Why Endpoint Values Matter
A reduction formula turns a high power of secant into a lower power, but a definite integral has an extra advantage: every reduction produces an endpoint term that can be evaluated immediately. Repeating the formula eventually reduces the problem to a simple base integral.
The recurrence is useful only when the integrand is defined throughout the interval. Since \(\sec x=1/\cos x\), any point where \(\cos x=0\) is a singularity. Before applying the formula, check that the interval does not include one of those points.
The Definite Secant-Power Reduction Formula
For \(n\geq 2\), the secant-power reduction formula expresses \(I_n(a,b)\) using \(I_{n-2}(a,b)\). The bracket notation means to evaluate the expression at the upper endpoint and subtract its value at the lower endpoint.
The formula lowers the exponent by two each time. Thus an even power eventually reaches \(I_0\), while an odd power eventually reaches \(I_1\). The endpoint term must be calculated as a signed difference; it is not the sum of the endpoint values, nor should either endpoint be discarded unless its value is actually zero.
Why the Recurrence Works
A short integration-by-parts argument gives the formula. Write \(\sec^n x=\sec^{n-2}x\sec^2x\), and use \(u=\sec^{n-2}x\) and \(dv=\sec^2x\,dx\). Then \(du=(n-2)\sec^{n-2}x\tan x\,dx\) and \(v=\tan x\). These functions are differentiable on the interval because the interval contains no point where cosine vanishes.
For the definite integral this gives \(I_n=[\sec^{n-2}x\tan x]_a^b-(n-2)\int_a^b\sec^{n-2}x\tan^2x\,dx\).
Since \(\tan^2x=\sec^2x-1\), the remaining integral is \(I_n-I_{n-2}\).
Substitution gives \(I_n=[\sec^{n-2}x\tan x]_a^b-(n-2)(I_n-I_{n-2})\). Hence \((n-1)I_n=[\sec^{n-2}x\tan x]_a^b+(n-2)I_{n-2}\), which gives the stated recurrence after division by \(n-1\).
The base case \(I_0=b-a\) follows by integrating \(1\). For \(I_1\), differentiate the proposed antiderivative on an interval where \(\cos x\neq0\): \(\frac{d}{dx}\ln|\sec x+\tan x|=\sec x\). Indeed, the derivative of \(\sec x+\tan x\) is \(\sec x\tan x+\sec^2x=\sec x(\tan x+\sec x)\), so division by \(\sec x+\tan x\) leaves \(\sec x\). This verifies the base case used for odd powers.
Worked Examples
Worked Example: A Definite Integral of \(\sec^2 x\)
Evaluate \(I_2(0,\pi/4)=\int_0^{\pi/4}\sec^2x\,dx\). The interval contains no zero of cosine, so the ordinary definite integral and recurrence apply. With \(n=2\), the lower-power coefficient is zero and \(\sec^{n-2}x=1\):
Check: an antiderivative is \(\tan x\), and its derivative is \(\sec^2x\). Evaluating it at the two endpoints gives \(\tan(\pi/4)-\tan(0)=1\), as required.
Worked Example: Reducing an Even Power
Evaluate \(\int_0^{\pi/4}\sec^4x\,dx\). The cosine is positive on this interval. The recurrence with \(n=4\) gives
At \(x=\pi/4\), \(\sec^2(\pi/4)=2\) and \(\tan(\pi/4)=1\), so the upper endpoint term is \(2/3\). At \(x=0\), \(\sec^2(0)\tan(0)=1\cdot0=0\). From the first example, \(I_2(0,\pi/4)=1\). Therefore
Check by differentiating the corresponding antiderivative \(\frac{1}{3}\sec^2x\tan x+\frac{2}{3}\tan x\). Its derivative is \(\frac{1}{3}(2\sec^2x\tan^2x+\sec^4x)+\frac{2}{3}\sec^2x\). Using \(\tan^2x=\sec^2x-1\), this becomes \(\frac{1}{3}(2\sec^4x-2\sec^2x+\sec^4x+2\sec^2x)=\sec^4x\). The antiderivative is valid, and its endpoint difference is \(4/3\).
