Integrating Products with Matching Powers

Why Matching Powers Simplify

Some trigonometric products look like mixed powers, but when sine and cosine have the same exponent they share a useful structure. The identity \(\sin(2x)=2\sin x\cos x\) combines the two factors into one sine, leaving an integral whose form is governed by the parity of a single exponent.

Throughout, let \(m\) be a positive integer and consider the indefinite integral \(\int \sin^m x\cos^m x\,dx\). The key is to rewrite the product before choosing an integration method. This makes the distinction between odd and even powers immediate.

Definition: A matching-power sine-cosine product is an expression of the form \(\sin^m x\cos^m x\), where the sine and cosine have the same exponent \(m\). In this tutorial, \(m\) is a positive integer.

The Double-Angle Reduction

Since \(\sin(2x)=2\sin x\cos x\), we have \(\sin x\cos x=\frac12\sin(2x)\). Raising both sides to the \(m\)th power gives the reduction that makes this integral manageable.

\sin^m x\cos^m x = (\sin x\cos x)^m = \frac{1}{2^m}\sin^m(2x).

To integrate, set \(u=2x\). Then \(du=2\,dx\), so \(dx=\frac12\,du\). The factor \(\frac12\) from the change of variable must be included in addition to the factor \(2^{-m}\) already present.

\int \sin^m x\cos^m x\,dx = \frac{1}{2^{m+1}}\int \sin^m u\,du, \qquad u=2x.

This equation is the working rule for matching powers. It reduces the original product to an ordinary sine-power integral, multiplied by a constant. The exponent \(m\) is unchanged; the angle doubles.

Key takeaway: First use \(\sin x\cos x=\frac12\sin(2x)\), then account for \(du=2\,dx\). Together, these steps give the factor \(2^{-(m+1)}\) in the integral.

Why the Reduction Works

The reduction follows directly from the double-angle identity and the substitution rule. The substitution is valid because \(u=2x\) is differentiable and has nonzero derivative \(du/dx=2\). The proof can be written as a short chain of equalities.

Combine the matching factors.
Since their exponents agree, \(\sin^m x\cos^m x=(\sin x\cos x)^m\).

Apply the double-angle identity.
Replacing \(\sin x\cos x\) by \(\frac12\sin(2x)\) gives \(2^{-m}\sin^m(2x)\).

Change variables.
With \(u=2x\), we have \(dx=du/2\), producing the additional factor \(1/2\).

Integrate the resulting sine power.
The remaining task is to integrate \(\sin^m u\), choosing a method according to whether \(m\) is odd or even.

Thus the integral becomes \(2^{-(m+1)}\int\sin^m u\,du\). No approximation or special assumption about \(x\) is involved; the identity holds for every real \(x\), and the substitution is valid on any interval.

Odd Exponents: Save a Sine Factor

When \(m\) is odd, write \(m=2k+1\), where \(k\) is a nonnegative integer. Separate one factor of sine and convert the remaining even power with \(\sin^2u=1-\cos^2u\). The substitution \(t=\cos u\), with \(dt=-\sin u\,du\), then turns the integral into a polynomial integral.

\int \sin^{2k+1}u\,du = -\int (1-t^2)^k\,dt = -\sum_{j=0}^{k}(-1)^j\binom{k}{j} \frac{\cos^{2j+1}u}{2j+1}+C.

Here the finite sum comes from expanding \((1-t^2)^k\) with the binomial theorem and integrating each power of \(t\). This method is suited to odd exponents because removing one sine leaves an even power that can be expressed using cosine.

Worked Example: A Matching Product with Exponent Three

Evaluate \(\int \sin^3x\cos^3x\,dx\). First apply the matching-power identity, and then set \(u=2x\):

\sin^3x\cos^3x=\frac18\sin^3(2x), \qquad \int\sin^3x\cos^3x\,dx = \frac1{16}\int\sin^3u\,du.

To integrate the odd power, write \(\sin^3u=\sin u(1-\cos^2u)\). Let \(t=\cos u\), so \(dt=-\sin u\,du\). Then

\int\sin^3u\,du = -\int(1-t^2)\,dt = -t+\frac{t^3}{3}+C = -\cos u+\frac{\cos^3u}{3}+C.

Multiplying by \(1/16\) and returning to \(u=2x\) gives

\int\sin^3x\cos^3x\,dx = -\frac{\cos(2x)}{16} +\frac{\cos^3(2x)}{48}+C.

To check, differentiate the proposed antiderivative. The chain rule gives

\frac{d}{dx}\left( -\frac{\cos(2x)}{16} +\frac{\cos^3(2x)}{48} \right) = \frac{\sin(2x)}8-\frac{\cos^2(2x)\sin(2x)}8 = \frac{\sin^3(2x)}8 = \sin^3x\cos^3x.

The last equality uses \(\sin(2x)=2\sin x\cos x\), so the check recovers the original integrand.

