Introduction
A mixed power of sine and cosine may look different when the exponents are exchanged. On the interval from \(0\) to \(\pi/2\), however, the two definite integrals have the same value. The reason is a reflection of the interval: replacing an angle by its distance from \(\pi/2\) interchanges sine and cosine.
This is a statement about definite integrals on a particular interval. It does not say that the two integrands are equal at every angle. Instead, the substitution \(x=\pi/2-u\) pairs each point of the interval with its reflected point, and the sine and cosine factors trade places under that reflection.
The Quarter-Period Symmetry
For nonnegative integers \(m\) and \(n\), consider the definite integral of the product \(\sin^m(x)\cos^n(x)\) over the quarter-period interval. The integrand is continuous there, so the integral exists. When an exponent is zero, the corresponding factor is understood to be \(1\), including at an endpoint.
The exponents need not be equal. For example, the theorem says that the integral with sine to the first power and cosine to the third power matches the integral with sine to the third power and cosine to the first power. It does not claim that the integrands match at the same value of \(x\).
Why Reflection Exchanges the Factors
The essential trigonometric identities are the cofunction identities \(\sin(\pi/2-u)=\cos u\) and \(\cos(\pi/2-u)=\sin u\). Under \(x=\pi/2-u\), the endpoints reverse: \(x=0\) corresponds to \(u=\pi/2\), and \(x=\pi/2\) corresponds to \(u=0\). Also, \(dx=-du\). Both the reversed bounds and the negative differential matter.
Set \(x=\pi/2-u\). Then \(\sin x=\cos u\) and \(\cos x=\sin u\).
The limits \(x=0\) and \(x=\pi/2\) become \(u=\pi/2\) and \(u=0\), respectively, while \(dx=-du\).
The negative differential changes the integral from bounds \(\pi/2\) to \(0\) into the same integral with bounds \(0\) to \(\pi/2\).
Proof of the Symmetry
Start with the integral whose sine exponent is \(m\) and whose cosine exponent is \(n\). Apply the change of variable \(x=\pi/2-u\). The trigonometric factors exchange, and the differential contributes a minus sign:
The minus sign can be absorbed by reversing the bounds. Since multiplication is commutative, the resulting integrand can be written with the sine factor first:
The variable \(u\) is a dummy variable of integration, so the final integral is the same quantity as the one on the right side of the theorem. This proves the equality. The proof relies on the interval being reflected onto itself and on the cofunction identities; the same argument does not automatically establish the equality on arbitrary limits.
Worked Example: First Powers Interchanged
Worked Example: First Powers Interchanged
Compare the two integrals with exponent pairs \((1,0)\) and \((0,1)\). The theorem predicts that they are equal.
For the interchanged exponents, integrate cosine:
Thus both integrals equal \(1\), as predicted. The antiderivatives check by differentiation: \(\frac{d}{dx}(-\cos x)=\sin x\), and \(\frac{d}{dx}(\sin x)=\cos x\). The endpoint evaluations also show why the limits are important to the result.
Worked Example: One Exponent Is Zero
Worked Example: One Exponent Is Zero
Compare the integral of \(\sin^2 x\) with the integral of \(\cos^2 x\). This is the theorem with \((m,n)=(2,0)\). Use the power-reduction identities \(\sin^2 x=(1-\cos 2x)/2\) and \(\cos^2 x=(1+\cos 2x)/2\).
For cosine squared, the corresponding calculation is:
The two values agree. Differentiating the first antiderivative gives \(1/2-\cos(2x)/2=\sin^2 x\), and differentiating the second gives \(1/2+\cos(2x)/2=\cos^2 x\). Thus both evaluations have the required integrands.
Worked Example: Unequal Positive Exponents
Worked Example: Unequal Positive Exponents
Now compare \(\int_0^{\pi/2}\sin^3 x\cos x\,dx\) and \(\int_0^{\pi/2}\sin x\cos^3 x\,dx\). Here both exponents are positive, and the theorem predicts equal values.
For the first integral, let \(v=\sin x\), so \(dv=\cos x\,dx\). The endpoint values are \(v=0\) when \(x=0\) and \(v=1\) when \(x=\pi/2\):
For the interchanged exponents, let \(w=\cos x\), so \(dw=-\sin x\,dx\). The endpoint values are \(w=1\) at \(x=0\) and \(w=0\) at \(x=\pi/2\):
The integrals are equal and each is \(1/4\). The antiderivative checks are \(\frac{d}{dx}(\sin^4 x/4)=\sin^3 x\cos x\) and \(\frac{d}{dx}(-\cos^4 x/4)=\sin x\cos^3 x\). In the second computation, the reversed substitution limits and the negative differential combine to give a positive value.
Common Pitfall: Losing a Minus Sign
A frequent error is to use \(x=\pi/2-u\) but keep the original order of the limits while also keeping the negative differential. That would incorrectly turn the transformed integral into the negative of the desired integral.
For instance, starting with \(\int_0^{\pi/2}\sin x\,dx\), the correct substitution gives:
The middle equality follows because \(-du\) together with bounds from \(\pi/2\) to \(0\) is equivalent to \(du\) with bounds from \(0\) to \(\pi/2\). A wrong calculation would keep the minus sign but not reverse the bounds, writing \(-\int_0^{\pi/2}\cos u\,du=-1\). That contradicts the direct evaluation \([-\cos x]_0^{\pi/2}=1\). Always transform both the differential and the limits.
When the Symmetry Is Useful
The equality lets us choose whichever ordering of the exponents is more convenient to evaluate. If one integral has a factor that makes a substitution immediate, the interchanged integral has the same value, even if its appearance suggests a different calculation. It also gives a consistency check: two exact evaluations with swapped exponents must agree when the limits are \(0\) and \(\pi/2\).
There are important limits to this conclusion. The reflection argument proves the equality for this quarter-period interval; it does not assert that the integrals over other intervals are equal. Nor does it say that \(\sin^m x\cos^n x=\sin^n x\cos^m x\) pointwise. The equality concerns accumulated area over the entire interval, established by a change of variable.
Check Your Understanding
Use the reflection argument and the computations in this tutorial to answer the following questions.
- Under \(x=\pi/2-u\), what do \(\sin x\), \(\cos x\), and \(dx\) become?
- What are the transformed limits when \(x\) runs from \(0\) to \(\pi/2\), and how does the minus sign affect the result?
- Use the theorem to state an integral equal to \(\int_0^{\pi/2}\sin^4 x\cos x\,dx\).
- Why does the theorem not imply that \(\sin^2 x\) and \(\cos^2 x\) are equal at every \(x\) in the interval?
- Identify the sign error in the claim \(\int_0^{\pi/2}\sin x\,dx=-\int_0^{\pi/2}\cos u\,du\).
