Difficulty: Intermediate | Prerequisites: Basic integration techniques, u-substitution, trigonometric identities (Pythagorean, double-angle).
Trigonometric substitution is an integration technique for handling integrals containing radical expressions of the form √(x² + a²), √(a² − x²), or √(x² − a²). These radicals resist ordinary u-substitution because no simple algebraic substitution eliminates the square root. By replacing x with a trigonometric function of θ, the Pythagorean identities collapse the radical into a clean trig expression you can integrate directly. This technique appears throughout Calculus 2 and resurfaces in multivariable calculus, differential equations, and physics whenever arc length, surface area, or force calculations produce square-root expressions.
When you see a square root of a sum or difference of squares inside an integral, substitute x with a trig function that turns the radical into a single trig term via a Pythagorean identity. Integrate in θ, then convert back to x using a reference triangle. The choice of sin, tan, or sec depends on which form of radical you have.
Trigonometric substitution (trig sub)
An integration technique that replaces an algebraic variable with a trigonometric expression to eliminate a radical. In simple terms, you swap x for something like a sin(θ) so that a nasty square root simplifies through a Pythagorean identity.
Radical expression
An expression under a square root sign, specifically one involving sums or differences of squares such as √(x² + a²). Think of it as the part of the integrand that makes direct integration difficult.
Reference triangle
A right triangle drawn after integrating in θ, used to convert the answer back into terms of x. You label the sides using your original substitution (e.g. if x = a tan(θ), the opposite side is x, the adjacent side is a, and the hypotenuse is √(x² + a²)).
Back-substitution
The final step where you replace θ and all trig functions with expressions in x, using the reference triangle or the original substitution equation. In simple terms, this is how you get your answer back into the variable you started with.
Completing the square
An algebraic technique that rewrites a quadratic expression ax² + bx + c into the form (x + d)² + e, which then matches one of the three standard trig sub patterns. Think of it as the preprocessing step that makes trig sub possible when the integrand does not already contain a clean sum or difference of squares.
Every trig sub problem begins with identifying which radical pattern you have. Each pattern has exactly one correct substitution, a restricted range for θ, and a known simplification.
Form 1: √(x² + a²), substitute x = a tan(θ)
Range: −π/2 < θ < π/2
Then dx = a sec²(θ) dθ
The radical becomes: √(a² tan²(θ) + a²) = a√(tan²(θ) + 1) = a sec(θ)
Use when the integrand contains a sum of squares under the root
Form 2: √(a² − x²), substitute x = a sin(θ)
Range: −π/2 ≤ θ ≤ π/2
Then dx = a cos(θ) dθ
The radical becomes: √(a² − a² sin²(θ)) = a√(1 − sin²(θ)) = a cos(θ)
Use when the integrand contains a difference with the constant term first (a² − x²)
Form 3: √(x² − a²), substitute x = a sec(θ)
Range: 0 ≤ θ < π/2 or π ≤ θ < 3π/2
Then dx = a sec(θ) tan(θ) dθ
The radical becomes: √(a² sec²(θ) − a²) = a√(sec²(θ) − 1) = a tan(θ)
Use when the integrand contains a difference with the variable term first (x² − a²)
Identify the radical form and choose the substitution
Compute dx in terms of dθ
Substitute everything into the integral (x, dx, and the radical)
Simplify using the Pythagorean identity
Integrate in θ (this may require further trig identities such as double-angle formulas)
Draw a reference triangle and back-substitute to return to x
Sometimes the expression x² + bx + c does not look like a standard form. In these cases, complete the square first to rewrite it as (x + d)² ± e², then apply trig sub to the new variable u = x + d. This is covered in the companion notes on completing the square with trig sub.
Expression | Substitution | dx | Radical becomes |
|---|---|---|---|
√(x² + a²) | x = a tan(θ) | a sec²(θ) dθ | a sec(θ) |
√(a² − x²) | x = a sin(θ) | a cos(θ) dθ | a cos(θ) |
√(x² − a²) | x = a sec(θ) | a sec(θ)tan(θ) dθ | a tan(θ) |
This matches Form 3 with a = 1.
Substitute x = sec(θ), so dx = sec(θ) tan(θ) dθ
The integral becomes ∫ sec²(θ) / √(sec²(θ) − 1) · sec(θ) tan(θ) dθ
Simplify: √(sec²(θ) − 1) = tan(θ), so the integrand reduces to ∫ sec²(θ) · sec(θ) tan(θ) / tan(θ) dθ = ∫ sec³(θ) dθ
The integral of sec³(θ) is a standard result (typically derived via integration by parts)
This matches Form 1 with a = 1.
Substitute x = tan(θ), so dx = sec²(θ) dθ
The integral becomes 6 ∫ sec²(θ) / [tan(θ) · √(tan²(θ) + 1)] dθ
Simplify: √(tan²(θ) + 1) = sec(θ), so we get 6 ∫ sec²(θ) / [tan(θ) · sec(θ)] dθ = 6 ∫ sec(θ) / tan(θ) dθ
Rewrite: sec(θ)/tan(θ) = [1/cos(θ)] / [sin(θ)/cos(θ)] = 1/sin(θ) = csc(θ)
Result: 6 ∫ csc(θ) dθ
This matches Form 3 with a = 2/5 (rewrite as √((5x)² − 4) and substitute accordingly).
