Integral of sin²(x): x/2 − sin(2x)/4 + C

The integral of sin²(x) is x/2 − sin(2x)/4 + C. You can also write it as (x − sin(x)cos(x))/2 + C. You solve it by applying the half-angle power-reduction identity. This identity is sin²(x) = (1 − cos(2x))/2. Then you integrate term by term.

The integral of sin²(x) appears frequently in single-variable calculus, physics wave calculations, electrical signal processing, and trigonometric substitution problems. It is a standard question on calculus exams because trying to use simple power rules or direct u-substitution fails. Lowering the power using a trigonometric identity transforms an impossible integrand into a straightforward linear integration problem.

Why does sin²(x) integrate to x/2 − sin(2x)/4 + C?

To understand the result, compare the graphs of y = sin(x) and y = sin²(x). You can intuitively see the difference. Sin(x) oscillates above and below the x-axis. It ranges between −1 and 1. Also, its period is 2π. However, squaring the function makes all negative values positive. The curve y = sin²(x) bounces smoothly between 0 and 1. It completes a full cycle every π radians, not every 2π.

Geometrically, the curve y = sin²(x) is a shifted and compressed cosine wave centered on the horizontal line y = 1/2. Its precise identity is 1/2 − (1/2)cos(2x). When you integrate this expression with respect to x, you are calculating the cumulative area beneath two separate components:

• The steady baseline term (1/2): The area accumulates linearly. It does so at a constant rate of 1/2 unit per distance x. This gives you the term x/2. As a result, over any long interval, the accumulated area under sin²(x) averages out to half the interval length.
• The oscillating wave term (−(1/2)cos(2x)): This term adds and subtracts area symmetrically. It does so around the baseline every π units. Because it oscillates uniformly above and below zero, its integral −sin(2x)/4 fluctuates. It boundedly fluctuates between −1/4 and 1/4. As a result, it does not grow endlessly over time.

How do you integrate sin²(x) step by step using power reduction?

The primary method for integrating sin²(x) uses the half-angle power-reduction identity. No basic derivative rule directly produces sin²(x). Therefore, you must convert the quadratic power into a first-power trigonometric expression.

Step 1: Derive the power-reduction identity

Start with the cosine double-angle formula from trigonometry:

$$\cos(2x) = 1 - 2\sin^2(x)$$

Rearrange this formula to isolate sin²(x):

$$2\sin^2(x) = 1 - \cos(2x)$$

$$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$

This identity substitutes a squared sine term. It uses a linear cosine expression instead. Also, the angle of this expression is doubled.

Step 2: Rewrite the integral

Substitute the identity into the original integral:

$$\int \sin^2(x) \, dx = \int \frac{1 - \cos(2x)}{2} \, dx$$

Factor out the constant denominator 1/2 or split the integrand into two distinct terms:

$$\int \sin^2(x) \, dx = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(2x) \, dx$$

Step 3: Integrate term by term

The first integral is standard: \(\int 1 \, dx = x\).

For the second integral, \(\int \cos(2x) \, dx\), apply a quick linear substitution. Let \(u = 2x\). This means \(du = 2 \, dx\), so \(dx = \frac{du}{2}\). Alternatively, you can use the reverse chain rule.

$$\int \cos(2x) \, dx = \frac{\sin(2x)}{2}$$

Combine both integrated pieces:

$$\frac{1}{2}(x) - \frac{1}{2}\left(\frac{\sin(2x)}{2}\right) + C = \frac{x}{2} - \frac{\sin(2x)}{4} + C$$

Step 4: Check your work by taking the derivative

Verify the antiderivative by differentiating \(\frac{x}{2} - \frac{\sin(2x)}{4} + C\) using the chain rule:

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

The derivative returns the original integrand, confirming the answer is correct.

How do you integrate sin²(x) using integration by parts?

If you do not remember the power-reduction identity during an exam, you can solve the integral using integration by parts. The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\).

Set up the integral \(I = \int \sin^2(x) \, dx\). First, split \(\sin^2(x)\) into \(\sin(x) \cdot \sin(x)\).

