Integrals of Powers of Sine and Cosine

11 min read

Textbook Reference

OpenStax Calculus Volume 2 is free and openly licensed. The strategies and example integrands below follow this section.

Key idea

The whole game is to turn a product of sines and cosines into something a single substitution can swallow.

You already know substitution: if you can spot a function and its derivative both sitting in the integrand, you rename the inside piece $u$ and the integral collapses. The derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. So sine and cosine are each other’s substitution partners. The difficulty with an integral like $\int \sin^m x \cos^n x\, dx$ is that there is too much sine and cosine mixed together for a clean substitution. The trick is to peel off exactly one factor to serve as the “$du$” and convert all the rest into the other function using the identity $\sin^2 x + \cos^2 x = 1$.

Here is the picture. Suppose one of the powers is odd, say the cosine power. Pull one cosine aside; that lone cosine will become $du$ when you let $u = \sin x$. The cosines left behind come in an even power, and any even power of cosine is a power of $\cos^2 x = 1 - \sin^2 x$, which is entirely sines. Now the whole integrand is sines times one stray cosine, which is exactly the shape substitution wants. If instead both powers are even, there is no factor to peel off and no clean substitution, so you take a different road: the power-reducing identities trade a squared sine or cosine for a plain cosine of a doubled angle, lowering the powers until the integral is easy.

This means: before you integrate, you look at the parity of the powers (whether each is odd or even). Odd power present means substitution after splitting off one factor. Both powers even means power-reducing identities. The parity tells you which tool to grab.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• Carry out a $u$-substitution, including changing $dx$ into $du$
• State the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, and rearrange it to $\sin^2 x = 1 - \cos^2 x$ or $\cos^2 x = 1 - \sin^2 x$
• Recall the power-reducing identities $\sin^2 x = \dfrac{1}{2} - \dfrac{1}{2}\cos 2x$ and $\cos^2 x = \dfrac{1}{2} + \dfrac{1}{2}\cos 2x$
• Integrate $\cos(2x)$ and $\sin(2x)$, remembering the $\frac{1}{2}$ from the chain rule
• Recognize odd versus even exponents

If substitution is shaky, review the Substitution lessons first. If the identities are unfamiliar, review the trigonometric-functions material.

Quick Reference

The problem: evaluate $\displaystyle\int \sin^m x \cos^n x\, dx$. Choose your strategy by the parity of $m$ (the sine power) and $n$ (the cosine power).

Identities you will use. \[ \sin^2 x + \cos^2 x = 1, \qquad \sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x, \qquad \cos^2 x = \frac{1}{2} + \frac{1}{2}\cos 2x. \]

If both powers are odd, either substitution works; split off whichever factor you like.

Key Concepts

1. The Odd-Power Strategy

When at least one power is odd, the integral is a substitution problem in disguise.

The method (cosine power odd).

1. Write $\cos^n x = \cos^{n-1} x \cdot \cos x$, peeling off one $\cos x$.
2. Since $n - 1$ is even, replace $\cos^{n-1} x = (\cos^2 x)^{(n-1)/2} = (1 - \sin^2 x)^{(n-1)/2}$.
3. Now everything is sines except the lone $\cos x$. Let $u = \sin x$, so $du = \cos x\, dx$.
4. Integrate the resulting polynomial in $u$, then substitute $u = \sin x$ back.

(If the sine power is odd instead, peel off one $\sin x$, convert the even cosines, and let $u = \cos x$, remembering $du = -\sin x\, dx$ carries a minus sign.)

Example 1. Find $\displaystyle\int \cos^2 x \sin^3 x\, dx$.

Goal and thought process. The sine power is $3$, which is odd, so peel off one $\sin x$ and convert the rest to cosines, then substitute $u = \cos x$.

