Checking Reduction Formulas by Differentiation

Why Differentiate a Reduction Formula?

A reduction formula expresses an integral in terms of an explicit function and a simpler integral. The explicit term, its sign, and the coefficient multiplying the simpler integral all matter: a small error in any of them can make the proposed recurrence false.

Differentiation provides a direct test. An indefinite integral represents a family of antiderivatives, and an expression is an antiderivative of an integrand precisely when its derivative equals that integrand. Thus, to check a proposed recurrence, replace the simpler integral by a primitive, differentiate the entire right-hand side, and simplify.

This test checks whether the formula is correct, not whether a particular method was used to derive it. It also makes errors easier to diagnose: an incorrect derivative often reveals whether the problem is a sign, a coefficient, or an omitted explicit term.

What Is Being Checked?

Suppose a proposed recurrence has the form

where \(I_n\) is intended to be a primitive of an integrand \(f_n\), \(B_n(x)\) is an explicit term, \(c_n\) is a coefficient, and \(I_{n-2}\) is a primitive of a simpler integrand \(f_{n-2}\). To test the recurrence, differentiate both sides. The right-hand side must have derivative \(f_n(x)\):

This equation is the practical check. It avoids needing to know a particular formula for \(I_{n-2}\), because its defining property is that \(I_{n-2}'(x)=f_{n-2}(x)\).

The check is performed on an interval where the expressions and derivatives involved are defined. For example, a recurrence involving \(\sec x\) must be checked on an interval that does not contain a point where \(\cos x=0\). A derivative identity on such an interval verifies the recurrence there; it does not automatically extend the formula through points where its terms are undefined.

If two differentiable expressions have the same derivative on an interval, their difference is constant there. Consequently, a derivative check verifies an indefinite-integral formula up to an additive constant. A proposed expression that differs only by a constant is not mathematically wrong; one that has the wrong derivative is.

Worked Example: Checking a Sine-Power Recurrence

A Complete Reason the Test Works

The derivative test follows directly from the definition of an antiderivative. More explicitly, suppose \(F\) and \(G\) are differentiable on an interval \(J\), and \(F'(x)=G'(x)\) for every \(x\in J\). Then

By the mean value theorem, for any \(a,b\in J\) with \(a<b\), there is a point \(c\in(a,b)\) such that

Thus \((F-G)(b)=(F-G)(a)\) for every such pair, so \(F-G\) is constant on \(J\). Conversely, if \(F-G\) is constant, then differentiating shows \(F'=G'\). Therefore, checking that a proposed right-hand side has the correct derivative is exactly what is needed to verify an indefinite-integral formula, with the usual freedom to add a constant.

Worked Example: Finding a Sign Error

Worked Example: Detecting a Missing Explicit Term

A Practical Checking Routine

When a recurrence looks complicated, organize the check instead of differentiating from memory. Keep the target integrand visible, and identify separately the derivative of the explicit term and the contribution from the reduced integral.

Common Pitfalls

A frequent mistake is to differentiate only the explicit term and forget that the reduced integral contributes its integrand. In the sine-power example, omitting that contribution would leave only the derivative of \(-\sin^{n-1}x\cos x/n\), which is not \(\sin^n x\) by itself.

Another common error is to alter a coefficient while simplifying. For example, if the coefficient \((n-1)/n\) in the verified sine recurrence were replaced by \(1/n\), the proposed formula would be \(\int\sin^n x\,dx=-\frac{\sin^{n-1}x\cos x}{n}+\frac1n\int\sin^{n-2}x\,dx\). Its derivative is

For \(n=2\) the factor in parentheses is \(\sin^2x+1-\cos^2x=2\sin^2x\), so the derivative is \(\sin^2x\); this special case happens to pass. For \(n=3\) and \(x=\pi/4\), however, the derivative is \(\frac{\sqrt2/2}{3}\left(\frac12+1-1\right)=\frac{\sqrt2}{12}\), whereas \(\sin^3(\pi/4)=\frac{\sqrt2}{4}\). A check at a single special exponent can therefore conceal a coefficient error in a general recurrence. Verify the symbolic formula for its stated range, not just one convenient case.

Finally, do not confuse a missing constant with a missing function of \(x\). Constants disappear under differentiation, so they do not affect validity. An omitted boundary term such as \(\sec x\tan x/2\), by contrast, changes the derivative and must be included.

Why This Check Matters

Reduction formulas can contain several linked pieces: an explicit term, a numerical coefficient, and a lower-power integral. Differentiation tests all of them at once. It is particularly useful after algebraic simplification, where a sign change or an identity applied incorrectly can make a previously valid recurrence fail.

The method is also independent of how the recurrence was obtained. Whether it arose through integration by parts or algebraic rearrangement, the same question decides its validity: does the derivative of the complete proposed right-hand side recover the original integrand on the stated interval?