• Let \(u = \sin(x)\), which gives \(du = \cos(x) \, dx\).
• Let \(dv = \sin(x) \, dx\), which gives \(v = -\cos(x)\).

Apply the integration by parts formula:

$$I = \sin(x)(-\cos(x)) - \int (-\cos(x))(\cos(x)) \, dx$$

$$I = -\sin(x)\cos(x) + \int \cos^2(x) \, dx$$

Next, use the fundamental Pythagorean identity \(\cos^2(x) = 1 - \sin^2(x)\) to rewrite the remaining integral:

$$I = -\sin(x)\cos(x) + \int (1 - \sin^2(x)) \, dx$$

$$I = -\sin(x)\cos(x) + \int 1 \, dx - \int \sin^2(x) \, dx$$

Notice that the original integral \(I = \int \sin^2(x) \, dx\) has reappeared. It is on the right-hand side. Also, it has a negative sign. So, add \(I\) to both sides of the equation. You can then solve for it algebraically:

$$I + I = -\sin(x)\cos(x) + x$$

$$2I = x - \sin(x)\cos(x)$$

$$I = \frac{x - \sin(x)\cos(x)}{2} + C$$

You can confirm this result. It matches the power-reduction output. First, recall the double-angle identity \(\sin(2x) = 2\sin(x)\cos(x)\). Next, replace \(\sin(x)\cos(x)\) with \(\frac{\sin(2x)}{2}\). This yields:

$$\frac{x - \frac{\sin(2x)}{2}}{2} + C = \frac{x}{2} - \frac{\sin(2x)}{4} + C$$

Which method should you use?

Both techniques yield identical mathematical results, but they differ in setup speed, trigonometric prerequisite knowledge, and algebraic complexity.

Comparison Factor	Power-Reduction Identity Method	Integration by Parts Method
Main Prerequisite	\(\sin^2(x) = \frac{1 - \cos(2x)}{2}\)	\(\int u \, dv = uv - \int v \, du\) and \(\sin^2(x) + \cos^2(x) = 1\)
Number of Steps	3 steps	5 steps
Result Format	\(\frac{x}{2} - \frac{\sin(2x)}{4} + C\)	\(\frac{x - \sin(x)\cos(x)}{2} + C\)
Best Used For	Quick manual evaluation and definite integrals	Proving equivalence or when trig identities are forgotten

How do you evaluate definite integrals of sin²(x)?

When evaluating definite integrals of sin²(x), chosen limits often simplify the oscillating sine term. Below are common interval computations evaluated on standard calculus problems.

Integrating sin²(x) from 0 to π

$$\int_{0}^{\pi} \sin^2(x) \, dx = \left[ \frac{x}{2} - \frac{\sin(2x)}{4} \right]_{0}^{\pi}$$

Substitute the upper limit \(\pi\) and lower limit \(0\):

$$\left(\frac{\pi}{2} - \frac{\sin(2\pi)}{4}\right) - \left(\frac{0}{2} - \frac{\sin(0)}{4}\right)$$

Since \(\sin(2\pi) = 0\) and \(\sin(0) = 0\), the oscillating term evaluates to zero at both endpoints:

$$\int_{0}^{\pi} \sin^2(x) \, dx = \frac{\pi}{2}$$

Integrating sin²(x) from 0 to 2π

For a full period from \(0\) to \(2\pi\):

$$\int_{0}^{2\pi} \sin^2(x) \, dx = \left( \frac{2\pi}{2} - \frac{\sin(4\pi)}{4} \right) - (0 - 0) = \pi$$

Here is a helpful shortcut for definite integrals. Use it over whole half-periods (intervals of length \(n\pi\)). Because \(\sin^2(x)\) and \(\cos^2(x)\) share identical area properties over symmetric periods. Also, \(\sin^2(x) + \cos^2(x) = 1\). So, each function accounts for exactly half the total area under \(y = 1\). This explains why the average value of \(\sin^2(x)\) over complete cycles is always \(1/2\).

