Difficulty: Intermediate to Advanced | Prerequisites: Calc 1 integration, basic trig identities (Pythagorean identities), u-substitution, familiarity with sec and tan derivatives.
Trigonometric integrals and trig substitution are two closely related techniques that handle integrals involving powers of trig functions and square roots of quadratic expressions. They typically appear in Calc 2 right alongside or shortly after partial fractions. The key insight is that Pythagorean identities let you trade one trig function for another, and trig substitution lets you remove awkward square roots by replacing x with a trig expression. If you are solid on the derivatives of sec, tan, sin, and cos, and know the identity tan²θ + 1 = sec²θ, you are ready.
To integrate products like sec θ tan²θ, use trig identities to rewrite the integrand in a form where standard techniques (integration by parts, known antiderivatives) apply. For square roots like √(x² + 2x), complete the square first, then use a trig substitution to eliminate the radical. The result of the trig integral often feeds directly into the substitution problem.
Trigonometric integral
An integral involving products or powers of trigonometric functions, such as ∫ sec θ tan²θ dθ. Think of it as: any integral where the main challenge is trig functions tangled together.
Trig substitution
A technique where you replace x (or a shifted version of x) with a trigonometric expression to eliminate a square root. In simple terms, you temporarily switch to an angle variable so that Pythagorean identities simplify a radical.
Completing the square
Rewriting a quadratic expression ax² + bx + c in the form a(x + h)² + k. This step is often required before a trig substitution, because the standard substitution forms require the expression under the radical to match x² + a², x² - a², or a² - x².
Integration by parts
The technique ∫ u dv = uv - ∫ v du. In the context of trig integrals, it is frequently needed for integrals like ∫ sec³θ dθ, which appears inside the solution to ∫ sec θ tan²θ dθ.
Use the identity tan²θ = sec²θ - 1 to rewrite:
∫ sec θ tan²θ dθ = ∫ sec θ (sec²θ - 1) dθ = ∫ sec³θ dθ - ∫ sec θ dθ
The integral ∫ sec θ dθ = ln|sec θ + tan θ| + C. This is a standard result worth memorising.
For ∫ sec³θ dθ, use integration by parts with u = sec θ and dv = sec²θ dθ:
du = sec θ tan θ dθ
v = tan θ
So ∫ sec³θ dθ = sec θ tan θ - ∫ sec θ tan²θ dθ.
Notice that ∫ sec θ tan²θ dθ has appeared on both sides. Call it I:
I = sec θ tan θ - I - ∫ sec θ dθ + ∫ sec θ dθ
More carefully: from the original split, I = ∫ sec³θ dθ - ∫ sec θ dθ, and ∫ sec³θ dθ = sec θ tan θ - I. Substituting:
I = (sec θ tan θ - I) - ln|sec θ + tan θ|
2I = sec θ tan θ - ln|sec θ + tan θ|
I = (1/2) sec θ tan θ - (1/2) ln|sec θ + tan θ| + C
This "solve for the integral" trick, where the same integral shows up on both sides, is a technique you will use repeatedly.
The problem says to use the previous result. First, complete the square:
x² + 2x = (x + 1)² - 1
So √(x² + 2x) = √((x + 1)² - 1).
Let u = x + 1, du = dx:
∫ √(u² - 1) du
Now apply trig substitution. For √(u² - 1), use u = sec θ, du = sec θ tan θ dθ:
√(u² - 1) = √(sec²θ - 1) = tan θ (taking the positive root)
The integral becomes:
∫ tan θ · sec θ tan θ dθ = ∫ sec θ tan²θ dθ
This is exactly the integral from Problem 3. Applying that result:
= (1/2) sec θ tan θ - (1/2) ln|sec θ + tan θ| + C
Finally, back-substitute. Since u = sec θ, we have sec θ = u = x + 1, and tan θ = √(u² - 1) = √(x² + 2x):
= (1/2)(x + 1)√(x² + 2x) - (1/2) ln|x + 1 + √(x² + 2x)| + C
The chain here is: complete the square, shift variable, trig sub, use a known trig integral, back-substitute. This multi-step pattern is very typical of exam problems.
