Why Odd Secant Powers Reduce by Two
Integrals of higher powers of secant can look increasingly complicated, but the reduction formula turns each one into a simpler integral whose exponent is two less. For an odd power, repeating that step eventually reaches the basic integral of secant. That base integral introduces a logarithm, while the preceding reduction steps contribute products of secant and tangent.
The key is to keep track of the coefficients at every step. A small coefficient error in one reduction changes the whole antiderivative, so each result below is checked by differentiation.
The Recurrence and Its Base Case
For integers \(n\ge 3\), the secant-power reduction formula is
The restriction \(n\ge 3\) matters: it ensures that \(n-2\) is a positive integer and that the lower-power integral in the formula is defined by our notation. The odd-power calculations use \(n=3,5,7,\ldots\), so each reduction remains within the odd powers and ultimately reaches \(I_1\).
The base case is
To check the base case, differentiate the logarithm on any interval where \(\cos x\ne 0\):
The logarithm is defined here because \(\sec x+\tan x\ne 0\) wherever \(\cos x\ne 0\). Indeed, \(\sec x+\tan x=(1+\sin x)/\cos x\), and \(1+\sin x=0\) only when \(\sin x=-1\), where \(\cos x=0\).
Why the Recurrence Works
The reduction follows from integration by parts and the identity \(\tan^2 x=\sec^2 x-1\). For \(n\ge 3\), write \(\sec^n x=\sec^{n-2}x\sec^2 x\). Choose \(u=\sec^{n-2}x\) and \(dv=\sec^2 x\,dx\), so that \(du=(n-2)\sec^{n-2}x\tan x\,dx\) and \(v=\tan x\). Integration by parts gives
Collect the \(I_n\) terms on the left:
Dividing by \(n-1\), which is nonzero for \(n\ge 3\), yields the stated recurrence. This derivation also explains the reduction: the tangent-squared factor becomes a difference involving the original integral and the integral two powers lower.
Worked Example: Integrating \(\sec^3 x\)
Worked Example: Integrating \(\sec^3 x\)
Apply the recurrence with \(n=3\). The lower integral is \(I_1\), the base case:
Check by differentiating. Since \(\frac{d}{dx}(\sec x\tan x)=\sec x\tan^2x+\sec^3x=2\sec^3x-\sec x\), the derivative is
Thus the displayed expression is an antiderivative of \(\sec^3x\) on any interval where \(\cos x\ne 0\).
Worked Example: Integrating \(\sec^5 x\)
Worked Example: Integrating \(\sec^5 x\)
First reduce \(I_5\) to \(I_3\), then substitute the result for \(I_3\):
For the check, the product rule and \(\tan^2x=\sec^2x-1\) give \(\frac{d}{dx}(\sec^3x\tan x)=4\sec^5x-3\sec^3x\) and \(\frac{d}{dx}(\sec x\tan x)=2\sec^3x-\sec x\). Therefore the derivative of the proposed antiderivative is
The lower-power terms cancel, leaving exactly the integrand.
Worked Example: Integrating \(\sec^7 x\)
Worked Example: Integrating \(\sec^7 x\)
Reduce \(I_7\) to \(I_5\), and then substitute the complete expression for \(I_5\):
To check, differentiate the products using \(\frac{d}{dx}(\sec^p x\tan x)=(p+1)\sec^{p+2}x-p\sec^p x\), which follows from the product rule and \(\tan^2x=\sec^2x-1\). The derivative is
Again, the terms introduced by differentiating the products cancel in pairs.
Continuing to Higher Odd Powers
For an odd exponent \(n=2m+1\), where \(m\) is a positive integer, each recurrence step reduces the exponent by two:
At each step, retain both parts of the recurrence: the product term involving \(\sec^{n-2}x\tan x\), and the coefficient multiplying the lower-power integral. Continue until every integral has been replaced by an explicit term, including the final logarithm from \(I_1\). This process gives a finite sum of secant-tangent terms and one logarithmic term.
The exponent \(n=2\) is not part of this odd-power chain. It can be integrated directly as \(\int\sec^2x\,dx=\tan x+C\). The odd-power recurrence above is stated for \(n\ge 3\), where its lower-power integral is well-defined.
A Common Pitfall: Using the Recurrence Outside Its Range
If the displayed recurrence is mistakenly applied with \(n=2\), it produces
The notation \(I_n\) here is defined only for positive integers, so \(I_0\) lies outside it, and the recurrence is proved above only for \(n\ge 3\). (If one extends the notation by \(I_0=\int dx=x+C\), the formula does happen to give the correct value \(I_2=\tan x+C\), but that is a separate check, not a use of the stated result.) The safe course is to use the recurrence only for integers \(n\ge 3\), as stated. For the separate even-power case \(n=2\), directly differentiate \(\tan x\):
Another frequent error is to reduce the power correctly but drop a coefficient. For example, the recurrence gives \(I_5=\frac14\sec^3x\tan x+\frac34 I_3\), not \(\frac14\sec^3x\tan x+I_3\). The coefficient \(\frac34\) must multiply every term of the \(I_3\) antiderivative, as the worked calculation shows.
Using the Method Reliably
The recurrence is useful because it replaces a higher odd power with a lower one while preserving an exact equality of antiderivatives. It also provides a predictable route to the answer: reduce by two repeatedly, use the secant base case, and then check by differentiation. Keeping the domain in view is important: these antiderivatives apply on intervals that do not contain a point where \(\cos x=0\).
Use the recurrence only when the exponent is an integer at least three.
Write the product term and the full coefficient of the lower-power integral.
Substitute each lower-power formula until the logarithmic base case appears.
Differentiate the result and simplify until the original secant power remains.
Check Your Understanding
Use the recurrence and the base case to answer these questions. Check any antiderivative you find by differentiation.
- What condition on \(n\) makes the stated recurrence valid with the definition of \(I_n\) used here?
- Write the first recurrence step for \(I_9\), including both coefficients.
- Which base integral is reached by repeatedly reducing \(I_7\), and what is its antiderivative?
- What is the coefficient of \(\ln|\sec x+\tan x|\) in the formula for \(I_5\)?
- Why is it invalid to apply this recurrence at \(n=2\) as written?
