Trigonometric substitution replaces algebraic expressions involving square roots with trig identities, turning ugly integrals into standard trig integrals you already know how to handle. The three cases map to three identities: use sin for √(a² − x²), tan for √(a² + x²), and sec for √(x² − a²). Once you pick the right substitution, simplify, integrate in trig-land, and convert back to x.
Difficulty: Intermediate | Prerequisites: basic integration techniques (u-substitution), trig identities (Pythagorean, double-angle), inverse trig functions.
Trigonometric substitution sits in the integration toolkit right after u-substitution and integration by parts. It is the go-to method whenever a square root of a quadratic expression appears in the integrand and basic u-sub does not simplify it. The technique exploits the Pythagorean identities (sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, sec²θ − 1 = tan²θ) to eliminate the square root entirely. If you are comfortable with u-substitution and know your trig identities cold, this topic is a matter of pattern recognition and careful algebra.
Trigonometric substitution
A technique that replaces an algebraic variable with a trigonometric expression (x = a sin θ, x = a tan θ, or x = a sec θ) to eliminate a square root involving a sum or difference of squares. Think of it as: "turn the square root into a trig identity so it cancels cleanly."
Pythagorean identity
The three forms: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and sec²θ − 1 = tan²θ. In simple terms, these are the engine behind every trig substitution, each one matched to one of the three substitution patterns.
Completing the square
Rewriting a quadratic ax² + bx + c into the form a(x − h)² + k so that a trig substitution can be applied. In simple terms, this is the pre-processing step you use when the expression under the root is not already in the form a² ± x².
Back-substitution
Converting the result from trig variables (θ) back to the original variable (x) using a reference triangle or inverse trig functions. Think of it as: "undoing the substitution to get an answer the original problem can use."
Reference triangle
A right triangle drawn from the substitution (e.g., if x = a sin θ, then the opposite side is x, the hypotenuse is a, and the adjacent side is √(a² − x²)). In simple terms, it is a visual tool that makes back-substitution straightforward.
The choice of substitution is driven entirely by the form of the expression under the square root (or raised to a fractional power).
√(a² − x²) → set x = a sin θ. The identity sin²θ + cos²θ = 1 turns a² − x² into a² cos²θ. Then dx = a cos θ dθ. The square root becomes a cos θ. (See Q2, Q3, Q4, Q5.)
√(a² + x²) → set x = a tan θ. The identity 1 + tan²θ = sec²θ turns a² + x² into a² sec²θ. Then dx = a sec²θ dθ. The square root becomes a sec θ. (See Q1, Q7.)
√(x² − a²) → set x = a sec θ. The identity sec²θ − 1 = tan²θ turns x² − a² into a² tan²θ. Then dx = a sec θ tan θ dθ. The square root becomes a tan θ. (See Q8, Q9.)
When the quadratic under the root does not match one of these three forms directly, complete the square first, then apply the appropriate substitution. (See Q10, where 15 + 2x − x² becomes 16 − (x − 1)².)
Example (Q3): Evaluate ∫₀¹ 2/√(4 − 3x²) dx.
Set √3 x = 2 sin θ, so √3 dx = 2 cos θ dθ. Then 4 − 3x² = 4 cos²θ. The limits change: x = 0 gives θ = 0, x = 1 gives θ = π/3. The integral simplifies to (2/√3) ∫₀^(π/3) dθ = 2π/(3√3).
Example (Q5): Evaluate ∫₀² (1 + √(4 − x²)) dx.
Split into two integrals. The first is trivial: ∫₀² 1 dx = 2. For the second, set x = 2 sin u, so dx = 2 cos u du and 4 − x² = 4 cos²u. The integral becomes 4 ∫₀^(π/2) cos²u du = 2 ∫₀^(π/2) (1 + cos 2u) du = π. Total: 2 + π.
Example (Q6): Determine ∫ x²/(9 − x²)^(3/2) dx.
Set x = 3 sin θ, so 9 − x² = 9 cos²θ and (9 − x²)^(3/2) = 27 cos³θ. The integrand reduces to ∫ tan²θ dθ = ∫ (sec²θ − 1) dθ = tan θ − θ + C. Back-substituting: x/√(9 − x²) − sin⁻¹(x/3) + C.
Example (Q1): Find ∫₂⁵ 1/(9 + (x − 2)²) dx.
Set 3 tan u = x − 2, so 9 + (x − 2)² = 9 sec²u and dx = 3 sec²u du. The integral simplifies to (1/3) ∫₀^(π/4) du = π/12.
Example (Q7): Evaluate ∫₀² 6/(x² + 4)^(3/2) dx.
Set x = 2 tan u, so (x² + 4)^(3/2) = 8 sec³u and dx = 2 sec²u du. The integral simplifies to (3/2) ∫₀^(π/4) cos u du = (3/2) sin u |₀^(π/4) = (3/4)√2.
Example (Q8): Determine ∫ (4 − x)/√(x² − 1) dx.
Set x = sec θ, so x² − 1 = tan²θ and dx = sec θ tan θ dθ. The integral becomes ∫ (4 sec θ − sec²θ) dθ = 4 ln|sec θ + tan θ| − tan θ + C. Back-substituting: 4 ln|x + √(x² − 1)| − √(x² − 1) + C.
Example (Q9): Evaluate ∫_(√2)^2 8/(x√(x² − 1)) dx.