Integrating sin²(x) over partial intervals \(0 \to π/4\)

When boundaries do not fall on whole multiples of \(\pi/2\), the oscillating sine term does not drop out:

$$\int_{0}^{\pi/4} \sin^2(x) \, dx = \left[ \frac{x}{2} - \frac{\sin(2x)}{4} \right]_{0}^{\pi/4} = \left(\frac{\pi}{8} - \frac{\sin(\pi/2)}{4}\right) - (0) = \frac{\pi}{8} - \frac{1}{4} \approx 0.1427$$

How does integrating sin²(x) compare with cos²(x)?

The integral of \(\cos^2(x)\) uses the corresponding power-reduction identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\). As a result, the antiderivative differs only by the operational sign in front of the sine term:

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

Adding the antiderivatives of \(\sin^2(x)\) and \(\cos^2(x)\) provides a quick sanity check:

$$\int (\sin^2(x) + \cos^2(x)) \, dx = \left(\frac{x}{2} - \frac{\sin(2x)}{4}\right) + \left(\frac{x}{2} + \frac{\sin(2x)}{4}\right) + C = x + C$$

Because \(\sin^2(x) + \cos^2(x) = 1\), the integral of \(1\) is \(x + C\), matching the sum of the two individual integrals.

Worked examples and variations

Example 1: Inner coefficient adjustment, sin²(3x)

Evaluate \(\int \sin^2(3x) \, dx\).

When an inner linear constant \(k\) is present, double the entire argument. Do this inside the power-reduction identity. For example, \(\sin^2(3x) = \frac{1 - \cos(6x)}{2}\).

$$\int \sin^2(3x) \, dx = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(6x) \, dx = \frac{x}{2} - \frac{\sin(6x)}{12} + C$$

The denominator \(12\) comes from multiplying \(2\) from the half-angle formula by \(6\) from the inner derivative coefficient.

Example 2: Half-angle coefficient, sin²(x/2)

Evaluate \(\int \sin^2(x/2) \, dx\).

You can double the argument \(x/2\). This gives \(x\). As a result, you simplify the identity to \(\sin^2(x/2) = \frac{1 - \cos(x)}{2}\).

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

Example 3: Higher even powers, sin⁴(x)

Evaluate \(\int \sin^4(x) \, dx\).

You can express \(\sin^4(x)\) as \((\sin^2(x))^2\). Then apply the power-reduction formula twice.

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

Apply power reduction again to \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\):

$$\sin^4(x) = \frac{1}{4} - \frac{\cos(2x)}{2} + \frac{1 + \cos(4x)}{8} = \frac{3}{8} - \frac{\cos(2x)}{2} + \frac{\cos(4x)}{8}$$

Now integrate term by term:

$$\int \sin^4(x) \, dx = \frac{3x}{8} - \frac{\sin(2x)}{4} + \frac{\sin(4x)}{32} + C$$

Example 4: Volume of a solid of revolution

You can find the volume of a solid. First, rotate one arch of \(y = \sin(x)\) around the x-axis. This arch goes from \(x = 0\) to \(x = \pi\).

Using the disk method, the cross-sectional area is \(A(x) = \pi r^2 = \pi \sin^2(x)\). So the volume formula is:

$$V = \pi \int_{0}^{\pi} \sin^2(x) \, dx$$

Using our previous definite integral result \(\int_{0}^{\pi} \sin^2(x) \, dx = \frac{\pi}{2}\):

$$V = \pi \cdot \left(\frac{\pi}{2}\right) = \frac{\pi^2}{2} \approx 4.935 \text{ cubic units}$$

Example 5: Product integral sin²(x)cos²(x)

Evaluate \(\int_0^{2\pi} \sin^2(x)\cos^2(x) \, dx\).

Combine terms using the identity \(\sin(x)\cos(x) = \frac{\sin(2x)}{2}\):

$$\sin^2(x)\cos^2(x) = \left(\frac{\sin(2x)}{2}\right)^2 = \frac{\sin^2(2x)}{4}$$

Over the full period \([0, 2\pi]\), the average value of \(\sin^2(2x)\) is \(1/2\). Multiply by the interval length \(2\pi\) and factor \(1/4\):

$$\int_0^{2\pi} \frac{\sin^2(2x)}{4} \, dx = \frac{1}{4} \cdot \left(\frac{1}{2}\right) \cdot 2\pi = \frac{\pi}{4}$$

You can verify these solutions step-by-step using an online integral calculator or third-party graphing calculator.