Key trig identities:
tan²θ = sec²θ - 1
sec²θ = 1 + tan²θ
sin²θ + cos²θ = 1
Standard trig substitution forms:
√(a² - x²): let x = a sin θ
√(a² + x²): let x = a tan θ
√(x² - a²): let x = a sec θ
Standard results worth memorising:
∫ sec θ dθ = ln|sec θ + tan θ| + C
∫ sec³θ dθ = (1/2) sec θ tan θ + (1/2) ln|sec θ + tan θ| + C
d/dθ (sec θ) = sec θ tan θ
d/dθ (tan θ) = sec²θ
Trig substitution appears in physics whenever you compute arc lengths, surface areas of revolution, or work done by variable forces that produce square-root expressions. It is also central to computing gravitational and electric potentials in physics and engineering, where integrals of the form ∫ 1/√(x² + a²) dx arise naturally.
Students frequently forget to complete the square before attempting a trig substitution. If the expression under the radical is x² + 2x rather than x² + 1 or (x + 1)² - 1, no standard substitution applies directly.
When using u = sec θ for √(u² - 1), students sometimes write √(sec²θ - 1) = sec θ tan θ. It should just be tan θ. The sec θ comes from the du = sec θ tan θ dθ part.
After the trig integral is done, students often forget (or struggle with) the back-substitution step. Draw a right triangle to read off the trig functions in terms of the original variable.
The "solve for the integral" trick in ∫ sec³θ dθ catches students off guard. If the same integral appears on both sides after integration by parts, collect it algebraically rather than starting over.
⚠️ ∫ sec³θ dθ is one of the most commonly tested integrals in Calc 2. Know the result by heart and understand how it is derived via integration by parts.
⚠️ Problems that chain completing the square into a trig substitution into a known trig integral (like Problems 3 and 4 here) are classic exam fare. Practise the full pipeline, not just isolated steps.
⚠️ Always double-check your back-substitution. A correct trig integral with a botched back-sub still loses most of the marks.
True or false: tan²θ = sec²θ + 1. (Answer: False. It is tan²θ = sec²θ - 1.)
Fill in the blank: For √(x² - a²), the appropriate trig substitution is x = _____. (Answer: a sec θ)
True or false: ∫ sec θ dθ = ln|sec θ| + C. (Answer: False. It is ln|sec θ + tan θ| + C.)
Fill in the blank: Before applying trig substitution to √(x² + 2x), you must first _____ ___ _____. (Answer: complete the square)
Q: Compute ∫ sec θ tan²θ dθ from scratch.
A: Rewrite tan²θ = sec²θ - 1. The integral becomes ∫ sec³θ dθ - ∫ sec θ dθ. Use integration by parts on ∫ sec³θ dθ (u = sec θ, dv = sec²θ dθ) and the "solve for the integral" trick. Final answer: (1/2) sec θ tan θ - (1/2) ln|sec θ + tan θ| + C.
Q: What trig substitution would you use for ∫ √(x² + 9) dx?
A: x = 3 tan θ, because the expression has the form x² + a² with a = 3.
Q: Compute ∫ √(x² + 2x) dx, stating each step.
A: Complete the square: x² + 2x = (x + 1)² - 1. Shift: u = x + 1. Sub: u = sec θ. This gives ∫ sec θ tan²θ dθ. Use the result from Problem 3 and back-substitute: (1/2)(x + 1)√(x² + 2x) - (1/2) ln|x + 1 + √(x² + 2x)| + C.
Q: Why does the integral ∫ sec³θ dθ require integration by parts rather than a simple substitution?
A: There is no clean u-substitution that simplifies sec³θ. Setting u = sec θ leaves du = sec θ tan θ dθ, which does not match the remaining integrand. Integration by parts with u = sec θ, dv = sec²θ dθ produces a solvable equation because the original integral reappears on the right-hand side.
The trig integral ∫ sec θ tan²θ dθ draws on integration by parts (a technique that appears throughout Calc 2) and feeds directly into trig substitution problems. Trig substitution itself is closely related to the geometry of arc length and surface area integrals, which you will encounter soon. The completing-the-square step connects back to algebra and also to the study of conic sections.
Trigonometric integrals, trig substitution, integration by parts, sec cubed integral, ∫ sec³θ dθ, ∫ sec θ tan²θ dθ, completing the square integration, √(x²+2x) integral, Pythagorean identity, Calc 2 trig sub, NYU Calculus II PS2, back-substitution trig, right triangle method