Step 1. Split the odd factor. Write $\sin^3 x = \sin^2 x \cdot \sin x$: \[ \int \cos^2 x \sin^3 x\, dx = \int \cos^2 x \cdot \sin^2 x \cdot \sin x\, dx. \]

Step 2. Convert the even sines to cosines using $\sin^2 x = 1 - \cos^2 x$: \[ = \int \cos^2 x (1 - \cos^2 x)\sin x\, dx. \]

Step 3. Substitute $u = \cos x$, so $du = -\sin x\, dx$, that is $\sin x\, dx = -du$: \[ = \int u^2 (1 - u^2)(-du) = -\int (u^2 - u^4)\, du. \]

Step 4. Integrate in $u$. \[ = -\left(\frac{u^3}{3} - \frac{u^5}{5}\right) + C = -\frac{u^3}{3} + \frac{u^5}{5} + C. \]

Step 5. Substitute back $u = \cos x$: \[ \boxed{\int \cos^2 x \sin^3 x\, dx = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C} \]

Check by differentiating. $\frac{d}{dx}\left[\frac{\cos^5 x}{5}\right] = \cos^4 x(-\sin x)$ and $\frac{d}{dx}\left[-\frac{\cos^3 x}{3}\right] = -\cos^2 x(-\sin x) = \cos^2 x \sin x$. Sum: $-\cos^4 x \sin x + \cos^2 x \sin x = \cos^2 x \sin x(1 - \cos^2 x) = \cos^2 x \sin x \cdot \sin^2 x = \cos^2 x \sin^3 x$. Correct.

Quick self-check. In $\displaystyle\int \cos^3 x\, dx$, the sine power is $0$ (even) and the cosine power is $3$ (odd). Which substitution should you use, and what is the lone factor you peel off?

Show answer

Peel off one $\cos x$ and use $u = \sin x$. Write $\cos^3 x = \cos^2 x \cos x = (1 - \sin^2 x)\cos x$. Then $u = \sin x$, $du = \cos x\, dx$, giving $\int (1 - u^2)\, du = u - \frac{u^3}{3} = \sin x - \frac{\sin^3 x}{3} + C$.

2. The Even-Power Strategy

When both powers are even, there is no odd factor to peel off, so substitution does not start. Instead, lower the powers with the power-reducing identities.

The method.

1. Replace each $\sin^2 x$ with $\frac{1}{2} - \frac{1}{2}\cos 2x$ and each $\cos^2 x$ with $\frac{1}{2} + \frac{1}{2}\cos 2x$.
2. Expand. You will get a constant, a $\cos 2x$ term, and possibly a $\cos^2 2x$ term.
3. If a $\cos^2 2x$ appears, reduce it again with the same identity (now at angle $2x$).
4. Integrate term by term, remembering the $\frac{1}{2}$ chain-rule factor on $\int \cos 2x\, dx = \frac{1}{2}\sin 2x$.

Example 2. Find $\displaystyle\int \sin^2 x\, dx$.

Goal and thought process. The sine power is $2$, even, and there is no cosine factor to peel off. Use the power-reducing identity.

Step 1. Reduce the power. \[ \int \sin^2 x\, dx = \int \left(\frac{1}{2} - \frac{1}{2}\cos 2x\right) dx. \]

Step 2. Integrate term by term. \[ = \frac{1}{2}x - \frac{1}{2}\cdot\frac{1}{2}\sin 2x + C = \frac{x}{2} - \frac{\sin 2x}{4} + C. \]

Boxed answer. \[ \boxed{\int \sin^2 x\, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C} \]

Check by differentiating. $\frac{d}{dx}\left[\frac{x}{2} - \frac{\sin 2x}{4}\right] = \frac{1}{2} - \frac{2\cos 2x}{4} = \frac{1}{2} - \frac{1}{2}\cos 2x = \sin^2 x$. Correct, using the identity in reverse.

Recap. The even case never uses substitution. It trades squared sines and cosines for plain cosines of a doubled angle, which integrate directly. The price is that the answer comes out in terms of $2x$.

3. Choosing the Strategy

A quick decision before you compute, based only on the parities:

• Either power odd: peel off one copy of the odd-powered function. If cosine is odd, save a $\cos x$ and substitute $u = \sin x$. If sine is odd, save a $\sin x$ and substitute $u = \cos x$.
• Both powers even: no peeling; apply power-reducing identities and integrate term by term.
• Both powers odd: either substitution works; pick the one that leaves less algebra.

The parity check costs a glance and tells you the whole plan. (A common first mistake is to reach for power-reducing identities when an odd power is present, which works but is far more labor than the one-line substitution.)

