Eliminate radicals of the form √(a²−x²), √(a²+x²), and √(x²−a²) using sine, tangent, and secant substitutions — the complete technique with worked examples and reference triangles.
Trigonometric substitution is the technique of choice whenever an integrand contains a radical whose expression under the square root is a difference or sum of squares. The three triggering patterns are:
√(a² − x²)
Looks like a circle or ellipse equation. The − sign between the squares is the key indicator.
√(a² + x²)
Looks like a Pythagorean sum. The + sign between the squares distinguishes this case.
√(x² − a²)
The variable term dominates. The x² is larger, and a² is subtracted. Compare to case 1.
Substitution
x = a sinθ
dx = a cosθ dθ
θ ∈ [−π/2, π/2]
Simplification
a² − x² = a² − a²sin²θ
= a²(1 − sin²θ)
= a²cos²θ
√(a²−x²) = a cosθ
REFERENCE TRIANGLE (Case 1)
Opposite = x Hypotenuse = a Adjacent = √(a²−x²)
sinθ = x/a cosθ = √(a²−x²)/a tanθ = x/√(a²−x²)
Substitution
x = a tanθ
dx = a sec²θ dθ
θ ∈ (−π/2, π/2)
Simplification
a² + x² = a² + a²tan²θ
= a²(1 + tan²θ)
= a²sec²θ
√(a²+x²) = a secθ
REFERENCE TRIANGLE (Case 2)
Opposite = x Adjacent = a Hypotenuse = √(a²+x²)
tanθ = x/a secθ = √(a²+x²)/a sinθ = x/√(a²+x²)
Substitution
x = a secθ
dx = a secθ tanθ dθ
θ ∈ [0, π/2) ∪ (π/2, π]
Simplification
x² − a² = a²sec²θ − a²
= a²(sec²θ − 1)
= a²tan²θ
√(x²−a²) = a tanθ
REFERENCE TRIANGLE (Case 3)
Hypotenuse = x Adjacent = a Opposite = √(x²−a²)
secθ = x/a tanθ = √(x²−a²)/a sinθ = √(x²−a²)/x
| Radical | Substitution | dx = | Radical becomes | Identity used |
|---|---|---|---|---|
| √(a²−x²) | x = a sinθ | a cosθ dθ | a cosθ | 1 − sin²θ = cos²θ |
| √(a²+x²) | x = a tanθ | a sec²θ dθ | a secθ | 1 + tan²θ = sec²θ |
| √(x²−a²) | x = a secθ | a secθ tanθ dθ | a tanθ | sec²θ − 1 = tan²θ |
Identify the radical pattern
Look at the expression under the radical. Match it to √(a²−x²), √(a²+x²), or √(x²−a²). If you see a general quadratic like √(x²+bx+c), complete the square first to expose the pattern.
Choose and apply the substitution
Write down x = a sinθ (or tanθ or secθ) and differentiate to find dx. Substitute for BOTH x and dx in the integral. The goal is to replace every occurrence of x with a trig function.
Simplify the integrand
Apply the appropriate Pythagorean identity to collapse the radical. Factor and cancel common terms. You should now have a pure trig integral — often powers of sinθ, cosθ, secθ, or tanθ.
Integrate with respect to θ
Use standard trig integrals: ∫sinⁿθ cosᵐθ dθ (use reduction formulas or half-angle identities), ∫secⁿθ dθ (use reduction formula or integration by parts), ∫tanⁿθ dθ (use tanⁿθ = tanⁿ⁻²θ sec²θ − tanⁿ⁻²θ).
Back-substitute using the reference triangle
Draw the reference triangle for your case. Label the three sides using your substitution (x, a, and the radical). Read off sinθ, cosθ, tanθ, secθ in terms of x and a. Replace every trig function in your answer with its algebraic equivalent.
Simplify and add the constant of integration
Combine terms and simplify. If you have a definite integral, apply the original x-limits after back-substitution (or convert the limits to θ at step 2 and skip back-substitution entirely). Add +C for indefinite integrals.
When the expression under the radical is a general quadratic — not already in the form a²±x² — you must complete the square first. This reduces the quadratic to a perfect square ± a constant, revealing the correct trig substitution.
Method: Complete the Square
ax² + bx + c
Step 1: Factor out a from the x-terms
= a(x² + (b/a)x) + c
Step 2: Add and subtract (b/2a)²
= a(x + b/2a)² − b²/4a + c
Step 3: Let u = x + b/2a, so du = dx
= a·u² + k where k = c − b²/4a
√(x² + 6x + 13)
x² + 6x + 13 = (x+3)² − 9 + 13
= (x+3)² + 4
Let u = x+3, a = 2
→ √(u² + 4): use u = 2 tanθ
√(−x² + 4x)
−x² + 4x = −(x²−4x)
= −(x²−4x+4) + 4
= 4 − (x−2)²
Let u = x−2, a = 2
→ √(4−u²): use u = 2 sinθ
After integrating in terms of θ, you need to convert the answer back to x. Build the reference triangle from your substitution and read off every trig ratio directly.
