Parity-Based Classification of Sine-Cosine Powers

Why Parity Guides the Method

Integrals involving powers of sine and cosine often have the form \(\int \sin^m(x)\cos^n(x)\,dx\). The exponents tell us which trigonometric identity or substitution is likely to make the integral simpler. In particular, an odd exponent provides a single sine or cosine factor that can be reserved for a substitution; when both exponents are even, power-reduction identities are the natural choice.

This is a way to choose a method, not a separate rule for evaluating every integral. Once the parity is identified, the remaining algebra still needs to be carried out carefully. We will use indefinite integrals, so each antiderivative includes an arbitrary constant \(C\).

Definition: An integer is even if it is divisible by \(2\), and odd if it is not. For this tutorial, \(m\) and \(n\) are nonnegative integers. An exponent of \(0\) is even, and a zeroth power is interpreted as \(1\).

The Four Parity Cases

There are four possibilities for the pair of exponents: sine odd and cosine even; sine even and cosine odd; both odd; or both even. The first three have an odd power available to supply a single sine or cosine factor. In the fourth case, neither exponent supplies such a factor, so we reduce the even powers instead.

Sine exponent \(m\)	Cosine exponent \(n\)	Method to try
Odd	Even	Save one factor of \(\sin x\); use \(u=\cos x\).
Even	Odd	Save one factor of \(\cos x\); use \(u=\sin x\).
Odd	Odd	Save either one sine factor or one cosine factor; either substitution can work.
Even	Even	Use power-reduction identities to rewrite the even powers.

The identity \(\sin^2 x+\cos^2 x=1\) lets us rewrite the remaining even powers when we save an odd factor. For example, \(\sin^{m-1}x=(1-\cos^2 x)^{(m-1)/2}\) when \(m\) is odd. In the even-even case, use the power-reduction identities

\sin^2 x=\frac{1-\cos(2x)}{2}, \qquad \cos^2 x=\frac{1+\cos(2x)}{2}.

These identities convert powers of sine and cosine into expressions involving multiple angles, which can then be integrated using standard antiderivatives. The parity classification identifies when that approach is needed; it does not itself carry out all the algebra in a higher-power case.

Why Saving an Odd Factor Works

The substitution method works because the derivative of cosine is \(-\sin x\), while the derivative of sine is \(\cos x\). When an exponent is odd, we can separate one matching factor and express all remaining powers using the identity \(\sin^2 x+\cos^2 x=1\). The factor left outside the rewritten power then matches the differential from the substitution.

Principle: If \(m\) is odd, write \(\sin^m x=\sin^{m-1}x\sin x\), rewrite \(\sin^{m-1}x\) in powers of \(\cos x\), and try \(u=\cos x\). If \(n\) is odd, write \(\cos^n x=\cos^{n-1}x\cos x\), rewrite \(\cos^{n-1}x\) in powers of \(\sin x\), and try \(u=\sin x\). When both exponents are odd, either choice is available.

Here is the substitution argument in full for an odd sine exponent. Since \(m-1\) is an even nonnegative integer, \((m-1)/2\) is a nonnegative integer. Therefore,

\sin^m x\cos^n x = \sin x\,(1-\cos^2 x)^{(m-1)/2}\cos^n x.

Set \(u=\cos x\). Then \(du=-\sin x\,dx\), so \(\sin x\,dx=-du\). Substituting both the remaining cosine powers and the differential gives

\int \sin^m x\cos^n x\,dx = -\int (1-u^2)^{(m-1)/2}u^n\,du.

The right-hand side is a polynomial integral because both exponents in the expression are nonnegative integers. This explains why saving one sine factor is effective. The odd-cosine case follows by the same reasoning with \(u=\sin x\) and \(du=\cos x\,dx\).

Worked Examples

Worked Example: Odd Sine and Even Cosine Powers

Classify and evaluate \(\int \sin^3 x\cos^2 x\,dx\). The sine exponent \(3\) is odd and the cosine exponent \(2\) is even, so reserve one sine factor and use \(u=\cos x\).

First rewrite the remaining sine square using \(\sin^2 x=1-\cos^2 x\):

\begin{aligned} \int \sin^3 x\cos^2 x\,dx &=\int \sin x\sin^2 x\cos^2 x\,dx\\ &=\int \sin x(1-\cos^2 x)\cos^2 x\,dx. \end{aligned}

Set \(u=\cos x\), so \(du=-\sin x\,dx\). Thus

\begin{aligned} \int \sin x(1-\cos^2 x)\cos^2 x\,dx &=-\int (1-u^2)u^2\,du\\ &=-\int (u^2-u^4)\,du\\ &=-\left(\frac{u^3}{3}-\frac{u^5}{5}\right)+C\\ &=-\frac{u^3}{3}+\frac{u^5}{5}+C\\ &=-\frac{\cos^3 x}{3}+\frac{\cos^5 x}{5}+C. \end{aligned}

Check by differentiating. The derivative is \(\cos^2 x\sin x-\cos^4 x\sin x=\sin x\cos^2 x(1-\cos^2 x)=\sin^3 x\cos^2 x\), as required.

Worked Example: Even Sine and Odd Cosine Powers

Classify and evaluate \(\int \sin^2 x\cos^3 x\,dx\). The sine exponent \(2\) is even and the cosine exponent \(3\) is odd, so reserve one cosine factor and use \(u=\sin x\).