What is the general strategy for powers of sines and cosines?

The integral of sin²(x) is a classic example. It belongs to the general class of trigonometric integrals. These have the form \(\int \sin^m(x) \cos^n(x) \, dx\). Follow this decision rule based on the powers \(m\) and \(n\):

• If the power of sine (\(m\)) is odd: You should save one \(\sin(x)\) factor for \(du\). Then, rewrite the remaining even power. Use the identity \(\sin^2(x) = 1 - \cos^2(x)\). After that, execute a u-substitution with \(u = \cos(x)\).
• If the power of cosine (\(n\)) is odd: You should save one \(\cos(x)\) factor for \(du\). Also, rewrite the remaining even power. Use the identity \(\cos^2(x) = 1 - \sin^2(x)\). Next, execute a \(u\)-substitution with \(u = \sin(x)\).
• If both powers (\(m\) and \(n\)) are even: You cannot use \(u\)-substitution directly. Instead, you must use power-reduction formulas. These convert all squared terms into first powers of cosine with doubled angles.

You can solve odd powers of sines or cosines more easily. This is because you can isolate a single trigonometric factor for substitution. However, even powers require power reduction.

Common pitfalls when integrating sin²(x)

1. Applying the basic power rule: Writing \(\frac{\sin^3(x)}{3} + C\) is incorrect. Taking the derivative of \(\frac{\sin^3(x)}{3}\) yields \(\sin^2(x)\cos(x)\). This happens due to the chain rule. It includes an extra \(\cos(x)\) term. This term is not present in the original integrand.
2. Misplacing formula signs: Confusing \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\) with \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\). Remember that \(\sin(0) = 0\), so at \(x = 0\) the expression \(\frac{1 - \cos(0)}{2} = \frac{1 - 1}{2} = 0\), confirming the minus sign belongs to sine.
3. Forgetting the inner linear factor during integration: Integrating \(\cos(2x)\) requires dividing by \(2\). Combining this with the front factor of \(1/2\) produces \(1/4\) in the second term (\(\frac{\sin(2x)}{4}\)). Writing \(\frac{\sin(2x)}{2}\) is a common arithmetic mistake.
4. Confusing sin²(x) with sin(x²): \(\sin^2(x)\) represents \((\sin(x))^2\). In contrast, \(\sin(x^2)\) squares the input angle. Then it takes the sine. The integral \(\int \sin(x^2) \, dx\) cannot use elementary functions. Instead, it requires Fresnel integrals.
5. Calculator mode set to degrees: You evaluate definite limits like \(\frac{\pi}{8} - \frac{1}{4}\). Your computing device must be in radian mode. Degree inputs yield incorrect results.

Where is the integral of sin²(x) used in real applications?

The integral of sin²(x) plays a central role across science and engineering disciplines:

• Alternating Current (AC) electricity: Electrical power dissipates in a resistor. The formula for this is \(P(t) = \frac{V(t)^2}{R}\). You must integrate \(\sin^2(\omega t)\). This calculates average power over a cycle. Also, this applies to sinusoidal voltages \(V(t) = V_{\text{peak}} \sin(\omega t)\). As a result, the average value of \(1/2\) leads directly to the Root Mean Square (RMS) formula \(V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}}\).
• Physics and wave optics: Wave energy density and light intensity are proportional to the square of wave amplitude. Integrating \(\sin^2(x)\) yields time-averaged intensity values in diffraction patterns and acoustic measurements.
• Fourier series expansions: You can approximate arbitrary periodic functions using sines and cosines. This requires computing orthogonality integrals. For example, consider \(\int_0^{2\pi} \sin(mx)\sin(nx) \, dx\). The \(m = n\) case reduces directly to \(\int \sin^2(mx) \, dx\).
• Trigonometric substitution: You need to substitute \(x = \sin(\theta)\) and \(dx = \cos(\theta) \, d\theta\). This happens when you evaluate integrals with terms like \(\sqrt{1 - x^2}\) in calculus. For example, problems like the integral of √(1−x²) convert into \(\int \cos^2(\theta) \, d\theta\). This uses the same power-reduction steps as \(\sin^2(x)\).

