Cosine-Power Recurrences and Endpoint Terms

Why Reduce a Cosine Power?

An integral such as \(\int \cos^6 x\,dx\) can be awkward to expand directly. A reduction formula replaces an integral with a high power by one with a lower power, so repeated application eventually brings the calculation to simple base integrals. The key is to keep track not only of the lower-power integral, but also of the term produced at the endpoints of integration by parts.

For cosine powers, that term contains \(\sin x\cos^{n-1}x\). Its form is different from the corresponding term in the sine-power formula. The difference follows from the different antiderivatives and derivatives used in the two integrations by parts.

Deriving the Cosine-Power Recurrence

Let \(n\ge 2\) be an integer. In \(\int \cos^n x\,dx\), separate one factor of cosine and use integration by parts with \(u=\cos^{n-1}x\) and \(dv=\cos x\,dx\). Then \(du=-(n-1)\cos^{n-2}x\sin x\,dx\) and \(v=\sin x\). The integration-by-parts formula, \(\int u\,dv=uv-\int v\,du\), gives

Use \(\sin^2 x=1-\cos^2 x\) in the remaining integral. The first resulting term is the integral with exponent \(n-2\); the second is the original integral. Writing \(I_n=\int\cos^n x\,dx\) for this calculation,

Collect the \(I_n\) terms on the left and divide by \(n\). This division is valid because \(n\ge2\), so \(n\ne0\). The result is the cosine-power reduction formula.

For indefinite integrals, the equality is understood up to an additive constant. In applying the formula, include a constant once in the final antiderivative; it need not be added separately at every reduction step. The formula itself does not evaluate the base cases. The simplest ones are \(\int 1\,dx=x+C\) and \(\int\cos x\,dx=\sin x+C\).

Why the Boundary Term Has This Form

The boundary term comes directly from \(uv\) in integration by parts. Here, the factor \(\cos^{n-1}x\) is chosen as \(u\), and an antiderivative of the remaining factor \(\cos x\) is \(\sin x\). Their product is \(\sin x\cos^{n-1}x\). The factor \(1/n\) appears only after collecting the two copies of the original integral and dividing by \(n\).

By contrast, the sine-power recurrence uses \(u=\sin^{n-1}x\) and \(dv=\sin x\,dx\). Since an antiderivative of \(\sin x\) is \(-\cos x\), its boundary term is negative. In the same notation, the sine-power formula is

Thus the boundary terms are not interchangeable: the cosine recurrence has \(+\sin x\cos^{n-1}x/n\), while the sine recurrence has \(-\cos x\sin^{n-1}x/n\). The sign and the trigonometric factors reflect the choices of \(u\), \(dv\), and their derivatives in each integration by parts.

If the integral is definite on \([a,b]\), the cosine boundary term is evaluated at both limits. With \(C_n(a,b)=\int_a^b\cos^n x\,dx\), the same calculation gives

Here \([g(x)]_a^b\) means \(g(b)-g(a)\). The endpoint contribution may be zero for particular limits, but it is part of the general formula and should be evaluated rather than silently discarded.

Worked Examples

A Common Sign and Factor Mistake

A frequent error is to stop after the first integration by parts and forget that \(\sin^2x\cos^{n-2}x\) contains a term equal to \(\cos^n x\). For example, an incorrect calculation might claim

This omits the negative \(-(n-1)\int\cos^n x\,dx\) contributed by \(\sin^2x=1-\cos^2x\). That term is precisely why the coefficient of the reduced integral is \((n-1)/n\), not \(n-1\), and why the boundary term is divided by \(n\). The correction is to retain both terms after substitution and collect the original integral before dividing.

When the Recurrence Is Useful

Each application lowers the exponent by two, so the formula is useful when an integral contains a pure integer power of cosine. It also makes the boundary contribution visible in definite integrals, where omitting or mis-evaluating it changes the answer. The recurrence applies for integer \(n\ge2\); it does not itself provide a reduction step for \(n=0\) or \(n=1\), which are handled directly.

A reliable procedure is to record the exponent used at each step, preserve the boundary term, and stop when the integral has reached a directly evaluable base case. For a definite integral, evaluate each bracket as upper-limit value minus lower-limit value. For an indefinite integral, differentiate the finished expression to check that it returns the original power of cosine.