Trig integrals handle powers of sine, cosine, tangent and secant by using Pythagorean identities to peel off factors. Trig substitution tackles integrands containing √(a² - b²x²), √(a² + b²x²), or √(b²x² - a²) by replacing x with a trig expression that eliminates the square root. Together they cover a large family of integrals that u-sub and integration by parts cannot simplify directly.
Difficulty: Intermediate to Advanced | Prerequisites: U-substitution, integration by parts, Pythagorean identities, unit circle fluency.
Many integrals in physics and geometry produce square roots of quadratic expressions or products of trig functions raised to powers. Neither u-sub nor integration by parts handles these cleanly on its own. Trig integrals and trig substitution are purpose-built techniques for these shapes. The key skill is pattern recognition: you look at the integrand, match it to one of a few standard forms, and apply the corresponding identity or substitution.
Trig integral
An integral of the form ∫ (trig function)^n dx, or a product of trig functions raised to powers. In simple terms, any integral where the main challenge is dealing with powers of sin, cos, tan, or sec.
Trig substitution
A technique where you replace x with a trig expression (x = (a/b) sin θ, x = (a/b) tan θ, or x = (a/b) sec θ) to eliminate a square root of a quadratic. Think of it as using the Pythagorean identity to turn a radical into a plain trig function.
Pythagorean identities
The three related identities: sin²x + cos²x = 1, tan²x + 1 = sec²x, and sec²x - 1 = tan²x. These are the engine behind both trig integrals and trig substitution.
Half-angle identities
sin²x = (1 - cos 2x)/2 and cos²x = (1 + cos 2x)/2. Used when both powers of sin and cos are even, since there is no odd power to peel off.
Reference triangle
A right triangle drawn to convert back from θ to x after a trig substitution. You label the sides using your substitution and read off the remaining trig functions from the triangle.
If one power is odd, peel off one factor of that trig function, use the Pythagorean identity to convert the rest, and u-sub the other trig function.
Odd power of sin: save one sin x, write sin²x = 1 - cos²x, set u = cos x.
Odd power of cos: save one cos x, write cos²x = 1 - sin²x, set u = sin x.
If both powers are even, use half-angle identities to reduce the powers.
sin²x = (1 - cos 2x)/2
cos²x = (1 + cos 2x)/2
Rewrite using the half-angle identity:
I = ∫₀^(π/4) (1 - cos 2x)/2 dx = (1/2)[x - sin(2x)/2]₀^(π/4)
= (1/2)(π/4 - sin(π/2)/2) - 0 = (1/2)(π/4 - 1/2) = π/8 - 1/4 = (1/4)(π/2 - 1).
Split into two integrals:
∫ 3 cos θ dθ = 3 sin θ
∫ cos³θ dθ: save one cos θ, write cos²θ = 1 - sin²θ, set u = sin θ.
∫ cos³θ dθ = ∫ (1 - sin²θ) cos θ dθ = sin θ - sin³θ/3.
Combine and evaluate from 0 to π/2:
I = [3 sin θ + sin θ - sin³θ/3]₀^(π/2) = 3 + 1 - 1/3 = 11/3.
Similar logic applies. Use tan²x = sec²x - 1 and the fact that d(sec x) = sec x tan x dx.
Rewrite as ∫ tan x sec x · tan²x sec²x dx.
Use tan²x = sec²x - 1, set u = sec x, du = sec x tan x dx.
I = ∫ (u² - 1) u² du = u⁵/5 - u³/3 + C = sec⁵x/5 - sec³x/3 + C.
The substitution depends on which quadratic form sits under (or inside) the radical.
a² - b²x² appears: set x = (a/b) sin θ. Then a² - b²x² = a² cos²θ, and the root becomes a cos θ.
a² + b²x² appears: set x = (a/b) tan θ. Then a² + b²x² = a² sec²θ, and the root becomes a sec θ.
b²x² - a² appears: set x = (a/b) sec θ. Then b²x² - a² = a² tan²θ, and the root becomes a tan θ.
This comes directly from the three Pythagorean identities:
sin²θ + cos²θ = 1, rearranged as 1 - sin²θ = cos²θ
tan²θ + 1 = sec²θ
sec²θ - 1 = tan²θ
Identify which quadratic form you have.
Make the substitution for x, compute dx.
Replace everything, simplify the radical using the identity.
Integrate in θ.
Draw a reference triangle and convert back to x.
This is the step students most often skip, and it causes sign errors. After substituting, you need to know whether tan θ is positive or negative over the interval, because |tan θ| = tan θ only when tan θ ≥ 0.
For definite integrals, convert your x-limits to θ-limits and check the sign of each trig function in that range.
Set x = (2/5) sec θ, so dx = (2/5) sec θ tan θ dθ.
25x² - 4 = 25 · (4/25) sec²θ - 4 = 4 sec²θ - 4 = 4 tan²θ, so √(25x² - 4) = 2|tan θ|.
Convert limits: x = -2/5 gives sec θ = -1, so θ = π. x = -4/5 gives sec θ = -2, so θ = 2π/3.
Since 2π/3 ≤ θ ≤ π, tan θ ≤ 0, so |tan θ| = -tan θ.