Try it yourself

Practice problems

Work through these problems to test your understanding before checking the solutions below.

1. \(\int \cos^2(2x) \, dx\)
2. \(\int_0^{\pi/2} \sin^2(x) \, dx\)
3. \(\int \sin^3(x) \, dx\)
4. \(\int_0^{\pi} \cos^2(x) \, dx\)
5. \(\int_0^{\pi/2} \sin^2(2x) \, dx\)
6. \(\int \cos^3(x) \, dx\)

Solutions:

1. \(\frac{x}{2} + \frac{\sin(4x)}{8} + C\) (using \(\cos^2(2x) = \frac{1 + \cos(4x)}{2}\)).

2. \(\frac{\pi}{4}\) (evaluated as \(\left[\frac{x}{2} - \frac{\sin(2x)}{4}\right]_0^{\pi/2} = \frac{\pi}{4} - 0\)).

3. \(-\cos(x) + \frac{\cos^3(x)}{3} + C\) (rewrite as \((1 - \cos^2(x))\sin(x)\) and substitute \(u = \cos(x)\)).

4. \(\frac{\pi}{2}\) (evaluated as \(\left[\frac{x}{2} + \frac{\sin(2x)}{4}\right]_0^{\pi} = \frac{\pi}{2}\)).

5. \(\frac{\pi}{4}\) (let \(u = 2x\), converting the integral to \(\frac{1}{2}\int_0^{\pi}\sin^2(u) \, du = \frac{1}{2} \cdot \frac{\pi}{2}\)).

6. \(\sin(x) - \frac{\sin^3(x)}{3} + C\) (rewrite as \((1 - \sin^2(x))\cos(x)\) and substitute \(u = \sin(x)\)).

Frequently asked questions

Is the integral of sin²(x) equal to sin³(x)/3?

No. Applying the power rule to trigonometric functions ignores the chain rule. As a result, differentiating \(\frac{\sin^3(x)}{3}\) gives \(\sin^2(x)\cos(x)\), which contains an extra cosine factor.

How do you integrate higher odd powers like sin³(x)?

For odd powers, split off one sine factor to write \(\sin^3(x) = (1 - \cos^2(x))\sin(x)\). Then perform \(u\)-substitution with \(u = \cos(x)\) and \(du = -\sin(x) \, dx\).

Is (x − sin(x)cos(x))/2 mathematically equivalent to x/2 − sin(2x)/4?

Yes, you can apply the double-angle formula \(\sin(2x) = 2\sin(x)\cos(x)\). As a result, the term \(\frac{\sin(2x)}{4}\) becomes \(\frac{2\sin(x)\cos(x)}{4}\). This simplifies to \(\frac{\sin(x)\cos(x)}{2}\). Therefore, both forms match exactly.

What is the integral of sin²(x) over a full 0 to 2π interval?

The definite integral from \(0\) to \(2\pi\) equals \(\pi\). Over any complete period, the oscillating sine component averages to zero, leaving half the interval length as the total area.

Why is a trigonometric identity necessary to integrate sin²(x)?

The term \(\sin^2(x)\) lacks an accompanying \(\cos(x)\) term. Therefore, you cannot use \(u\)-substitution. The power-reduction identity replaces the squared exponent. It uses a linear first-power cosine function. You can integrate this directly.

Related calculus guides: derivative of cos²x, integral of tan x.

Further reading

• Integrals Involving Trig Functions (Paul’s Online Math Notes) — This comprehensive guide covers strategy. Also, it addresses odd and even powers of sines and cosines.
• Average Function Value (Paul’s Online Math Notes) — Detailed mathematical derivation for the mean value of squared trigonometric functions.