Eliminating square roots by replacing x with a trig function, and the Pythagorean identity does the heavy lifting.
3Substitution Cases
12Worked Examples
10Quiz Questions
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§ 01When Trig Substitution Applies
Standard integration methods break down in the face of expressions like √(a² − x²), √(a² + x²), and √(x² − a²). Trig substitution is the tool that removes those square roots entirely.
The core idea is borrowed from right-triangle geometry. Every Pythagorean identity says that some combination of trig functions squared always collapses to a perfect square, and therefore its square root is just another trig function: no radical required.
The technique works in three situations, each corresponding to one of the three Pythagorean identities:
Case 1 — Sine Sub√(a² − x²)
Use x = a sin θ. The identity 1 − sin²θ = cos²θ clears the root.
Case 2 — Tangent Sub√(a² + x²)
Use x = a tan θ. The identity 1 + tan²θ = sec²θ clears the root.
Case 3 — Secant Sub√(x² − a²)
Use x = a sec θ. The identity sec²θ − 1 = tan²θ clears the root.
When the expression is not under a square root
Trig substitution also works when you see (a² − x²)ⁿ, (a² + x²)ⁿ, or (x² − a²)ⁿ for any n — including positive integers. The square root is the most common scenario, but the same substitution clears the radical-free version too.
After substituting, you integrate in terms of θ, then convert back to x using a reference triangle or inverse trig functions. The four-step workflow is: choose the substitution → find dx → substitute everything → back-convert.
§ 02Case 1 — The Sine Substitution for √(a² − x²)
Whenever you see a² − x² under or raised to a power, let x = a sin θ.
Case 1 Substitution
x = a sin θ, dx = a cos θ dθ
√(a² − x²) = a cos θ (for θ ∈ [−π/2, π/2])
Verification: a² − a²sin²θ = a²(1 − sin²θ) = a²cos²θ. Taking the positive root gives a cos θ.
Reference Triangle for Case 1
Draw a right triangle with hypotenuse a, opposite side x, and adjacent side √(a² − x²). Then sin θ = x/a and cos θ = √(a² − x²)/a. This triangle is essential for converting back at the end.
Key back-conversion identities — Case 1
sin θ = x/a, cos θ = √(a² − x²)/a, tan θ = x/√(a² − x²)
Also useful: θ = arcsin(x/a), so any leftover θ in the answer becomes arcsin(x/a).
Example 1∫ √(9 − x²) dx
Identify
We have √(a² − x²) with a = 3. Use x = 3 sin θ.
Substitute
dx = 3 cos θ dθ√(9 − x²) = √(9 − 9sin²θ) = 3 cos θ
Integrate
∫ 3 cos θ · 3 cos θ dθ = 9∫ cos²θ dθ
Apply the half-angle identity cos²θ = (1 + cos 2θ)/2:
= 9/2 ∫(1 + cos 2θ) dθ = 9/2 (θ + sin 2θ/2) + C= 9θ/2 + 9 sin 2θ/4 + C
Back-convert
θ = arcsin(x/3). For sin 2θ use sin 2θ = 2 sin θ cos θ = 2(x/3)(√(9−x²)/3):
Write sec θ/tan²θ = (1/cos θ)·(cos²θ/sin²θ) = cos θ/sin²θ. Let u = sin θ:
= (1/4) ∫ u⁻² du = (1/4)(−1/u) + C = −1/(4 sin θ) + C
Back-convert
sin θ = x/√(4+x²) from the reference triangle:
Answer
= −√(4 + x²)/(4x) + C
§ 07Quick-Reference Table
This table summarises all three cases. Keep it in mind whenever you spot a quadratic expression under a radical.
Expression
Substitution
dx
Simplified radical
Identity used
√(a² − x²)
x = a sin θ
a cos θ dθ
a cos θ
1 − sin²θ = cos²θ
√(a² + x²)
x = a tan θ
a sec²θ dθ
a sec θ
1 + tan²θ = sec²θ
√(x² − a²)
x = a sec θ
a sec θ tan θ dθ
a tan θ
sec²θ − 1 = tan²θ
The general workflow always follows four stages:
The Four-Step Process
1. Identify the form — which of the three cases does the integral match?
2. Substitute — replace x and dx, and simplify the radical using the relevant identity.
3. Integrate in θ — the θ-integral should involve only standard trig integrals.
4. Back-convert to x — draw the reference triangle and replace every θ-expression with an x-expression.
§ 08Common Mistakes
Trig substitution has predictable pitfalls. Here are the ones that cost the most marks.
Mistake 1 — Choosing the wrong case
Confusing √(a² + x²) with √(a² − x²) is very common under time pressure. √(a² − x²) has the minus sign inside, so it matches the sine sub. √(a² + x²) has the plus sign, so it matches the tangent sub. If you pick the wrong one, the Pythagorean identity will not simplify cleanly.
Mistake 2 — Forgetting to change dx
You must differentiate your substitution and replace dx. In x = a sin θ: dx = a cos θ dθ. Forgetting the dθ factor, or computing the derivative of the wrong function, ruins the whole integral.
Mistake 3 — Incorrect back-conversion
Never write "θ + C" as your final answer — θ is not the original variable. Always draw the reference triangle and convert every θ-term back to an expression in x. Writing sin θ = x/a for a case-2 (tan) substitution is a frequent triangle error.
Mistake 4 — Sign errors with √(x² − a²) = a tan θ
The secant substitution gives √(x² − a²) = a|tan θ|. When θ is in [0, π/2) the tangent is non-negative and |tan θ| = tan θ. For θ in (π/2, π] the tangent is non-positive and you need to track the sign carefully. On most exam problems x ≥ a > 0 so θ ∈ [0, π/2) and the issue does not arise, but be aware of it.
Mistake 5 — Applying trig sub when simpler methods work
If the integrand is ∫ x/√(1 − x²) dx, a simple u-substitution u = 1 − x² is faster than trig sub. Always check whether an algebraic substitution or direct formula applies before committing to the more involved trig procedure.
Pro tip — Definite integrals: change limits, don't convert back
For definite integrals, converting the limits of integration from x to θ (when x = a) is usually faster than back-converting the antiderivative. Once you have θ-limits, evaluate directly and skip the back-conversion step entirely.
Ten questions covering all three substitution cases, completing the square, back-conversion, and common mistakes. Work through each one before checking the explanation.
Trigonometric Substitution Quiz
Question 1 of 10
Which substitution should you use to evaluate ∫ 1/√(9 − x²) dx?
Question 2 of 10
After applying x = 5 tan θ to an integral involving √(25 + x²), what does √(25 + x²) simplify to?
Question 3 of 10
Which substitution is correct for ∫ √(x² − 16) dx?
Question 4 of 10
When using x = a sin θ, what is dx in terms of θ?
Question 5 of 10
Using the sine substitution x = 2 sin θ in the reference triangle, what is tan θ in terms of x?
Question 6 of 10
To evaluate ∫ 1/(x² + 6x + 10) dx using trig substitution, what is the first step?
Question 7 of 10
Which Pythagorean identity is used to simplify the radical in the tangent substitution?
Question 8 of 10
What is ∫₀^(√3) x³/√(x² + 1) dx? (Hint: use x = tan θ, change limits.)
Question 9 of 10
After using x = 3 sec θ to compute ∫ √(x² − 9)/x dx, you obtain √(x²−9) − 3 arcsec(x/3) + C. What is the critical final step before writing this answer?
Question 10 of 10
Which of the following integrals does NOT require trigonometric substitution — a simpler method is more efficient?