Substitute x = (2/5) sec(θ), so dx = (2/5) sec(θ) tan(θ) dθ
The radical becomes √(4 sec²(θ) − 4) = 2√(sec²(θ) − 1) = 2 tan(θ)
After simplification: 2 ∫ tan²(θ) dθ = 2 ∫ (sec²(θ) − 1) dθ = 2(tan(θ) − θ) + C
Back-substitute using a reference triangle: tan(θ) = √(25x² − 4) / 2 and θ = sec⁻¹(5x/2)
Final answer: √(25x² − 4) − 2 sec⁻¹(5x/2) + C
Trig substitution is not just an exam trick. Any time you compute the arc length of a curve, the surface area of a solid of revolution, or the work done by a variable force, you will likely encounter integrals with radical expressions. In physics, these appear when calculating electric fields around charged objects, gravitational potential energy, and orbital mechanics. Engineers use them when modelling stress distributions in beams with curved cross-sections.
Students often confuse which substitution goes with which radical. The mnemonic: "tan for plus, sin for minus (constant first), sec for minus (variable first)" keeps them straight.
Students sometimes forget to change dx when substituting. Every trig sub requires replacing both x and dx. Missing the dx term produces a wrong integral.
Students often assume the radical simply "disappears." It does simplify, but you still need to carry the constant factor a that comes out of the identity.
Students frequently skip drawing the reference triangle and try to back-substitute by memory. This leads to sign errors, especially with the sec substitution where the range of θ matters.
⚠️ Trig sub is almost guaranteed to appear on the M408L exam, often as a standalone integral or embedded in an arc-length or volume problem.
⚠️ You will be expected to identify the correct substitution quickly. Practise pattern recognition with all three forms.
⚠️ Completing the square combined with trig sub is a common exam question. If the expression under the radical is a general quadratic, expect to complete the square first.
⚠️ Always simplify fully and back-substitute. Partial answers left in terms of θ typically receive partial credit at best.
True or False: For √(9 − x²), the correct substitution is x = 3 tan(θ).
Answer: False. The correct substitution is x = 3 sin(θ), because the form is √(a² − x²).
Fill in the blank: When substituting x = a sec(θ), the differential dx = ___.
Answer: a sec(θ) tan(θ) dθ.
True or False: After performing trig substitution, you can leave your final answer in terms of θ.
Answer: False. You must back-substitute to express the answer in terms of x.
Fill in the blank: The identity that makes trig sub work for √(x² + a²) is 1 + tan²(θ) = ___.
Answer: sec²(θ).
True or False: Trig substitution can only be used when the radical expression is in the denominator.
Answer: False. It works regardless of whether the radical is in the numerator, denominator, or both.
Q: Evaluate ∫ √(9 − x²) dx.
A: Substitute x = 3 sin(θ), dx = 3 cos(θ) dθ. The integral becomes 9 ∫ cos²(θ) dθ = (9/2)(θ + sin(θ) cos(θ)) + C. Back-substitute: θ = sin⁻¹(x/3), sin(θ) = x/3, cos(θ) = √(9 − x²)/3. Final answer: (9/2) sin⁻¹(x/3) + (x√(9 − x²))/2 + C.
Q: What substitution would you use for ∫ dx / √(x² + 4), and why?
A: Use x = 2 tan(θ) because the expression under the radical is x² + a² with a = 2. This turns √(x² + 4) into 2 sec(θ).
Q: Evaluate ∫ √(25x² − 4) / x dx.
A: Substitute x = (2/5) sec(θ). After simplification, the integral reduces to 2 ∫ (sec²(θ) − 1) dθ = 2(tan(θ) − θ) + C. Back-substitute to get √(25x² − 4) − 2 sec⁻¹(5x/2) + C.
Q: You encounter ∫ dx / (x² + 2x + 5). What is your first step before applying trig sub?
A: Complete the square: x² + 2x + 5 = (x + 1)² + 4. Then let u = x + 1 and apply trig sub with u = 2 tan(θ). The final answer is (1/2) tan⁻¹((x + 1)/2) + C.
Q: After substituting x = a sin(θ), how do you determine the bounds of θ for a definite integral?
A: Solve for θ at each original x-bound. If x = 0, then θ = sin⁻¹(0) = 0. If x = a, then θ = sin⁻¹(1) = π/2. The restricted range −π/2 ≤ θ ≤ π/2 ensures θ is uniquely determined.
Trig sub connects directly to integration by parts, because integrals like ∫ sec³(θ) dθ that arise after substitution require IBP to evaluate. It also connects to partial fractions: some problems can be solved by either method, and recognising which is more efficient saves time on exams.
This technique reappears in Calculus 3 when computing arc length of space curves and surface integrals, and in differential equations when solving certain ODEs that produce radical expressions during separation of variables.
Trigonometric substitution, trig sub, radical integrals, sqrt(x^2 + a^2), sqrt(a^2 - x^2), sqrt(x^2 - a^2), Pythagorean identity, reference triangle, back-substitution, completing the square, integration techniques, Calc 2, M408L, UT Austin, integral calculus, sec substitution, sin substitution, tan substitution, inverse trig functions, arc length integrals