Set x = sec u, so the integrand simplifies to 8 du. The limits change: x = √2 gives u = π/4, x = 2 gives u = π/3. Result: 8(π/3 − π/4) = 2π/3.
Example (Q10): Evaluate ∫₁³ 2/√(15 + 2x − x²) dx.
Complete the square: 15 + 2x − x² = 16 − (x − 1)². Now set x − 1 = 4 sin u, so dx = 4 cos u du. The limits become u = 0 to u = π/6. The integral simplifies to ∫₀^(π/6) 2 du = π/3.
Expression | Substitution | Identity Used | dx becomes | Root becomes |
|---|---|---|---|---|
√(a² − x²) | x = a sin θ | 1 − sin²θ = cos²θ | a cos θ dθ | a cos θ |
√(a² + x²) | x = a tan θ | 1 + tan²θ = sec²θ | a sec²θ dθ | a sec θ |
√(x² − a²) | x = a sec θ | sec²θ − 1 = tan²θ | a sec θ tan θ dθ | a tan θ |
Useful trig integrals to have memorised:
∫ sin²θ dθ = θ/2 − sin(2θ)/4 + C
∫ cos²θ dθ = θ/2 + sin(2θ)/4 + C
∫ tan²θ dθ = tan θ − θ + C
∫ sec θ dθ = ln|sec θ + tan θ| + C
∫ sec²θ dθ = tan θ + C
∫ sec³θ dθ = (1/2)(sec θ tan θ + ln|sec θ + tan θ|) + C
Trigonometric substitution appears whenever a model involves circular or elliptical geometry. Calculating the arc length of an ellipse, the area of a region bounded by a conic section, or the gravitational field of a uniform disc all require integrals of the form √(a² − x²) or √(a² + x²). In physics and engineering, the Pappus theorem application (Q11, computing the volume of a torus) connects trig substitution directly to computing volumes of revolution for curved cross-sections.
Students often confuse which substitution to use. If you see a² − x², use sine, not tangent. Tangent is for a² + x². Mixing these up gives an expression that does not simplify.
Students frequently forget to change the limits of integration when evaluating a definite integral. If you substitute x = a sin θ, the limits must be converted to θ-values. Leaving them as x-values and evaluating in θ produces a wrong answer.
A common error is dropping the absolute value when back-substituting with secant. Since sec θ can be negative in certain quadrants, |sec θ + tan θ| matters. For definite integrals on a chosen interval this usually resolves, but for indefinite integrals the absolute value must stay.
Students sometimes forget to complete the square first. If the expression under the root is 15 + 2x − x², you cannot apply a trig substitution directly. Rewrite it as 16 − (x − 1)² first, then substitute.
⚠️ Trig substitution problems are a near-certainty on any integral calculus exam. Expect at least one definite integral and one indefinite integral requiring trig sub.
⚠️ Problems that combine completing the square with trig substitution (like Q10) are common exam curveballs. Practise recognising when the quadratic needs rewriting first.
⚠️ The volume-of-revolution application (Q11, the torus via Pappus' theorem) is a classic exam problem. Know that volume = 2π²Rr² for a torus with centre distance R and tube radius r.
⚠️ Back-substitution mistakes are the most common source of lost marks. Draw the reference triangle every time.
True or False: To evaluate ∫ x²/√(1 − x²) dx, you should set x = tan θ. (False: x = sin θ, because the form is √(a² − x²).)
Fill in the blank: When you substitute x = a sec θ, then dx = ______. (a sec θ tan θ dθ)
True or False: ∫₀^(π/2) cos²u du = π/2. (False: it equals π/4.)
Fill in the blank: Before applying trig substitution to ∫ 1/√(5 + 4x − x²) dx, the first step is ______. (Complete the square: rewrite as 9 − (x − 2)².)
True or False: The volume of a torus formed by rotating x² + (y − R)² = r² about the x-axis is 2π²Rr². (True.)
Q: Evaluate ∫₂⁵ 1/(9 + (x − 2)²) dx.
A: Set 3 tan u = x − 2. The integral becomes (1/3) ∫₀^(π/4) du = π/12.
Q: What trig substitution reduces ∫ x²/√(1 − x²) dx, and what does it simplify to?
A: Set x = sin θ, dx = cos θ dθ. The integral reduces to ∫ sin²θ dθ.
Q: Evaluate ∫₀¹ 2/√(4 − 3x²) dx.
A: Set √3 x = 2 sin θ. The integral simplifies to (2/√3) ∫₀^(π/3) dθ = 2π/(3√3).
Q: Evaluate ∫₀² 6/(x² + 4)^(3/2) dx.
A: Set x = 2 tan u. The integral simplifies to (3/2)[sin u]₀^(π/4) = (3/4)√2.
Q: Evaluate ∫₁³ 2/√(15 + 2x − x²) dx.
A: Complete the square to get 16 − (x − 1)². Set x − 1 = 4 sin u. The integral equals [2u]₀^(π/6) = π/3.
Trig substitution builds directly on u-substitution (the trig sub is itself a specialised change of variable). It connects forward to integration by partial fractions, which handles rational functions, as the two techniques together cover most of the integrands you will face on exams. The volume-of-revolution problems (Q5, Q11) tie this material to the applications-of-integration unit, specifically the disc/washer method and Pappus' theorem.
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