Worked Example: Reducing an Odd Power
Evaluate \(\int_{-\pi/4}^{\pi/4}\sec^5x\,dx\). Cosine is nonzero throughout the interval. Apply the recurrence once with \(n=5\):
At the upper endpoint, \(\sec^3(\pi/4)\tan(\pi/4)=2\sqrt{2}\), so the upper boundary value is \(\sqrt{2}/2\). At the lower endpoint, \(\sec^3(-\pi/4)\tan(-\pi/4)=-2\sqrt{2}\), so the lower boundary value is \(-\sqrt{2}/2\). Their difference is \(\sqrt{2}\).
Now reduce \(I_3\):
The boundary values of \(\sec x\tan x/2\) are \(\sqrt{2}/2\) and \(-\sqrt{2}/2\), so their difference is \(\sqrt{2}\). For the remaining base integral,
because \((\sqrt{2}+1)(\sqrt{2}-1)=1\). Thus \(I_3=\sqrt{2}+\ln(1+\sqrt{2})\). Substituting this into the first reduction gives
Check: the recurrence is derived from integration by parts on this interval, and the base logarithmic expression differentiates to \(\sec x\). The two recurrence steps therefore give an antiderivative of \(\sec^5x\); its endpoint differences are exactly the boundary values and base integral computed above.
Endpoint Order and Singularities
The recurrence preserves the orientation of the original integral. In particular, \([F(x)]_a^b\) always means \(F(b)-F(a)\). If the limits are reversed, both the definite integral and each endpoint difference change sign. A calculation that uses \(F(a)-F(b)\) while keeping the original order of the integral changes the answer.
For example, \(\int_0^\pi\sec^2x\,dx\) cannot be evaluated by the ordinary endpoint formula because \(x=\pi/2\) lies inside the interval. It is not legitimate to substitute \(0\) and \(\pi\) into a formula that assumes continuity throughout \([0,\pi]\). The singularity must be examined using limits, and in this case the integral diverges.
A Common Endpoint Error
Consider again the boundary term for \(I_4(0,\pi/4)\). A common mistake is to subtract in the wrong order and write
This is the lower endpoint minus the upper endpoint. The correct bracket is upper minus lower:
Using the incorrect value would give \(-2/3+(2/3)(1)=0\), contradicting the positive integrand \(\sec^4x\) on an interval of positive length. This sign check is a useful way to catch endpoint-order errors, though the defining calculation remains upper value minus lower value.
Using the Recurrence Efficiently
For a definite integral, keep track of the endpoint contribution at each reduction and reduce the exponent by two until reaching its base case. This avoids unnecessary algebra and makes it easier to locate a sign or arithmetic error.
- For an even exponent, continue to \(I_0(a,b)=b-a\).
- For an odd exponent, continue to \(I_1(a,b)=[\ln|\sec x+\tan x|]_a^b\).
- Evaluate each bracket as upper endpoint minus lower endpoint.
- Check that no point \(x=\pi/2+k\pi\) lies in the interval before using the ordinary recurrence.
Definite reduction formulas are valuable because they combine a general method with direct endpoint evaluation. Each step replaces a difficult power by a simpler one, while the assumptions and endpoint signs remain visible throughout the calculation.
Check Your Understanding
Use the recurrence and the interval conditions to answer these questions.
- State the reduction formula for \(I_n(a,b)\), including the condition on \(\cos x\).
- For an even power, which base integral does repeated reduction reach? What is its value?
- For an odd power, which base integral does repeated reduction reach, and what endpoint expression evaluates it?
- In \([F(x)]_a^b\), which endpoint value is subtracted from which?
- Why can the ordinary recurrence not be applied across an interval containing \(x=\pi/2\)?