Rewrite \(\cos^2 x=1-\sin^2 x\), then substitute \(u=\sin x\), \(du=\cos x\,dx\):

\begin{aligned} \int \sin^2 x\cos^3 x\,dx &=\int \sin^2 x\cos^2 x\cos x\,dx\\ &=\int \sin^2 x(1-\sin^2 x)\cos x\,dx\\ &=\int u^2(1-u^2)\,du\\ &=\int (u^2-u^4)\,du\\ &=\frac{u^3}{3}-\frac{u^5}{5}+C\\ &=\frac{\sin^3 x}{3}-\frac{\sin^5 x}{5}+C. \end{aligned}

Check: differentiating gives \(\sin^2 x\cos x-\sin^4 x\cos x=\sin^2 x\cos x(1-\sin^2 x)=\sin^2 x\cos^3 x\).

Worked Example: Both Exponents Odd

Classify and evaluate \(\int \sin^3 x\cos^3 x\,dx\). Both exponents are odd. Choose to save one sine factor and set \(u=\cos x\):

\begin{aligned} \int \sin^3 x\cos^3 x\,dx &=\int \sin x(1-\cos^2 x)\cos^3 x\,dx\\ &=-\int (1-u^2)u^3\,du\\ &=-\int (u^3-u^5)\,du\\ &=-\frac{u^4}{4}+\frac{u^6}{6}+C\\ &=-\frac{\cos^4 x}{4}+\frac{\cos^6 x}{6}+C. \end{aligned}

The minus sign comes from \(du=-\sin x\,dx\). To check, differentiate the result: \(\cos^3 x\sin x-\cos^5 x\sin x=\sin x\cos^3 x(1-\cos^2 x)=\sin^3 x\cos^3 x\).

Worked Example: Both Exponents Even

Classify and evaluate \(\int \sin^2 x\cos^2 x\,dx\). Both exponents are even, so use power reduction rather than reserving an odd sine or cosine factor. The identities give

\begin{aligned} \sin^2 x\cos^2 x &=\left(\frac{1-\cos(2x)}{2}\right) \left(\frac{1+\cos(2x)}{2}\right)\\ &=\frac{1-\cos^2(2x)}{4}\\ &=\frac{1}{4}\sin^2(2x)\\ &=\frac{1}{8}\bigl(1-\cos(4x)\bigr). \end{aligned}

Therefore,

\begin{aligned} \int \sin^2 x\cos^2 x\,dx &=\int \frac{1}{8}\bigl(1-\cos(4x)\bigr)\,dx\\ &=\frac{x}{8}-\frac{\sin(4x)}{32}+C. \end{aligned}

For the last term, \(\int \cos(4x)\,dx=\frac14\sin(4x)\), so its coefficient is \(-\frac18\cdot\frac14=-\frac1{32}\). Check: differentiating gives \(\frac18-\frac18\cos(4x)=\frac18(1-\cos(4x))=\sin^2 x\cos^2 x\).

A Reliable Decision Process

For a product \(\sin^m x\cos^n x\), check the exponents before manipulating the integrand. This quick classification helps prevent choosing a substitution that has no matching differential factor.

Check the sine exponent.
If \(m\) is odd, separate one factor of \(\sin x\), rewrite the remaining sine powers using \(\sin^2 x=1-\cos^2 x\), and try \(u=\cos x\).

Check the cosine exponent.
If \(n\) is odd, separate one factor of \(\cos x\), rewrite the remaining cosine powers using \(\cos^2 x=1-\sin^2 x\), and try \(u=\sin x\).

Choose if both exponents are odd.
Either odd-factor substitution is available. Choose the one that leaves the simpler polynomial after rewriting.

Reduce if both exponents are even.
Use the power-reduction identities and simplify before integrating.

A Common Sign Pitfall

When using \(u=\cos x\), the differential is \(du=-\sin x\,dx\), not \(du=\sin x\,dx\). Omitting this minus sign reverses the sign of the antiderivative. For example, in the first worked example the transformed integral is \(-\int(u^2-u^4)\,du\), not \(\int(u^2-u^4)\,du\). The incorrect sign would produce \(\frac{\cos^3 x}{3}-\frac{\cos^5 x}{5}\); differentiating gives \(-\sin^3 x\cos^2 x\), the negative of the integrand. Including the minus sign gives an antiderivative whose derivative matches the original integrand.

A second useful check is to verify that the substitution has removed the trigonometric factor that supplied its differential. If using \(u=\cos x\), reserve \(\sin x\,dx\); if using \(u=\sin x\), reserve \(\cos x\,dx\). Rewriting the other powers then gives an expression in \(u\) alone.

Why the Classification Matters

Parity is a compact way to inspect the structure of a trigonometric integrand before calculating. If at least one exponent is odd, the corresponding factor can match a derivative under substitution. If both are even, repeatedly trying those same substitutions will not produce the same direct simplification: there is no single odd sine or cosine factor to reserve. Power reduction provides a systematic alternative.

The classification does not guarantee that every resulting expression is equally simple, and it does not replace checking the calculation. It does, however, suggest a sound first move. After integrating, differentiate the proposed answer and simplify using \(\sin^2 x+\cos^2 x=1\) when necessary. That final check catches sign errors and lost powers.

Check Your Understanding

For each question, identify the parity case before deciding on a method.

For \(\int \sin^5 x\cos^4 x\,dx\), which factor should be reserved, and what substitution should be tried?
For \(\int \sin^4 x\cos^3 x\,dx\), which substitution matches the odd factor?
If both exponents are odd, why are two substitutions available?
Which method is indicated for \(\int \sin^4 x\cos^2 x\,dx\), and why is an odd-factor substitution not the first choice?
When \(u=\cos x\), what is \(du\), and what sign must be included when replacing \(\sin x\,dx\)?