I = -2 ∫ tan²θ dθ = -2 ∫ (sec²θ - 1) dθ = -2[tan θ - θ] evaluated from 2π/3 to π.
= -2[(0 - π) - (√3 - 2π/3)] = 2√3 - 2π/3.
Set x = sin θ, dx = cos θ dθ. Then (1 - x²)^(3/2) = cos³θ.
I = 4 ∫ cos θ / cos³θ dθ = 4 ∫ sec²θ dθ = 4 tan θ.
Evaluate from θ = 0 to θ = π/4 (converting x-limits): I = 4[tan(π/4) - tan(0)] = 4.
Pythagorean identities
sin²θ + cos²θ = 1
tan²θ + 1 = sec²θ
sec²θ - 1 = tan²θ
Half-angle (power-reduction) identities
sin²θ = (1 - cos 2θ)/2
cos²θ = (1 + cos 2θ)/2
Trig substitution table
√(a² - b²x²): set x = (a/b) sin θ, dx = (a/b) cos θ dθ
√(a² + b²x²): set x = (a/b) tan θ, dx = (a/b) sec²θ dθ
√(b²x² - a²): set x = (a/b) sec θ, dx = (a/b) sec θ tan θ dθ
Useful derivatives
d/dθ (tan θ) = sec²θ
d/dθ (sec θ) = sec θ tan θ
Trig substitution is how you compute arc lengths of curves, surface areas of solids of revolution, and gravitational potentials in physics. Whenever a geometry or physics problem produces an integrand with √(a² - x²) (think: circles, ellipses) or √(x² + a²) (think: distances in a coordinate plane), trig substitution is the standard route.
Students often confuse which substitution goes with which quadratic form. The mnemonic: "minus means sine" (a² - b²x² uses sin), "plus means tangent" (a² + b²x² uses tan), "x leads means secant" (b²x² - a² uses sec).
Students frequently forget to check the sign of tan θ or other trig functions when the integration interval lands in a quadrant where they are negative. This causes sign errors in definite integrals.
When both sin and cos powers are even, students sometimes try the odd-power peeling strategy and get stuck. Use half-angle identities instead.
Students forget to convert dx when substituting. If x = (a/b) sin θ, then dx = (a/b) cos θ dθ, not just dθ.
⚠️ The trig substitution table is almost always on the exam. Know all three forms cold.
⚠️ Expect a definite integral where you must track the sign of tan θ or convert limits carefully. The worked example with √(25x² - 4)/x in the source notes is exactly this type.
⚠️ Trig integral problems with odd powers of sin or cos are high-frequency exam questions. Practice peeling off one factor and converting with the Pythagorean identity until it is automatic.
⚠️ Half-angle identity problems (even powers of sin and cos) are common on midterms. They are mechanical but error-prone if you rush the algebra.
Fill in the blank: To integrate ∫ sin⁵x cos²x dx, save one _______ factor, convert the remaining sin² to 1 - cos²x, and set u = cos x.
True or false: For ∫ sin²x cos²x dx, you should use half-angle identities because both powers are even.
Fill in the blank: If the integrand contains √(a² + b²x²), substitute x = _______.
True or false: After trig substitution, you can skip drawing a reference triangle if it is a definite integral with converted limits.
Fill in the blank: sec²θ - 1 = _______.
Answers: 1. sin x. 2. True. 3. (a/b) tan θ. 4. True (you evaluate in θ and never need to convert back). 5. tan²θ.
Q: Evaluate ∫₀^(π/4) sin²x dx.
A: Use the half-angle identity: sin²x = (1 - cos 2x)/2. Then I = (1/2)[x - sin(2x)/2]₀^(π/4) = (1/2)(π/4 - 1/2) = (1/4)(π/2 - 1).
Q: Evaluate ∫₀^(π/2) (3 cos θ + cos³θ) dθ.
A: The first term gives 3 sin θ. For cos³θ, save one cos θ, write cos²θ = 1 - sin²θ, set u = sin θ. The result is sin θ - sin³θ/3. Evaluating the combined expression from 0 to π/2 gives 3 + 1 - 1/3 = 11/3.
Q: What substitution would you use for ∫ √(9 - 4x²) dx?
A: The form is a² - b²x² with a = 3, b = 2. Set x = (3/2) sin θ.
Q: Evaluate ∫ tan³x sec³x dx.
A: Rewrite as ∫ tan x sec x · tan²x sec²x dx. Substitute tan²x = sec²x - 1, set u = sec x, du = sec x tan x dx. This gives ∫ (u² - 1)u² du = sec⁵x/5 - sec³x/3 + C.
Q: Why do you need to check the range of tan θ in the √(25x² - 4)/x example?
A: Because |tan θ| = tan θ only when tan θ ≥ 0. The given x-limits map to an interval where tan θ ≤ 0, so |tan θ| = -tan θ, and missing this flips the sign of the entire answer.
Trig integrals build directly on the Pythagorean identities from precalculus. If those feel shaky, revisit the unit circle and practice converting between sin²θ and cos²θ before attempting these integrals.
Trig substitution connects to analytic geometry: the expressions a² - x², a² + x², and x² - a² arise from circles, hyperbolas, and distance formulas. You will see trig substitution again in multivariable calculus when converting to polar or cylindrical coordinates.
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