Why a Full Period Can Have Integral Zero
A sine curve takes positive values on part of a period and negative values on another part. When the integrand is an odd power of sine, those values remain equal in magnitude and opposite in sign under a suitable reflection. The integral records signed area, so the positive and negative contributions cancel.
This cancellation is not a claim that the function is zero, or that every part of its graph has zero area. It is a claim about the net signed area across a complete cycle. The reflection that pairs the contributions makes the reason precise.
Here \(m\) is an integer exponent. The result below concerns positive odd integers, such as \(1,3,5,\dots\), not arbitrary real exponents.
The Reflection Behind the Cancellation
Start with the standard period from \(0\) to \(2\pi\). Reflection across the vertical line \(x=\pi\) sends \(x\) to \(2\pi-x\). The sine identity for this reflection is
Raising both sides to an odd positive integer power preserves the minus sign. Thus, if \(m\) is odd,
Each point \(x\) in the first half of the period is paired with \(2\pi-x\) in the second half. The paired values have equal magnitude and opposite signs. The following proof expresses that pairing directly in the integral.
Let \(f(x)=\sin^m x\), where \(m\) is positive and odd. Write the integral as \(\int_0^\pi f(x)\,dx+\int_\pi^{2\pi}f(x)\,dx\).
In the second integral, use \(u=2\pi-x\). Then \(dx=-du\); when \(x=\pi\), \(u=\pi\), and when \(x=2\pi\), \(u=0\). Therefore \(\int_\pi^{2\pi}f(x)\,dx=\int_\pi^0 f(2\pi-u)(-du)=\int_0^\pi f(2\pi-u)\,du\).
The reflection identity gives \(f(2\pi-u)=\sin^m(2\pi-u)=-\sin^m u=-f(u)\), so the second-half integral is \(-\int_0^\pi f(u)\,du\).
The first-half integral and the second-half integral cancel, giving \(\int_0^{2\pi}f(x)\,dx=\int_0^\pi f(x)\,dx-\int_0^\pi f(x)\,dx=0\).
This proves the result on \([0,2\pi]\). To see why the starting point of a full period does not matter, let \(f(x)=\sin^m x\), which is continuous and \(2\pi\)-periodic, and define \(F(a)=\int_a^{a+2\pi}f(x)\,dx\). By the Fundamental Theorem of Calculus and the chain rule, \(F'(a)=f(a+2\pi)-f(a)=0\). Hence \(F\) is constant in \(a\); its value is the one already found on \([0,2\pi]\), namely zero. This establishes the claim for every interval of length \(2\pi\).
Worked Examples
Worked Example: The First Power
Evaluate \(\int_0^{2\pi}\sin x\,dx\) directly, and show how the two half-period contributions cancel.
An antiderivative of \(\sin x\) is \(-\cos x\). On the first half-period,
On the second half-period,
Adding the two values gives \(2+(-2)=0\). Check: evaluating the antiderivative over the whole period gives \([-\cos x]_0^{2\pi}=-1-(-1)=0\), in agreement with the half-period calculation.
Worked Example: The Third Power
Evaluate \(\int_0^{2\pi}\sin^3 x\,dx\), showing the values on both halves. Rewrite one factor using \(\sin^2 x=1-\cos^2 x\):
An antiderivative is \(-\cos x+\frac{\cos^3 x}{3}\), since its derivative is \(\sin x-\cos^2 x\sin x=\sin^3 x\). The first half gives
The second half is
Thus the full-period value is \(\frac43-\frac43=0\). As a check, the same antiderivative evaluated from \(0\) to \(2\pi\) has value \(\left(-1+\frac13\right)-\left(-1+\frac13\right)=0\).
Worked Example: The Fifth Power
Evaluate \(\int_0^{2\pi}\sin^5 x\,dx\). This example shows the same cancellation for a higher odd power. Write
An antiderivative is \(-\cos x+\frac23\cos^3 x-\frac15\cos^5 x\). Differentiating gives \(\sin x-2\cos^2x\sin x+\cos^4x\sin x=\sin x(1-\cos^2x)^2=\sin^5x\), as required. On \([0,\pi]\), the endpoint values are \(8/15\) at \(\pi\) and \(-8/15\) at \(0\), so
On \([\pi,2\pi]\), the endpoint values occur in the reverse order, giving
The two contributions sum to \(\frac{16}{15}-\frac{16}{15}=0\). The antiderivative also has the same value at \(0\) and \(2\pi\), so evaluating over the full interval confirms the answer.
Why Oddness Is Essential
The key algebraic step in the proof was \((-\sin x)^m=-\sin^m x\). This is true precisely when \(m\) is odd. For an even exponent, \((-\sin x)^m=\sin^m x\), so reflection gives equal values, not opposite values. The cancellation argument therefore does not establish a zero integral for even powers. Their full-period integrals are a different question.
The result also distinguishes a full period from a half-period. For example, the first-half integrals in the worked examples are \(2\), \(4/3\), and \(16/15\), respectively—not zero. The full-period result follows because the second half contributes the negative of the first.
Common Sign Mistake
A frequent error is to use \(\sin(2\pi-x)=\sin x\) when reflecting across \(x=\pi\). For \(m=1\), that incorrect identity would say that the second-half integral equals the first-half integral, leading to the wrong calculation \(2+2=4\). The correct identity is \(\sin(2\pi-x)=-\sin x\), so the second-half integral is \(-2\), and the full-period value is \(2+(-2)=0\).
It is also important not to confuse a sign change after a shift by \(\pi\) with a period of \(\pi\). In fact, \(\sin(x+\pi)=-\sin x\), so an odd power changes sign after that shift; it is \(2\pi\)-periodic, not \(\pi\)-periodic. Keeping track of this sign prevents errors when selecting or splitting integration intervals.
When the Result Is Useful
Before calculating an antiderivative for an odd power of sine, check whether the definite integral covers a full period. If it does, reflection may give the value immediately. This can save work and provide a useful check on a longer calculation. It does not mean the corresponding indefinite integral is zero: an antiderivative must still differentiate to \(\sin^m x\), and the theorem concerns only the definite integral over a complete period.
Check Your Understanding
Use the reflection argument and the worked calculations to answer these questions.
- What point does \(x\) reflect to under the map \(x\mapsto 2\pi-x\), and what happens to \(\sin^m x\) when \(m\) is odd?
- Why does the substitution \(u=2\pi-x\) reverse the limits in the second-half integral?
- What are the values of \(\int_0^\pi\sin^3x\,dx\) and \(\int_\pi^{2\pi}\sin^3x\,dx\), and why do they cancel?
- Does the theorem imply that \(\int_0^\pi\sin^5x\,dx=0\)? Explain.
- What incorrect full-period answer might result from forgetting the negative sign in \(\sin(2\pi-x)=-\sin x\)?