Triangle sides:
All six ratios:
Triangle sides:
All six ratios:
Triangle sides:
All six ratios:
Radical: √(a²−x²) with a = 3. Use x = 3 sinθ.
Substitution: x = 3 sinθ dx = 3 cosθ dθ
Radical: √(9 − 9sin²θ) = √(9cos²θ) = 3 cosθ
∫√(9−x²) dx = ∫(3 cosθ)(3 cosθ dθ) = 9∫cos²θ dθ
Use half-angle: cos²θ = (1 + cos2θ)/2
= 9∫(1 + cos2θ)/2 dθ = (9/2)∫(1 + cos2θ) dθ
= (9/2)[θ + (1/2)sin2θ] + C
Apply sin2θ = 2 sinθ cosθ:
= (9/2)θ + (9/4)sin2θ + C = (9/2)θ + (9/2)sinθ cosθ + C
Back-substitute (Case 1 triangle: sinθ = x/3, cosθ = √(9−x²)/3):
θ = arcsin(x/3)
sinθ cosθ = (x/3)(√(9−x²)/3) = x√(9−x²)/9
= (9/2)arcsin(x/3) + (x/2)√(9−x²) + C
Classic formula derivation: ∫dx/(4+x²). Radical form: √(a²+x²) with a = 2.
Use x = 2 tanθ.
Substitution: x = 2 tanθ dx = 2 sec²θ dθ
Denominator: 4 + x² = 4 + 4tan²θ = 4sec²θ
∫dx/(4+x²) = ∫(2 sec²θ dθ)/(4 sec²θ)
= ∫(2 sec²θ)/(4 sec²θ) dθ = (1/2)∫dθ = θ/2 + C
Back-substitute: θ = arctan(x/2)
= (1/2)arctan(x/2) + C
General formula: ∫dx/(a²+x²) = (1/a)arctan(x/a) + C
Radical: √(a²+x²) with a = 4. Use x = 4 tanθ.
Substitution: x = 4 tanθ dx = 4 sec²θ dθ
Radical: √(16 + 16tan²θ) = 4 secθ
∫√(x²+16) dx = ∫(4 secθ)(4 sec²θ dθ) = 16∫sec³θ dθ
Use reduction: ∫sec³θ dθ = (1/2)(secθ tanθ + ln|secθ + tanθ|) + C
= 16 · (1/2)(secθ tanθ + ln|secθ + tanθ|) + C
= 8(secθ tanθ + ln|secθ + tanθ|) + C
Back-substitute (Case 2 triangle: tanθ = x/4, secθ = √(x²+16)/4):
secθ tanθ = (√(x²+16)/4)(x/4) = x√(x²+16)/16
ln|secθ + tanθ| = ln|(√(x²+16)+x)/4|
= (x/2)√(x²+16) + 8 ln|(x + √(x²+16))/4| + C
Often written: (x/2)√(x²+16) + 8 ln|x + √(x²+16)| + C₁ (absorbing −8 ln 4 into C)
Radical: √(x²−a²) with a = 5. Use x = 5 secθ.
Substitution: x = 5 secθ dx = 5 secθ tanθ dθ
Radical: √(25sec²θ − 25) = 5 tanθ
∫√(x²−25)/x dx = ∫(5 tanθ)/(5 secθ) · 5 secθ tanθ dθ
= 5∫tan²θ dθ
Use: tan²θ = sec²θ − 1
= 5∫(sec²θ − 1) dθ = 5(tanθ − θ) + C
Back-substitute (Case 3: secθ = x/5, tanθ = √(x²−25)/5):
tanθ = √(x²−25)/5 θ = arcsec(x/5)
= √(x²−25) − 5 arcsec(x/5) + C
General quadratic under radical — complete the square first.
Complete the square: x²+4x+8 = (x+2)²+4
Let u = x+2, du = dx
∫dx/√(x²+4x+8) = ∫du/√(u²+4)
Now Case 2 with a = 2: let u = 2 tanθ, du = 2 sec²θ dθ
√(u²+4) = 2 secθ
= ∫(2 sec²θ dθ)/(2 secθ) = ∫secθ dθ
= ln|secθ + tanθ| + C
Back-substitute (tanθ = u/2, secθ = √(u²+4)/2):
= ln|(√(u²+4) + u)/2| + C
= ln|√(u²+4) + u| + C₁
Replace u = x+2:
= ln|√(x²+4x+8) + (x+2)| + C
Case 1. Convert the x-limits to θ-limits to avoid back-substitution.
