain: [-1, 1]
shade:
  expr: arcsin(x)
  from: 0
  to: 1
points:
  - [1, 1.5708, '(1, π/2)']
  - [-1, -1.5708, '(−1, −π/2)']
```

## How to integrate arcsin x step by step

Use [integration by parts](/rules/integration-by-parts-rule), $\int u\,dv = uv - \int v\,du$.

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

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

The two fractions cancel, leaving only $\arcsin x$. This uses the [derivative of arcsin x](/derivatives/arcsin-x), $\frac{1}{\sqrt{1-x^2}}$.

## Definite integrals

**From 0 to 1:**

$$\int_0^1\arcsin x\,dx = \left(1\cdot\frac{\pi}{2} + 0\right) - (0 + 1) = \frac{\pi}{2} - 1 \approx 0.571$$

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:**

$$\frac{1}{2}\cdot\frac{\pi}{6} + \sqrt{\frac{3}{4}} - 1 = \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1 \approx 0.128$$

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

## Worked examples

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

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

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 $u = \frac{x}{2}$, $dx = 2\,du$:

$$2\left(u\arcsin u + \sqrt{1-u^2}\right) + C = x\arcsin\frac{x}{2} + \sqrt{4 - x^2} + C$$

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](/integrals/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)](/integrals/1-over-sqrt-1-minus-x-squared). The [arcsine](/functions/arcsine) page has more on the function itself, and the [integral of arctan x](/integrals/arctan-x) uses the same method.

## Frequently asked questions

### What is the integral of arcsin x?

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

### What is the integral of sin⁻¹ x?

$\sin^{-1}x$ is another name for $\arcsin x$, so the integral is $x\sin^{-1}x + \sqrt{1-x^2} + C$. It is not $\int \csc x\,dx$.

### What is the integral of arccos x?

It is $x\arccos x - \sqrt{1-x^2} + C$. The steps are the same as for arcsin, but $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}$ flips the sign of the root term.

### What is the integral of arcsin(x/2)?

It is $x\arcsin\frac{x}{2} + \sqrt{4-x^2} + C$. Substitute $u = \frac{x}{2}$, $dx = 2\,du$, and use the basic formula.