Full-Period Integrals of Even Powers

Why the Constant Term Determines a Full-Period Integral

A power such as \(\cos^6 x\) is nonnegative and varies throughout a cycle, so its integral is not found by treating the power as though it were a constant. Power-reduction expansions turn the power into a constant term plus cosine terms of higher frequencies. Over a full period, those oscillating terms contribute zero net area. The remaining constant term therefore determines the integral.

This method gives exact values for integrals such as \(\int_0^{2\pi}\sin^{2m}x\,dx\) and \(\int_0^{2\pi}\cos^{2m}x\,dx\), where \(m\) is a nonnegative integer. Here, a full period means an interval of length \(2\pi\), the period of sine and cosine.

The Power-Reduction Expansions

For a nonnegative integer \(m\), the binomial coefficient \(\binom{2m}{k}\) counts the ways to choose \(k\) objects from \(2m\). The expansions we need are

and

Both expansions have the same constant term, \(\binom{2m}{m}/4^m\). Their oscillatory terms differ in sign, but that difference does not affect a full-period integral: each cosine term has zero integral over \([0,2\pi]\).

Why the Expansions Have This Form

A short derivation from the binomial theorem explains both the coefficients and the shared constant term. Euler's formulas express cosine and sine using complex exponentials:

For cosine, expand \((e^{ix}+e^{-ix})^{2m}/2^{2m}\). Choosing \(k\) factors of \(e^{-ix}\) gives the term \(\binom{2m}{k}e^{i(2m-2k)x}/4^m\). The term is constant precisely when \(2m-2k=0\), or \(k=m\), so its coefficient is \(\binom{2m}{m}/4^m\). For each positive frequency \(2j\), the terms with exponents \(2j\) and \(-2j\) have equal coefficients. Combining them uses \(e^{i2jx}+e^{-i2jx}=2\cos(2jx)\), giving the cosine expansion above.

For sine, expand \((e^{ix}-e^{-ix})^{2m}/(2i)^{2m}\). The same central choice \(k=m\) gives the constant term. For a frequency \(2j\), pairing the terms with exponents \(2j\) and \(-2j\) introduces the factor \((-1)^j\). This gives the stated sine expansion. In particular, the different signs of the oscillatory terms do not change the constant coefficient.

To see why the other terms vanish on a full period, integrate a typical term directly. For any positive integer \(j\),

The constant term integrates to itself times the interval length. Thus, integrating either expansion leaves only its constant term multiplied by \(2\pi\).

Worked Examples

The Role of the Full-Period Condition

The cancellation depends on the limits. For the term \(\cos(2jx)\), the antiderivative is \(\sin(2jx)/(2j)\). Its endpoint values agree on \([0,2\pi]\), so the definite integral is zero. The same reasoning applies on any interval of length \(2\pi\), since \(\sin(2j(x+2\pi))=\sin(2jx+4\pi j)=\sin(2jx)\).

By contrast, a cosine term need not have zero integral on an interval that is not a full period. For instance, \(\int_0^\pi\cos(2x)\,dx=0\), but this happens because the endpoints also give the same sine value; it is not a reason to discard every oscillatory term for arbitrary limits. The safe procedure is to check the limits or use the full-period formula only when the interval has length \(2\pi\).

The constant coefficient also has an interpretation: over \([0,2\pi]\), the average value of a function \(f\) is \(\frac{1}{2\pi}\int_0^{2\pi}f(x)\,dx\). Therefore, the average value of either \(\sin^{2m}x\) or \(\cos^{2m}x\) over a full period is \(\binom{2m}{m}/4^m\). Since the integral is the average value multiplied by the interval length, the theorem follows in another way.

Common Mistake: Dropping Oscillatory Terms Without Checking

A tempting but incomplete calculation is to write \(\cos^4x=3/8+(1/2)\cos(2x)+(1/8)\cos(4x)\), discard the cosine terms, and conclude that the integral is \(3/8\). This is only the constant coefficient, not the integral. The constant must still be multiplied by the interval length.

On \([0,2\pi]\), the correct calculation is

There are two checks to make: verify that the limits make the oscillatory terms vanish, and multiply the constant term by the length of the interval. This keeps the coefficient of the constant term distinct from the integral itself.

Using the Result Efficiently

Once the expansion is understood, there is no need to write every oscillatory term when only a full-period integral is requested. Identify the exponent \(2m\), find the central binomial coefficient \(\binom{2m}{m}\), and multiply by \(2\pi/4^m\). Writing out the expansion remains useful when checking a particular case or explaining why the formula applies.

The method works equally for sine and cosine because their even-power expansions have the same constant term. The signs of the oscillatory terms differ, but on a full period every one of those terms contributes zero net area.