Why Mixed Powers Need Two Recurrences
A mixed trigonometric power integral has both a sine factor and a cosine factor raised to powers. Integration by parts can reduce either exponent, but the resulting formulas have different boundary terms. Keeping those terms and signs straight is essential: omitting one can produce an expression whose derivative is not the original integrand.
In this tutorial the exponents \(m\) and \(n\) are nonnegative integers, and \(x\) is the variable of integration. We will derive one formula that lowers \(m\), the sine exponent, and another that lowers \(n\), the cosine exponent. Each recurrence lowers its chosen exponent by two while leaving the other exponent unchanged.
The formulas below apply when the exponent being reduced is at least two. Their denominators involve \(m+n\), which is positive under those conditions. The remaining integral after a reduction has a smaller exponent, so repeated use moves toward a simpler case.
Reducing the Sine Exponent
To lower the sine exponent, take one factor of \(\sin x\) together with \(\cos^n x\) as the part to integrate, and leave \(\sin^{m-1}x\) as the part to differentiate. The resulting integration-by-parts expression contains a factor of \(\cos^{n+2}x\). Rewriting \(\cos^2 x\) as \(1-\sin^2 x\) then lets us isolate the original integral.
Here is a derivation that also explains the coefficient and the sign of the boundary term. Integration by parts first gives an identity with \(I_{m,n}\) on both sides. Rearranging that identity yields the displayed recurrence.
Set \(u=\sin^{m-1}x\) and \(dv=\sin x\cos^n x\,dx\). Then \(du=(m-1)\sin^{m-2}x\cos x\,dx\) and \(v=-\cos^{n+1}x/(n+1)\).
Since \(\cos x\,dx\) integrates to \(-\cos^{n+1}x/(n+1)\) in the chosen \(dv\), the result is \(I_{m,n}=-\sin^{m-1}x\cos^{n+1}x/(n+1)+(m-1)J/(n+1)\), where \(J=\int\sin^{m-2}x\cos^{n+2}x\,dx\).
Using \(\cos^2 x=1-\sin^2 x\), we have \(J=I_{m-2,n}-I_{m,n}\).
Substitute for \(J\), multiply by \(n+1\), and collect the \(I_{m,n}\) terms. This gives \((m+n)I_{m,n}=-\sin^{m-1}x\cos^{n+1}x+(m-1)I_{m-2,n}\). Dividing by \(m+n\) proves the recurrence.
Reducing the Cosine Exponent
The other recurrence follows by a parallel choice of parts: retain \(\cos^{n-1}x\), and integrate \(\sin^m x\cos x\). In the integration-by-parts result, rewrite \(\sin^2x\) as \(1-\cos^2x\). This time the isolated integral has cosine exponent \(n-2\).
For the derivation, let \(u=\cos^{n-1}x\) and \(dv=\sin^m x\cos x\,dx\). Then \(du=-(n-1)\sin x\cos^{n-2}x\,dx\) and \(v=\sin^{m+1}x/(m+1)\). Integration by parts gives \(I_{m,n}=\sin^{m+1}x\cos^{n-1}x/(m+1)+(n-1)K/(m+1)\), where \(K=\int\sin^{m+2}x\cos^{n-2}x\,dx\). From \(\sin^2x=1-\cos^2x\), it follows that \(K=I_{m,n-2}-I_{m,n}\). Substitution and rearrangement give \((m+n)I_{m,n}=\sin^{m+1}x\cos^{n-1}x+(n-1)I_{m,n-2}\), which proves the formula.
Both derivations use the same underlying idea: integration by parts creates a nearby power integral, and a Pythagorean identity expresses that integral in terms of the original one and a lower-power integral. The boundary term is part of the recurrence, not an optional extra.
Worked Examples
Worked Example: Reduce \(I_{3,2}\)
Since the sine exponent is \(3\), use the sine-reduction recurrence with \(m=3\), \(n=2\). Substitution gives
For the remaining integral, let \(u=\cos x\), so \(du=-\sin x\,dx\). Then \[ I_{1,2}=\int\sin x\cos^2x\,dx =-\int u^2\,du =-\frac{u^3}{3} =-\frac{\cos^3x}{3}. \] Substituting this value gives \[ I_{3,2}=-\frac{\sin^2x\cos^3x}{5}-\frac{2\cos^3x}{15}+C. \]
To check, differentiate the result: \[ \frac{d}{dx}\left(-\frac{\sin^2x\cos^3x}{5}-\frac{2\cos^3x}{15}\right) =-\frac{2\sin x\cos^4x-3\sin^3x\cos^2x}{5} +\frac{2\sin x\cos^2x}{5}. \] Factoring \(\sin x\cos^2x\) and using \(1-\cos^2x=\sin^2x\) reduces this to \(\sin^3x\cos^2x\), as required.
