Source: University of Texas at Austin, Calc 2 lecture notes
Tags: secant, tangent, sec tan integrals, trig integrals, u-substitution, Calc 2, integral calculus
Difficulty: Intermediate Prerequisites: Part 1 (Trig Identities for Integration), u-substitution, familiarity with derivatives of sec x and tan x.
After learning to integrate powers of sine and cosine, the next family of trig integrals involves products of secant and tangent: ∫ secⁿx tanᵐx dx. These come up less intuitively than sine/cosine integrals, but they follow a clear decision tree once you know the four cases. This topic builds directly on the Pythagorean identity sec²x = 1 + tan²x and on your ability to recognise which substitution (u = tan x or u = sec x) each situation calls for. If you are comfortable with Part 1, you are ready.
Integrals of secⁿx tanᵐx dx break into four cases depending on the exponents. Two key substitutions drive everything: u = tan x (when you can extract sec²x dx as du) and u = sec x (when you can extract sec x tan x dx as du). Whichever factors you cannot use for du, convert using sec²x = 1 + tan²x.
sec²x dx as du
When u = tan x, the derivative is du = sec²x dx. Any integral where you can isolate a sec²x factor paired with dx is a candidate for this substitution. Think of it as: if you see sec²x sitting next to dx, let u = tan x.
sec x tan x dx as du
When u = sec x, the derivative is du = sec x tan x dx. Any integral where you can isolate one sec x and one tan x paired with dx is a candidate. Think of it as: if you can peel off a sec x tan x pair, let u = sec x.
Converting between sec²x and tan²x
The identity sec²x = 1 + tan²x (equivalently, tan²x = sec²x – 1) lets you rewrite leftover factors after choosing your substitution. In simple terms, whatever you cannot absorb into du, you rewrite using this identity so everything is in terms of u.
The strategy depends on the exponents n (on secant) and m (on tangent). Work through these cases in order until one matches.
Case 1: n = 2 (you have exactly sec²x)
Set u = tan x, du = sec²x dx. The integral becomes ∫ uᵐ du, a straightforward power rule.
This is the simplest case. If you spot sec²x dx, grab it.
Case 2: n is even (n ≥ 2)
Peel off a sec²x to pair with dx (for du = sec²x dx with u = tan x). Convert the remaining secant factors using sec²x = 1 + tan²x. Everything is now in terms of tan x = u.
Case 3: m is odd (and m ≥ 1)
Peel off one sec x tan x to pair with dx (for du = sec x tan x dx with u = sec x). Convert the remaining tangent factors using tan²x = sec²x – 1. Everything is now in terms of sec x = u.
Case 4: m is odd and can be combined with Case 2
When both n is even and m is odd, either substitution works. Pick whichever looks cleaner.
Decision summary:
See sec²x dx available → u = tan x
See sec x tan x dx available → u = sec x
Leftover sec² or tan² factors → convert with sec²x = 1 + tan²x
Example 1: ∫ sec⁶x tan⁵x dx
The tangent power (5) is odd, so use Case 3: peel off sec x tan x for du.
Rewrite: ∫ sec⁴x · tan⁵x · sec²x dx → but it is easier to think of it as peeling sec x tan x:
∫ sec⁵x · tan⁴x · sec x tan x dx
Set u = sec x, du = sec x tan x dx. Convert tan⁴x = (sec²x – 1)² = (u² – 1)².
The integral becomes ∫ u⁵ (u² – 1)² du. Expand and integrate term by term.
Example 2: ∫ tan³x dx
Rewrite tan³x = tan x · tan²x = tan x · (sec²x – 1).
Split: ∫ tan x sec²x dx – ∫ tan x dx.
For the first part, u = tan x, du = sec²x dx → ∫ u du = u²/2 = tan²x / 2.
For the second part, ∫ tan x dx = –ln|cos x| = ln|sec x|.
Result: tan²x / 2 + ln|sec x| + C.
Example 3: ∫ sec x tan x dx
Convert to sine and cosine: (1/cos x) · (sin x / cos x) = sin x / cos²x.
Set u = cos x, du = –sin x dx.
∫ –1/u² du = 1/u = 1/cos x = sec x + C.
Example 4: ∫ sec x tan³x dx
Use tan²x = sec²x – 1:
∫ sec x · (sec²x – 1) · tan x dx = ∫ (sec²x – 1) · sec x tan x dx.
Set u = sec x, du = sec x tan x dx:
∫ (u² – 1) du = u³/3 – u + C = sec³x / 3 – sec x + C.
