Choosing Which Trigonometric Exponent to Reduce

Why the Choice of Exponent Matters

Reduction formulas turn an integral with large exponents into integrals with smaller exponents. But when a product contains two trigonometric powers, there may be more than one exponent to reduce. A formula can be valid and still be an inefficient choice: it may create a complicated endpoint term or lead to a base integral that is no easier than the original.

The practical goal is to choose a path that makes progress toward an integral you can evaluate directly. Exponent parity is a useful guide. An odd power often leaves a factor whose derivative supplies a convenient substitution; an even power can often be lowered repeatedly by a recurrence. The best choice also depends on which base integral the recurrence reaches.

Sine and Cosine: Two Recurrence Choices

For nonnegative integers \(m,n\), write

When \(m\ge 2\), one recurrence lowers the sine exponent by two. When \(n\ge 2\), another lowers the cosine exponent by two. Both formulas are valid, but they lead to different lower-exponent integrals.

Why the sine recurrence works

For \(m\ge2\), integrate by parts with \(u=\sin^{m-1}x\) and \(dv=\sin x\cos^n x\,dx\). Then \(du=(m-1)\sin^{m-2}x\cos x\,dx\) and \(v=-\cos^{n+1}x/(n+1)\). Substituting these expressions gives

Use \(\cos^{n+2}x=\cos^n x(1-\sin^2x)\). The remaining integral is \(I_{m-2,n}-I_{m,n}\). Thus

Collecting the \(I_{m,n}\) terms on the left and multiplying by \((n+1)/(m+n)\) yields the stated sine recurrence. The cosine recurrence follows by the same integration-by-parts process, choosing \(u=\cos^{n-1}x\) and \(dv=\cos x\sin^m x\,dx\).

Use Parity to Choose a Route

If one exponent is odd, separating one factor of that trigonometric function may make a substitution available. For example, with an odd sine exponent, leave one \(\sin x\) factor and convert the remaining sine powers using \(\sin^2x=1-\cos^2x\); then \(u=\cos x\) uses \(du=-\sin x\,dx\). With an odd cosine exponent, the corresponding choice is \(u=\sin x\).

This observation does not mean that the odd exponent must always be the one you reduce by recurrence. Often the more efficient choice is to lower the even exponent until it reaches zero, then use the remaining odd-power substitution. The two choices in the theorem let you compare these paths.

Tangent and Secant: Check Which Factor Is Available

For nonnegative integers \(m,n\), define \(J_{m,n}=\int\tan^m x\sec^n x\,dx\), on an interval where \(\cos x\ne0\). The substitution choices are governed by the derivatives \(d(\tan x)/dx=\sec^2x\) and \(d(\sec x)/dx=\sec x\tan x\).

If \(n\) is even and \(n\ge2\), set aside a factor of \(\sec^2x\,dx\). The remaining secant power can be rewritten using \(\sec^2x=1+\tan^2x\), so \(u=\tan x\) produces a polynomial in \(u\). If \(m\) is odd and \(m\ge1\), set aside a factor of \(\sec x\tan x\,dx\). Rewrite the other tangent factors using \(\tan^2x=\sec^2x-1\), and use \(u=\sec x\).

When a Recurrence Is More Useful Than a Substitution

Sometimes neither parity shortcut immediately gives a simple polynomial substitution, or the integral already has the form of a secant power. A secant-exponent recurrence can then be useful. For \(m\ge0\) and \(n\ge2\),

To see the identity, differentiate \(\tan^{m+1}x\sec^{n-2}x\). The product rule gives \((m+1)\tan^m x\sec^n x+(n-2)\tan^{m+2}x\sec^{n-2}x\). Rewrite the second term using \(\tan^2x=\sec^2x-1\). The derivative becomes \((m+n-1)\tan^m x\sec^n x-(n-2)\tan^m x\sec^{n-2}x\). Rearranging and integrating gives the recurrence.

A Practical Decision Process

Common Pitfall: Reducing an Exponent Without Checking the Endpoint

A recurrence is not a license to substitute negative exponents into its lower-index integral. For example, the sine-reduction formula requires \(m\ge2\). Applying it to \(I_{1,2}\) and writing \(I_{-1,2}\) is invalid: the recurrence was not established for that lower index, and it does not lead to the intended nonnegative-power base integral. The correct choice is the cosine recurrence, since \(n=2\), or a direct substitution.

For the cosine recurrence, the coefficient of the reduced integral is \((n-1)/(m+n)\), not \(n/(m+n)\). In the same example, the correct calculation is \(I_{1,2}=\sin^2x\cos x/3+I_{1,0}/3\). Replacing \(n-1\) by \(n\) would give \(2I_{1,0}/3\), which is wrong. Check the coefficient against the stated recurrence before simplifying.

Why a Good Choice Matters

Reduction formulas can produce equivalent antiderivatives with different-looking expressions, as the examples show. The point of choosing carefully is not to discover a uniquely correct route; it is to avoid needless algebra and to reach an integral whose evaluation is clear. Parity indicates which factors can support a substitution, while the remaining exponents reveal whether recurrence is likely to reach a simple base integral.

Keep the limits of a shortcut in view. An odd exponent or an even secant exponent can make a substitution convenient, but not every parity pattern immediately gives a polynomial integral. In those cases, use a valid recurrence and track its indices until a manageable base case is reached.