# Integral of sin² x: Formula, Derivation and Examples

> The integral of sin² x is x/2 - sin(2x)/4 + C. See the power-reducing identity step by step, a check by differentiating, sin²(ax) and definite integrals.

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

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

## At a glance

- **Antiderivative:** $\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$
- **Identity used:** $\sin^2 x = \dfrac{1 - \cos 2x}{2}$
- **Equivalent form:** $\dfrac{x - \sin x\cos x}{2} + C$
- **From 0 to π:** $\dfrac{\pi}{2}$
- **Average value:** $\dfrac{1}{2}$ over a full period

## Key takeaways

- The integral of $\sin^2 x$ is $\frac{x}{2} - \frac{\sin 2x}{4} + C$.
- Use the power-reducing identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ first. There is no direct substitution.
- Over any full period, $\sin^2 x$ averages to $\frac{1}{2}$, so $\int_0^{\pi}\sin^2 x\,dx = \frac{\pi}{2}$.
- It is not $\frac{\sin^3 x}{3}$. The power rule needs a $\cos x$ factor that is not there.

The **integral of $\sin^2 x$ is $\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$**. The key is a power-reducing identity that turns the square into something you can integrate term by term.

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

## Graph of sin² x and its antiderivative

```graph
title: 'Area under y = sin² x from 0 to π, with the antiderivative y = x/2 - sin(2x)/4'
caption: 'The shaded area is $\frac{\pi}{2} \approx 1.571$, exactly half of the $\pi$ by $1$ rectangle. The dashed antiderivative wobbles around the line $y = \frac{x}{2}$.'
x: [-0.5, 6.8]
y: [-0.5, 3.5]
functions:
  - expr: sin(x)^2
    label: y = sin² x
  - expr: x/2 - sin(2x)/4
    label: y = x/2 - sin(2x)/4 (antiderivative)
    dashed: true
shade: {expr: sin(x)^2, from: 0, to: 3.14159}
points:
  - [3.14159, 1.5708, '(π, π/2)']
```

The graph of $\sin^2 x$ oscillates between $0$ and $1$ with period $\pi$, centred on $y = \frac{1}{2}$. That is why the antiderivative grows on average like $\frac{x}{2}$.

## Derivation, step by step

1. Start from the double-angle identity $\cos 2x = 1 - 2\sin^2 x$.
2. Solve for $\sin^2 x$: $\sin^2 x = \dfrac{1 - \cos 2x}{2}$.
3. Substitute into the integral:
$$\int\sin^2 x\,dx = \frac{1}{2}\int(1 - \cos 2x)\,dx$$
4. Integrate each term. $\int 1\,dx = x$ and $\int\cos 2x\,dx = \frac{\sin 2x}{2}$, from the [integral of cos x](/integrals/cos-x) with $a = 2$.
5. Combine: $\frac{1}{2}\left(x - \frac{\sin 2x}{2}\right) + C = \frac{x}{2} - \frac{\sin 2x}{4} + C$.

Using $\sin 2x = 2\sin x\cos x$, you can also write the answer as $\frac{1}{2}(x - \sin x\cos x) + C$.

## Check by differentiating

$$\frac{d}{dx}\left(\frac{x}{2} - \frac{\sin 2x}{4}\right) = \frac{1}{2} - \frac{2\cos 2x}{4} = \frac{1 - \cos 2x}{2} = \sin^2 x \checkmark$$

The forward direction is on the [derivative of sin² x](/derivatives/sin-squared-x) page: $\frac{d}{dx}\sin^2 x = \sin 2x$.

## Definite integrals

| Integral | Value |
| --- | --- |
| $\int_0^{\pi/2}\sin^2 x\,dx$ | $\frac{\pi}{4}$ |
| $\int_0^{\pi}\sin^2 x\,dx$ | $\frac{\pi}{2}$ |
| $\int_0^{2\pi}\sin^2 x\,dx$ | $\pi$ |
| $\int_0^{\pi/4}\sin^2 x\,dx$ | $\frac{\pi}{8} - \frac{1}{4}$ |

For the last row: $\frac{\pi/4}{2} - \frac{\sin(\pi/2)}{4} = \frac{\pi}{8} - \frac{1}{4} \approx 0.143$.

A shortcut: over a full period, $\sin^2 x$ and $\cos^2 x$ have the same area, and together they integrate $1$. So each one gets half of the interval length.

## Substitution variants

For $\sin^2(ax)$, use $\sin^2(ax) = \frac{1 - \cos 2ax}{2}$:

$$\int\sin^2(ax)\,dx = \frac{x}{2} - \frac{\sin(2ax)}{4a} + C$$

So $\int\sin^2(3x)\,dx = \frac{x}{2} - \frac{\sin 6x}{12} + C$.

When a $\cos x$ factor is present, skip the identity and substitute $u = \sin x$:

$$\int\sin^2 x\cos x\,dx = \frac{\sin^3 x}{3} + C$$

## Worked examples

**Example 1: average value.** The average of $\sin^2 x$ on $[0, 2\pi]$ is $\frac{1}{2\pi}\cdot\pi = \frac{1}{2}$. This is why AC power uses the factor $\frac{1}{2}$ (and the RMS uses $\frac{1}{\sqrt{2}}$).

**Example 2: volume.** Rotating one arch of $y = \sin x$ about the $x$-axis gives a solid with volume $\pi\int_0^\pi\sin^2 x\,dx = \frac{\pi^2}{2}$.

**Example 3: sin⁴ x.** Square the identity: $\sin^4 x = \frac{(1 - \cos 2x)^2}{4} = \frac{1}{4} - \frac{\cos 2x}{2} + \frac{\cos^2 2x}{4}$, then reduce $\cos^2 2x$ again. The result is $\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$.

## Common mistakes

1. **Answering $\frac{\sin^3 x}{3}$.** Its derivative is $\sin^2 x\cos x$, not $\sin^2 x$.
2. **Answering $-\cos^2 x$.** Its derivative is $2\sin x\cos x = \sin 2x$.
3. **Using the wrong identity sign.** For $\sin^2$ it is $1 - \cos 2x$. For $\cos^2$ it is $1 + \cos 2x$.
4. **Forgetting to divide $\sin 2x$ by $2$** when integrating $\cos 2x$, which gives $\frac{\sin 2x}{2}$ instead of $\frac{\sin 2x}{4}$ at the end.

Next: the [integral of cos² x](/integrals/cos-squared-x) uses the same idea. Practise on [trig integral problems](/practice-problems/trig-integrals).

## Frequently asked questions

### What is the integral of sin² x?

It is $\frac{x}{2} - \frac{\sin 2x}{4} + C$, which can also be written $\frac{1}{2}(x - \sin x\cos x) + C$.

### What is the integral of sin² x from 0 to π?

It is $\frac{\pi}{2} - \frac{\sin 2\pi}{4} - 0 = \frac{\pi}{2}$.

### What is the integral of sin²(3x)?

It is $\frac{x}{2} - \frac{\sin 6x}{12} + C$. Use $\sin^2(3x) = \frac{1 - \cos 6x}{2}$.

### Why can't I use u = sin x?

Then $du = \cos x\,dx$, but there is no $\cos x$ in $\int\sin^2 x\,dx$ to absorb. That substitution only works for integrals like $\int\sin^2 x\cos x\,dx$.