Use x = 3 sinθ dx = 3 cosθ dθ
Change limits: x=0 → sinθ=0 → θ=0 x=3 → sinθ=1 → θ=π/2
∫₀^(π/2) (9sin²θ)/(3 cosθ) · 3 cosθ dθ = 9∫₀^(π/2) sin²θ dθ
Use: sin²θ = (1 − cos2θ)/2
= 9 · (1/2)∫₀^(π/2) (1−cos2θ) dθ
= (9/2)[θ − (1/2)sin2θ]₀^(π/2)
At θ = π/2: (π/2) − (1/2)sin(π) = π/2 − 0 = π/2
At θ = 0: 0 − 0 = 0
= (9/2)(π/2) = 9π/4
Powers of Sine and Cosine
∫sin²θ dθ = θ/2 − sin(2θ)/4 + C
∫cos²θ dθ = θ/2 + sin(2θ)/4 + C
∫sin³θ dθ = −cosθ + cos³θ/3 + C
∫cos³θ dθ = sinθ − sin³θ/3 + C
Secant and Tangent
∫secθ dθ = ln|secθ + tanθ| + C
∫sec²θ dθ = tanθ + C
∫tan²θ dθ = tanθ − θ + C
∫sec³θ dθ = (1/2)(secθ tanθ
+ ln|secθ+tanθ|) + C
Half-Angle Identities (essential for even powers)
sin²θ = (1 − cos2θ)/2
cos²θ = (1 + cos2θ)/2
Use trigonometric substitution when your integrand contains one of three radical expressions: √(a²−x²), √(a²+x²), or √(x²−a²), or when completing the square on a quadratic under a radical produces one of these forms. The technique converts an algebraic radical into a trigonometric expression that can be simplified using Pythagorean identities, making the integral tractable.
The three standard substitutions are: (1) For √(a²−x²): let x = a sinθ, so dx = a cosθ dθ and √(a²−x²) = a cosθ. (2) For √(a²+x²): let x = a tanθ, so dx = a sec²θ dθ and √(a²+x²) = a secθ. (3) For √(x²−a²): let x = a secθ, so dx = a secθ tanθ dθ and √(x²−a²) = a tanθ. Each substitution uses a Pythagorean identity to eliminate the radical.
After integrating with respect to θ, convert back to x using a reference triangle. Draw a right triangle with angle θ and label the sides based on your substitution. For x = a sinθ: opposite side = x, hypotenuse = a, adjacent = √(a²−x²). Read sinθ, cosθ, tanθ, secθ directly from the triangle. For example, if you get cos(2θ) in your answer, use cos(2θ) = 1 − 2sin²θ = 1 − 2x²/a² to express the result in x.
Each substitution exploits one of the three Pythagorean identities: (1) sin²θ + cos²θ = 1 is used when x = a sinθ, because a²−x² = a²−a²sin²θ = a²(1−sin²θ) = a²cos²θ. (2) 1 + tan²θ = sec²θ is used when x = a tanθ, because a²+x² = a²+a²tan²θ = a²(1+tan²θ) = a²sec²θ. (3) sec²θ − 1 = tan²θ is used when x = a secθ, because x²−a² = a²sec²θ−a² = a²(sec²θ−1) = a²tan²θ.
Completing the square rewrites a quadratic ax²+bx+c into the form a(x−h)²+k, which may match one of the three radical forms needed for trig substitution. For example, x²+6x+13 = (x+3)²+4 = u²+4 where u = x+3 and a = 2. This expression then has the form u²+a², so you apply the substitution u = 2 tanθ. Always complete the square when you see a general quadratic under a square root.
When you substitute x = a sinθ, differentiate both sides: dx = a cosθ dθ. For x = a tanθ: dx = a sec²θ dθ. For x = a secθ: dx = a secθ tanθ dθ. You must replace every x and every dx in the integrand before integrating. The combination of the substituted radical and the dx term typically produces a power of cosθ or secθ that simplifies the integrand considerably.
Restricting θ ensures the substitution is invertible (one-to-one) and the radical simplification is valid. For x = a sinθ: restrict θ to [−π/2, π/2] so cosθ ≥ 0 and √(a²cos²θ) = a cosθ (not −a cosθ). For x = a tanθ: restrict θ to (−π/2, π/2) so secθ > 0. For x = a secθ: restrict θ to [0, π/2) ∪ (π/2, π] so tanθ has the correct sign. Without these restrictions, you can introduce sign errors in back-substitution.
Yes. Trig substitution also works on integrands of the form 1/(a²+x²)ⁿ or (a²−x²)ⁿ for integer n ≥ 1 — even when no square root appears. For instance, ∫1/(a²+x²) dx is handled with x = a tanθ, giving ∫(1/a²sec²θ)(a sec²θ dθ) = (1/a)∫dθ = θ/a + C = (1/a)arctan(x/a) + C. This is one of the standard integration formulas derived from trig substitution.
Integrating powers of secant is a common challenge after trig substitution with x = a secθ. Key results: ∫secθ dθ = ln|secθ + tanθ| + C. ∫sec²θ dθ = tanθ + C. ∫sec³θ dθ = (1/2)(secθ tanθ + ln|secθ + tanθ|) + C (derived by integration by parts). Higher powers use the reduction formula: ∫secⁿθ dθ = [secⁿ⁻²θ tanθ]/(n−1) + [(n−2)/(n−1)]∫secⁿ⁻²θ dθ.
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