Common Errors Summary

Common Misconceptions

Leveled Practice

Level 1 -- Direct Application

Problem 1. Find $\displaystyle\int \sin^3 x\, dx$.

Problem 2. Find $\displaystyle\int \cos^3 x\, dx$.

Problem 3. Without computing, state which strategy fits $\displaystyle\int \sin^4 x \cos^2 x\, dx$ and why.

Level 2 -- Mixed Powers

Problem 4. Find $\displaystyle\int \cos^5 x \sin^2 x\, dx$.

Problem 5. Find $\displaystyle\int \cos^2 x\, dx$.

Level 3 -- Reasoning

Problem 6. Explain why an odd power anywhere in $\int \sin^m x \cos^n x\, dx$ guarantees that a single substitution will finish the integral, while two even powers do not.

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• Decide the strategy from the parity of the powers before computing
• Carry out the odd-power case: split off one factor, convert with the Pythagorean identity, substitute
• Handle the minus sign correctly when $u = \cos x$
• Carry out the even-power case with the power-reducing identities
• Integrate $\cos 2x$ and $\sin 2x$ with the correct $\frac{1}{2}$ factor
• Evaluate $\int \cos^3 x\, dx$, $\int \sin^3 x\, dx$, $\int \sin^2 x\, dx$, and $\int \cos^2 x\, dx$ from scratch
• Check an antiderivative by differentiating it back

Mental Model

Sorting a sine-cosine integral is like reading the parity of the exponents as a road sign.

An odd exponent is a green light for substitution. The odd power hands you a spare factor to use as $du$, and the Pythagorean identity converts everything else into the partner function. One clean substitution and you are done. The work is short and mechanical.

Two even exponents is a detour sign. There is no spare factor, so substitution cannot start. You take the longer road of the power-reducing identities, trading each squared function for a plain cosine of twice the angle and integrating those. It is more steps, but it always gets through.

So the first thing you do is not compute. You look at the two exponents and read the sign: any odd power means substitute, both even means reduce. The strategy is decided before you write a single line of integration.

Connections

Within Techniques of Integration (MATH162)

• Built directly on: substitution and the trigonometric identities from precalculus. This section is where those identities pay off inside integrals.
• Integration by parts (Section 7.1): reduction formulas that lower large even powers of sine or cosine in fewer steps are derived using integration by parts, tying this section back to the previous one.
• Powers of secant and tangent: the next topics extend the same parity-based reasoning to $\int \sec^m x \tan^n x\, dx$, using the identity $\sec^2 x = 1 + \tan^2 x$ in place of the Pythagorean identity.
• Trigonometric substitution (Section 7.3): the comfort with sine-cosine powers built here is exactly what trigonometric substitution relies on after a change of variable.

Toward Later Mathematics

• Fourier series: integrals of products of sines and cosines over a period are the orthogonality relations that make Fourier analysis work; the even-power and product cases computed here are the building blocks.
• Probability and physics: averages of $\sin^2$ and $\cos^2$ over a period (each equal to $\frac{1}{2}$) appear in computing the average power of an alternating signal, which the even-power result makes precise.

Audience Notes

For students who find math intimidating: You only decide between two recipes, and a single glance at the exponents tells you which. Odd power: save one factor, swap the rest using $\sin^2 + \cos^2 = 1$, and substitute. Both even: use the two power-reducing identities and integrate. Watch the minus sign when you let $u = \cos x$, and the $\frac{1}{2}$ when you integrate $\cos 2x$.

For students interested in proof: The parity argument is a small but complete proof that the odd case always reduces to integrating a polynomial. Notice that the method depends on a single algebraic fact, $\sin^2 x + \cos^2 x = 1$, applied to an even leftover power. The even-even case has no such reduction to a polynomial, which is why it requires the different power-reducing route.

For students interested in careers: These integrals are the computational core of signal processing and electrical engineering. Computing the energy or average power of a waveform, or projecting a signal onto its frequency components, comes down to exactly these products and powers of sine and cosine.

Back to Techniques of Integration | Previous: Repeated Integration by Parts | Next: Integrals of Powers of Secant and Tangent