Worked Example: Reduce \(I_{2,3}\)
Here the cosine exponent is \(3\), so use the cosine-reduction recurrence with \(m=2\), \(n=3\): \[ I_{2,3}=\frac{\sin^3x\cos^2x}{5}+\frac{2}{5}I_{2,1}. \] The remaining integral is found by substituting \(u=\sin x\), \(du=\cos x\,dx\): \[ I_{2,1}=\int\sin^2x\cos x\,dx=\int u^2\,du=\frac{\sin^3x}{3}. \] Therefore \[ I_{2,3}=\frac{\sin^3x\cos^2x}{5}+\frac{2\sin^3x}{15}+C. \]
Differentiate to check: \[ \frac{d}{dx}\left(\frac{\sin^3x\cos^2x}{5}+\frac{2\sin^3x}{15}\right) =\frac{3\sin^2x\cos^3x-2\sin^4x\cos x}{5} +\frac{2\sin^2x\cos x}{5}. \] Factoring \(\sin^2x\cos x\) leaves the bracket \((3\cos^2x-2\sin^2x+2)/5=\cos^2x\). Thus the derivative is \(\sin^2x\cos^3x\), the original integrand.
Worked Example: Apply the Sine Recurrence Twice
Consider \(I_{4,1}\). Applying the sine recurrence with \(m=4\), \(n=1\) gives \[ I_{4,1}=-\frac{\sin^3x\cos^2x}{5}+\frac{3}{5}I_{2,1}. \] Apply it again to \(I_{2,1}\): \[ I_{2,1}=-\frac{\sin x\cos^2x}{3}+\frac{1}{3}I_{0,1}. \] Since \(I_{0,1}=\int\cos x\,dx=\sin x\), substitution into the first equation yields \[ I_{4,1}=-\frac{\sin^3x\cos^2x}{5} -\frac{\sin x\cos^2x}{5} +\frac{\sin x}{5}+C. \]
For a direct check, differentiate. The derivative is \[ -\frac{3\sin^2x\cos^3x-2\sin^4x\cos x}{5} -\frac{\cos^3x-2\sin^2x\cos x}{5} +\frac{\cos x}{5}. \] After factoring \(\cos x/5\), the bracket becomes \(-3\sin^2x\cos^2x+2\sin^4x-\cos^2x+2\sin^2x+1\). Using \(\cos^2x=1-\sin^2x\), this bracket simplifies to \(5\sin^4x\). The derivative is therefore \(\sin^4x\cos x\), as required.
Parity and Choosing an Exponent
Each recurrence subtracts two from the exponent it targets. Consequently, that exponent's parity is preserved: an even exponent descends through even values to zero, while an odd exponent descends through odd values to one. The other exponent stays fixed in the remaining integral. This tells you what kind of terminal case to expect and helps you choose which recurrence to apply.
If one exponent is odd, reducing that exponent reaches the value one. At that stage a single sine or cosine factor may make substitution convenient. If both exponents are even, either recurrence preserves even parity in its chosen exponent; you can choose the one whose successive boundary terms and remaining integral are more useful for the calculation. This is a practical choice, not a claim that the formulas are different in validity: whenever its stated condition holds, either recurrence is correct.
A recurrence does not by itself evaluate every terminal integral. It organizes the reduction and exposes a simpler integral, which can then be handled with an appropriate basic method or identity. In particular, do not assume that reducing one exponent has reduced the other as well.
Why the Recurrences Are Useful
These formulas turn an integral with large exponents into an exact relation involving a lower-power integral and an explicit trigonometric term. That structure is useful both for evaluating particular integrals and for organizing families of related calculations. It also gives a built-in check: the recurrence can be differentiated, and the resulting terms must combine to the original mixed power.
Before applying a recurrence, check three details: the exponent being reduced is at least two; the denominator \(m+n\) is nonzero; and the boundary term has the correct sign and powers. Then track parity as the chosen exponent decreases by two. These checks prevent the most common errors and make the reduction predictable.
Check Your Understanding
Use the two reduction formulas and the conditions stated with them to answer the following questions.
- Write the sine-exponent recurrence for \(I_{5,2}\), including its boundary term and coefficient.
- Write the cosine-exponent recurrence for \(I_{2,4}\).
- When the sine exponent begins at seven and is reduced repeatedly by the sine recurrence, what is the final sine exponent in the remaining integral?
- Why does the sine-reduction recurrence have a negative boundary term?
- What condition must the exponent being reduced satisfy for the corresponding recurrence to apply?
