Why Reserve a Secant-Squared Factor?
Integrals containing powers of secant can often be made manageable by recognizing a derivative inside the integrand. The derivative of \(\tan x\) is \(\sec^2 x\). Thus, when an even power of secant is present, one factor of \(\sec^2 x\) can serve as the differential for the substitution \(u=\tan x\).
The remaining secant factors can then be rewritten in terms of \(\tan x\), using \(\sec^2 x=1+\tan^2 x\). The result is a polynomial integral in \(u\), rather than an integral involving powers of secant.
Conditions and the Basic Substitution
Throughout this tutorial, \(k\) is a positive integer. The substitution \(u=\tan x\) is used on an interval on which \(\cos x\ne 0\). On such an interval, \(\tan x\) and \(\sec x\) are defined, and
The identity needed to rewrite the other secant factors follows from \(\sin^2 x+\cos^2 x=1\): dividing by \(\cos^2 x\) gives \(\tan^2 x+1=\sec^2 x\).
Separate one factor of \(\sec^2 x\), rewrite the remaining factors, and substitute:
Since \(k-1\) is a nonnegative integer, the binomial theorem expands \((1+u^2)^{k-1}\) as a finite polynomial. Each term can be integrated using the power rule. The result is an antiderivative expressed as a polynomial in \(\tan x\), plus a constant.
A General Formula and Why It Works
The binomial theorem states that \((1+z)^n=\sum_{j=0}^{n}\binom{n}{j}z^j\) for a nonnegative integer \(n\). Apply it with \(z=u^2\) and \(n=k-1\), then integrate each power of \(u\).
To see why the formula gives an antiderivative, differentiate the sum. For each index \(j\), the derivative of \(\tan^{2j+1}x/(2j+1)\) is \(\tan^{2j}x\,\sec^2 x\). Therefore,
The first equality uses the chain rule, and the second uses the binomial theorem. The last two equalities use \(1+\tan^2 x=\sec^2 x\). This verifies the formula directly and shows that the substitution has not changed the target integrand.
A Reliable Calculation Process
Write \(\sec^{2k}x=(\sec^2 x)^{k-1}\sec^2 x\), leaving \(\sec^2 x\,dx\) intact.
Replace \((\sec^2 x)^{k-1}\) by \((1+\tan^2 x)^{k-1}\).
Set \(u=\tan x\), so that \(du=\sec^2 x\,dx\), expand the polynomial, and integrate each term.
Replace \(u\) by \(\tan x\), include \(C\), and differentiate to confirm that the result is \(\sec^{2k}x\).
Worked Examples
Worked Example: Integrating \(\sec^2 x\)
Here \(k=1\). There are no remaining secant factors after reserving \(\sec^2 x\,dx\). Set \(u=\tan x\), so \(du=\sec^2 x\,dx\). Then
Check by differentiation:
The derivative matches the integrand, so the antiderivative is correct on any interval where \(\cos x\ne 0\).
Worked Example: Integrating \(\sec^4 x\)
Factor out the secant-squared differential, then use the identity before substituting:
With \(u=\tan x\) and \(du=\sec^2 x\,dx\), this becomes
Differentiate the answer to check it:
This equals the original integrand, as required.
Worked Example: Integrating \(\sec^6 x\)
There are three factors of \(\sec^2 x\). Reserve one for \(du\), and rewrite the other two together:
Set \(u=\tan x\). Expanding before integrating gives
For the check, differentiate term by term and factor out \(\sec^2 x\):
The derivative is the given integrand, confirming the result.
A Common Mistake: Losing Track of the Differential
The substitution does not mean that every \(\sec^2 x\) in the integrand can be replaced by \(du\). Only the factor paired with \(dx\) becomes \(du\). The other factors must first be rewritten in terms of \(u\).
For instance, an incorrect calculation for \(\int \sec^4 x\,dx\) might replace every \(\sec^2 x\) by \(1+u^2\), and then write
This has treated the entire integrand as if it were multiplied by \(du\), without accounting for the original \(dx\). It produces \(u+\frac{2u^3}{3}+\frac{u^5}{5}+C\). Differentiating that expression with \(u=\tan x\) gives \(\sec^2 x(1+\tan^2 x)^2=\sec^6 x\), not \(\sec^4 x\).
The correction is to keep one factor of \(\sec^2 x\,dx\) as the differential and rewrite only the other factor: \(\sec^4 x\,dx=(1+\tan^2 x)\sec^2 x\,dx\). Then the integral becomes \(\int(1+u^2)\,du\), giving \(\tan x+\frac{\tan^3 x}{3}+C\), whose derivative is \(\sec^4 x\).
When the Method Applies
The method works directly for \(\sec^{2k}x\) because removing one factor of \(\sec^2 x\) leaves an integer power of \(\sec^2 x\). That remaining power can be rewritten entirely as a polynomial in \(\tan x\). The substitution then has exactly the differential it needs.
The restriction on the interval matters: \(\tan x\) is undefined where \(\cos x=0\). The antiderivative formula is valid separately on each interval that contains no such point. An arbitrary constant may be chosen on each interval. Also, the binomial expansion is finite because \(k\) is an integer; the polynomial integration described here is specifically for these even powers.
Check Your Understanding
For each question, show how the reserved \(\sec^2 x\,dx\) factor becomes \(du\). Check any proposed antiderivative by differentiation.
- Rewrite \(\int \sec^8 x\,dx\) as an integral in \(u=\tan x\), before expanding.
- Evaluate \(\int \sec^4 x\,dx\) by tangent substitution and verify the result.
- Expand \((1+u^2)^3\) and use it to find \(\int \sec^8 x\,dx\).
- Why must the calculation be restricted to an interval on which \(\cos x\ne 0\)?
- A student replaces all four factors in \(\sec^8 x\) with \(1+u^2\), then appends \(du\). Explain the error and state which factor should supply \(du\).
