Choosing a Method for Mixed Power Integrals

Why the Choice of Method Matters

A mixed power integral has the form \(\int \sin^m x\cos^n x\,dx\), where \(m\) and \(n\) are nonnegative integers. The same integral can sometimes be approached in more than one way, but the exponent pattern often points to a shorter route.

The first question is whether either exponent is odd. An odd power leaves one sine or cosine factor available to pair with the differential of the other function. If both exponents are even, there is no such leftover factor, so power-reduction identities are usually the natural starting point. Product-to-sum can be useful when those identities produce products of different-frequency terms.

Odd Exponents Suggest Substitution

Suppose \(m\) is odd. Write \(m=2k+1\), where \(k\) is a nonnegative integer, and separate one sine factor: \(\sin^{2k+1}x=\sin^{2k}x\sin x\). Use \(\sin^2x+\cos^2x=1\) to rewrite the remaining even power as \(\sin^{2k}x=(1-\cos^2x)^k\). Then \(u=\cos x\) gives \(du=-\sin x\,dx\), turning the integral into a polynomial integral in \(u\).

If instead \(n\) is odd, write \(\cos^n x=\cos^{2j}x\cos x\), use \(\cos^{2j}x=(1-\sin^2x)^j\), and set \(u=\sin x\), so \(du=\cos x\,dx\). Either route applies on any interval where the functions are defined; the substitution identities themselves hold for all real \(x\).

When both exponents are odd, either factor can be reserved. It is worth checking both possibilities: one may produce a simpler polynomial than the other. The method does not depend on choosing a unique substitution; it depends on matching the reserved factor to the differential of the chosen variable.

When Both Exponents Are Even

If \(m\) and \(n\) are both even, there is no single leftover sine or cosine factor to pair with a differential. Instead, use the power-reduction identities \(\sin^2x=(1-\cos 2x)/2\) and \(\cos^2x=(1+\cos 2x)/2\), repeatedly as needed. These identities replace even powers by sums involving multiples of the angle. The resulting terms are constants or trigonometric functions whose antiderivatives are direct.

For a simple mixed even-power integral, a double-angle identity can shorten the calculation. The identity \(\sin 2x=2\sin x\cos x\) implies \(\sin^2x\cos^2x=\frac14\sin^2 2x\). Reducing the remaining sine square then gives a constant term and one cosine term.

Where Product-to-Sum Fits

Power reduction may leave products of cosines with different angles. Product-to-sum identities turn such products into sums, which can be integrated term by term. In particular, \(\cos A\cos B=\frac12[\cos(A-B)+\cos(A+B)]\). This follows by adding the cosine addition identities \(\cos(A-B)=\cos A\cos B+\sin A\sin B\) and \(\cos(A+B)=\cos A\cos B-\sin A\sin B\), then dividing by \(2\).

This is not always shorter than continuing with power reduction. It is a choice to consider when different frequencies appear together; after applying the identity, check that the resulting sum is simpler. The following calculation illustrates the choice for an all-even mixed power.

A Practical Decision Process

Common Pitfall: Forgetting the Sign in the Substitution

If the reserved factor is \(\sin x\,dx\) and the substitution is \(u=\cos x\), then \(du=-\sin x\,dx\). Omitting that minus sign reverses the entire antiderivative. For instance, the first worked example would incorrectly produce \(+\cos^3x/3-2\cos^5x/5+\cos^7x/7\) if the sign were lost. Differentiating that expression gives \(-\sin^5x\cos^2x\), the negative of the required integrand. Keeping \(du=-\sin x\,dx\) in the substitution line prevents this error.

A second useful check is not to assume product-to-sum is automatically the best route for every even-power integral. If power reduction already produces a short sum, integrate it directly. The goal is an efficient valid method, not the greatest possible number of identities.