What you need | Formula | Purpose |
|---|---|---|
Derivative of tan x | d/dx tan x = sec²x | Tells you du when u = tan x |
Derivative of sec x | d/dx sec x = sec x tan x | Tells you du when u = sec x |
Pythagorean (sec/tan) | sec²x = 1 + tan²x | Convert leftover factors |
Rearranged | tan²x = sec²x – 1 | Convert tangent powers to secant |
∫ tan x dx | ln | sec x |
Integrals of secant and tangent appear in arc-length calculations for curves, particularly when computing the length of a curve y = ln(cos x) or similar logarithmic-trig functions. They also arise in physics when calculating work done by a variable force along a curved path, and in engineering when analysing the bending moment in certain beam configurations.
Students often confuse which substitution to use. Remember: sec²x dx pairs with u = tan x, and sec x tan x dx pairs with u = sec x. Mixing these up produces integrals that do not simplify.
When peeling off sec x tan x, some students accidentally peel off just tan x. You need both sec x and tan x together for du = sec x tan x dx.
Students sometimes write ∫ tan x dx = tan²x / 2. That is wrong. ∫ tan x dx = ln|sec x| + C. The power rule does not apply to tan x on its own.
Forgetting to convert all remaining factors into one variable after substitution. If u = sec x, every tan² must become sec² – 1 = u² – 1. Leaving a mix of sec and tan after substituting defeats the purpose.
⚠️ Exam problems typically give you a specific ∫ secⁿx tanᵐx dx and expect you to identify the correct case before computing. Showing you know the decision tree is half the marks.
⚠️ ∫ tan³x dx is a classic exam question because it requires splitting into two separate integrals with different techniques.
⚠️ Professors often test ∫ sec x tan x dx as a warm-up. The answer is simply sec x + C, but students who overthink it waste time.
⚠️ Harder problems combine high powers (e.g. ∫ sec⁶x tan⁵x dx) to test whether you can set up the substitution and expand correctly. Practice the algebra.
True or false: To integrate ∫ sec⁴x tan³x dx, you should set u = tan x. (False. The tangent power is odd, so u = sec x is the better choice, peeling off sec x tan x.)
Fill in the blank: d/dx sec x = ______. (sec x tan x)
True or false: tan²x = sec²x + 1. (False. tan²x = sec²x – 1.)
Fill in the blank: ∫ tan x dx = ______. (ln|sec x| + C)
True or false: If n is even in ∫ secⁿx tanᵐx dx, you can always set u = tan x. (True, because you can extract sec²x dx as du and convert the remaining even secant factors.)
Q: Evaluate ∫ tan³x dx step by step.
A: Rewrite tan³x = tan x (sec²x – 1). Split into ∫ tan x sec²x dx – ∫ tan x dx. First integral: u = tan x, du = sec²x dx → tan²x / 2. Second integral: ln|sec x|. Answer: tan²x / 2 + ln|sec x| + C.
Q: Evaluate ∫ sec x tan x dx.
A: Recognise this as the derivative of sec x. Answer: sec x + C. Alternatively, convert to sin/cos: ∫ sin x / cos²x dx, set u = cos x, and get 1/cos x = sec x + C.
Q: Evaluate ∫ sec x tan³x dx.
A: Use tan²x = sec²x – 1. The integral becomes ∫ (sec²x – 1) sec x tan x dx. Set u = sec x, du = sec x tan x dx. Integrate ∫ (u² – 1) du = u³/3 – u + C = sec³x / 3 – sec x + C.
Q: For ∫ sec⁶x tan⁵x dx, which substitution do you use and why?
A: The tangent power (5) is odd, so peel off sec x tan x dx and set u = sec x. Convert the remaining tan⁴x = (sec²x – 1)² = (u² – 1)². The integral becomes ∫ u⁵(u² – 1)² du.
Q: A student writes ∫ tan x dx = tan²x / 2 + C. What went wrong?
A: They incorrectly applied the power rule as if tan x were xⁿ. The antiderivative of tan x is ln|sec x| + C (or equivalently –ln|cos x| + C), found by rewriting tan x = sin x / cos x and substituting u = cos x.
This connects to trigonometric substitution, where after substituting x = a tan θ or x = a sec θ you end up needing to integrate products of secant and tangent of the new variable. It also connects to integration by parts, which is the standard approach for integrals like ∫ sec³x dx that do not fall neatly into the four cases above. If you continue to differential equations, sec x integrals appear in solving certain first-order linear ODEs with trig coefficients.
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