Integrating Odd Powers of Cosine

Why Odd Powers of Cosine Are Convenient

Integrals of powers of trigonometric functions are easier when the exponent has a useful parity. For a positive odd power of cosine, one factor of cosine can be kept aside, while the remaining power is even. The Pythagorean identity then rewrites that even power using sine, making the substitution \(u=\sin x\) available.

The method works for every positive odd integer exponent. Its main advantage is that it converts the trigonometric integral into a polynomial integral, which can be evaluated term by term.

The Method

Suppose \(n\) is a positive odd integer. Write it as \(n=2m+1\), where \(m\) is a nonnegative integer. Then

Now let \(u=\sin x\). Its differential is \(du=\cos x\,dx\), so the integral becomes

The expression \((1-u^2)^m\) is a polynomial in \(u\). It can be expanded and integrated using the power rule. The answer is then written back in terms of \(x\), replacing \(u\) by \(\sin x\). As with any indefinite integral, include an arbitrary constant of integration.

Why the Substitution Works

The substitution is justified by the chain rule. If \(F\) is an antiderivative of \((1-u^2)^m\), then \(F'(u)=(1-u^2)^m\). Applying the chain rule to \(F(\sin x)\) gives

The final equality uses \(1-\sin^2x=\cos^2x\). Thus the function produced by integrating the polynomial in \(u\), and then substituting \(u=\sin x\), really is an antiderivative of the original integrand.

The binomial theorem gives a general expression for this polynomial method. Expanding \((1-u^2)^m\) and integrating each term yields

Here \(\binom{m}{k}\) is the binomial coefficient. This formula is a compact version of the same procedure; in practice, expanding the polynomial for the particular exponent often makes the calculation clearer.

Worked Examples

A Common Substitution Mistake

A substitution changes both the trigonometric expressions and the differential. It is not enough to replace \(\sin x\) by \(u\) while leaving powers of cosine unchanged or treating them as powers of \(u\). For example, the following calculation is incorrect:

The error is that \(u=\sin x\) does not mean \(\cos x=u\). Also, the relation between \(dx\) and \(du\) is \(du=\cos x\,dx\), not \(du=dx\). Differentiating the incorrect answer exposes the problem:

which is not \(\cos^3x\). The correct calculation first reserves one factor \(\cos x\,dx\), then rewrites the other two cosine factors as \(1-\sin^2x\):

Keeping track of the differential is an essential part of substitution, not a separate bookkeeping detail.

How to Recognize When This Method Applies

The method applies directly when the integrand is a positive odd integer power of cosine, \(\cos^{2m+1}x\). The odd exponent ensures that after setting aside one cosine factor, the remaining exponent is even and can be expressed entirely in terms of \(\sin x\). The remaining cosine factor then supplies exactly the differential needed for \(u=\sin x\).

This parity check is useful before doing any algebra. If the cosine exponent is not odd, separating one cosine factor does not leave an even power in this same form. Do not force this particular procedure onto an integrand that does not meet its condition; the treatment of other exponent patterns requires different choices.

The resulting polynomial integral is also a practical check on the work. Its degree is finite, each term is integrated with the ordinary power rule, and differentiating the answer reverses the steps: the polynomial in \(\sin x\) becomes a polynomial in \(\sin x\) multiplied by \(\cos x\), which recombines into the original odd power of cosine.

Summary

For a positive odd exponent, the integral of a cosine power can be reduced to a polynomial integral by separating one cosine factor and expressing the other factors through \(\cos^2x=1-\sin^2x\). The substitution \(u=\sin x\) then turns the remaining factor \(\cos x\,dx\) into \(du\). After integrating, replace \(u\) by \(\sin x\) and include \(C\).