Deriving the Sine-Power Reduction Formula

Introduction

A power of sine can look like a new integration problem for every exponent. Integration by parts reveals a systematic relationship: an integral with exponent \(n\) can be reduced to one with exponent \(n-2\), together with an explicit trigonometric term.

The reduction is useful because repeated application eventually reaches one of two simple integrals. Its derivation also illustrates how to handle an integral that reappears during integration by parts: collect that term and solve for it.

Definition and scope

Throughout, \(n\) is a nonnegative integer and \(x\) is real. Write \(I_n\) for an antiderivative of \(\sin^n x\); when a formula is stated for \(I_n\), an arbitrary additive constant is understood. In particular, the integrand for \(n=0\) is the constant function \(1\).

We use integration by parts in the form \(\int u\,dv=uv-\int v\,du\), and the identity \(\cos^2 x=1-\sin^2 x\). The integer and exponent conditions matter: this recurrence is being derived for nonnegative integer powers, with \(n\ge 2\), so that the lower-power integral \(I_{n-2}\) is among the defined cases.

Deriving the recurrence

To apply integration by parts, separate one factor of \(\sin x\) and use the remaining power as \(u\). This makes \(dv=\sin x\,dx\) easy to integrate. The resulting integral contains a factor of \(\cos^2 x\), which can be rewritten to produce both \(I_{n-2}\) and \(I_n\). The latter is the original integral, so the key final step is to collect it.

Writing out the algebra makes the signs and coefficients explicit:

Add \((n-1)I_n\) to both sides, then divide by \(n\). The result is the sine-power reduction formula:

The derivation explains why the method works: integration by parts creates a copy of the original integral, and the identity for \(\cos^2x\) expresses that copy alongside the lower-power integral. Rearranging isolates the desired antiderivative. The coefficient \(1/n\) is not guessed; it comes from adding the \(n-1\) copies on the right to the one copy on the left.

The explicit term and endpoint contribution

The term \(-\sin^{n-1}x\cos x/n\) is the explicit term in the recurrence. Before dividing by \(n\), the integration-by-parts boundary expression is \(-\sin^{n-1}x\cos x\). For a definite integral over an interval \([a,b]\), such a boundary expression is evaluated at the endpoints, as \([-\sin^{n-1}x\cos x]_a^b\). For an indefinite integral there are no fixed endpoints; the same expression appears as part of the antiderivative.

It is important to keep this term as well as the reduced integral. Dropping it would generally make the proposed antiderivative have the wrong derivative. The examples below check the complete expressions by differentiation.

Initial cases

The recurrence lowers the exponent by two, so it needs two starting values, one for each parity of the exponent. These are obtained directly:

For example, reducing an even exponent eventually reaches exponent zero, while reducing an odd exponent eventually reaches exponent one. The recurrence itself applies only when the current exponent is at least two; it is not a formula for \(n=0\) or \(n=1\). In using it repeatedly, each equality is understood modulo an additive constant, so the final antiderivative still includes an arbitrary \(C\).

Worked applications

A common sign mistake

When expanding \(\cos^2x\), it is easy to replace it incorrectly by \(1+\sin^2x\). That changes the sign of the copy of \(I_n\) and produces a false recurrence. For instance, starting from the correct integration-by-parts line at \(n=2\), the incorrect identity would lead to

and hence to \(0=-\sin x\cos x+I_0\), which cannot be an identity of antiderivatives. The correction is \(\cos^2x=1-\sin^2x\), so the correct line is

Now collecting \(I_2\) gives \(2I_2=-\sin x\cos x+I_0\), exactly the recurrence result. Checking the trigonometric identity before collecting terms prevents this error.

Why the recurrence matters

The formula converts a family of integrals into a controlled sequence of simpler ones. Each use lowers the exponent by two, multiplies the remaining integral by a known coefficient, and adds a term that is explicit in \(x\). The base cases make this process terminate. The endpoint contribution also records the boundary expression produced by integration by parts, which is essential when the same calculation is used on a fixed interval.

As a practical check, differentiate a proposed result after applying the recurrence. This verifies both the reduced integral and the explicit term, including its sign. It also catches mistakes in distributing the coefficient \((n-1)/n\) during repeated reductions.