Even Exponents: Use Power Reduction

If \(m\) is even, the odd-power substitution does not leave a single sine factor. Instead, use the power-reduction identity \(\sin^2u=\frac12(1-\cos(2u))\), or its repeated forms for higher even powers. These identities express an even sine power as a sum of a constant and cosine terms, each of which can be integrated directly.

A standard reduction formula gives the same parity-based process. For \(n\ge2\), integration by parts yields

\int\sin^n u\,du = -\frac{\sin^{n-1}u\cos u}{n} + \frac{n-1}{n}\int\sin^{n-2}u\,du.

Repeated use lowers an even exponent by two until the base integral \(\int 1\,du=u\) is reached. For the examples below, direct power reduction keeps the calculations especially transparent.

Worked Example: Matching Squares

Evaluate \(\int\sin^2x\cos^2x\,dx\). The matching-power identity gives

\sin^2x\cos^2x=\frac14\sin^2(2x).

Apply \(\sin^2\theta=\frac12(1-\cos(2\theta))\) with \(\theta=2x\):

\sin^2x\cos^2x = \frac18(1-\cos(4x)).

Integrate the constant and cosine terms separately. Since \(\int\cos(4x)\,dx=\frac14\sin(4x)\), the result is

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

Differentiate to verify:

\frac{d}{dx}\left(\frac{x}{8}-\frac{\sin(4x)}{32}\right) = \frac18-\frac{\cos(4x)}8 = \frac18(1-\cos(4x)) = \sin^2x\cos^2x.

Thus the derivative agrees with the original product.

Worked Example: Matching Fourth Powers

Evaluate \(\int\sin^4x\cos^4x\,dx\). First reduce the matching product:

\sin^4x\cos^4x=\frac1{16}\sin^4(2x).

Use the fourth-power identity \(\sin^4u=\frac18(3-4\cos(2u)+\cos(4u))\). With \(u=2x\), this becomes

\sin^4x\cos^4x = \frac1{128}\bigl(3-4\cos(4x)+\cos(8x)\bigr).

Integrating each term gives

\int\sin^4x\cos^4x\,dx = \frac{3x}{128} -\frac{\sin(4x)}{128} +\frac{\sin(8x)}{1024}+C.

For the check, differentiate term by term:

\frac{d}{dx}\left( \frac{3x}{128} -\frac{\sin(4x)}{128} +\frac{\sin(8x)}{1024} \right) = \frac3{128}-\frac{\cos(4x)}{32}+\frac{\cos(8x)}{128} = \frac{3-4\cos(4x)+\cos(8x)}{128}.

This is exactly the power-reduced form of \(\sin^4x\cos^4x\), confirming the antiderivative.

A Common Factor Error

The most frequent mistake is to use the double-angle identity but omit the factor from the substitution \(u=2x\). For example, the incorrect calculation below keeps the \(1/4\) from the squared identity but forgets that \(dx=du/2\):

\text{Incorrect:}\qquad \int\sin^2x\cos^2x\,dx \overset{\text{wrong}}{=} \frac14\int\sin^2u\,du, \qquad u=2x.

This coefficient is too large by a factor of two. The correct transformation is

\int\sin^2x\cos^2x\,dx = \frac14\int\sin^2(2x)\,dx = \frac18\int\sin^2u\,du.

The derivative check exposes the error: an antiderivative obtained with the incorrect coefficient would differentiate to twice the desired integrand. Always write \(du=2\,dx\), or equivalently \(dx=du/2\), explicitly before integrating.

Choosing the Method and Checking the Result

The matching-power identity handles the product, while parity determines how to integrate the resulting sine power. For an odd exponent, reserve one sine factor and substitute cosine. For an even exponent, use power reduction or repeatedly lower the exponent by two. In both cases, preserve the constants from the identity and the change of variable.

Exponent \(m\)	Form after reduction	Useful method
Odd	\(2^{-(m+1)}\int\sin^m u\,du\)	Separate one sine factor; use \(t=\cos u\)
Even	\(2^{-(m+1)}\int\sin^m u\,du\)	Apply power reduction or the reduction formula

These methods matter because they turn a product of two trigonometric powers into one standard sine-power integral without expanding the original product unnecessarily. A final derivative check is especially useful: it tests both the trigonometric identity and every constant introduced by substitution.

Check Your Understanding

Use the matching-power identity first, and check any antiderivative you find by differentiation.

Rewrite \(\sin^5x\cos^5x\) as a constant multiple of a sine power with a doubled angle.
In the substitution \(u=2x\), what differential relation must be used?
Which method is suited to integrating the resulting sine power when \(m=7\), and why?
Use power reduction to express \(\sin^2x\cos^2x\) as a constant plus a cosine term.
Explain why omitting the factor \(1/2\) from \(dx=du/2\) gives an incorrect antiderivative.