The three forms
| If the integral has | Substitute | dx | Identity | The root becomes |
|---|---|---|---|---|
| √(a² − x²) | x = a sin θ, −π/2 ≤ θ ≤ π/2 | a cos θ dθ | 1 − sin²θ = cos²θ | a cos θ |
| √(a² + x²) | x = a tan θ, −π/2 < θ < π/2 | a sec²θ dθ | 1 + tan²θ = sec²θ | a sec θ |
| √(x² − a²) | x = a sec θ, θ in [0, π/2) or (π/2, π] | a sec θ tan θ dθ | sec²θ − 1 = tan²θ | a |tan θ| |
The θ restrictions make the substitution reversible and fix the sign of the root. For the third form, tan θ is positive when x ≥ a and negative when x ≤ −a, so handle each case separately.
The steps
- Check first whether a simpler method works, such as a basic formula or u-substitution.
- Match the root to a form and substitute for x and dx.
- Simplify with the identity so the root disappears.
- Integrate the trig integral.
- Convert back to x using a right triangle built from the substitution.
Example 1: ∫ √(9 − x²) dx
- Form √(a² − x²) with a = 3: let x = 3 sin θ, dx = 3 cos θ dθ. Then √(9 − x²) = 3 cos θ.
- The integral becomes ∫ 3 cos θ · 3 cos θ dθ = ∫ 9 cos²θ dθ.
- Use cos²θ = (1 + cos 2θ)/2: (9/2)θ + (9/4) sin 2θ + C.
- Convert back. θ = sin−1(x/3); sin 2θ = 2 sin θ cos θ = 2 · (x/3) · √(9 − x²)/3.
- Answer: (9/2) sin−1(x/3) + (x/2)√(9 − x²) + C.
Example 2: ∫ dx / √(1 + x²)
- Form √(a² + x²) with a = 1: let x = tan θ, dx = sec²θ dθ. Then √(1 + x²) = sec θ.
- The integral becomes ∫ sec²θ / sec θ dθ = ∫ sec θ dθ = ln|sec θ + tan θ| + C. (See the integral of sec x.)
- Convert back: tan θ = x and sec θ = √(1 + x²).
- Answer: ln|√(1 + x²) + x| + C.
Example 3: ∫ √(x² − 25) / x dx, for x ≥ 5
- Form √(x² − a²) with a = 5: let x = 5 sec θ, dx = 5 sec θ tan θ dθ. For x ≥ 5, √(x² − 25) = 5 tan θ.
- The integral becomes ∫ (5 tan θ / 5 sec θ) · 5 sec θ tan θ dθ = ∫ 5 tan²θ dθ.
- Use tan²θ = sec²θ − 1: 5 ∫ (sec²θ − 1) dθ = 5 tan θ − 5θ + C.
- Convert back: 5 tan θ = √(x² − 25) and θ = sec−1(x/5).
- Answer: √(x² − 25) − 5 sec−1(x/5) + C.
Example 4: a definite integral
The area under y = √(x² − 9) from x = 3 to x = 5 uses x = 3 sec θ. Working it through gives an exact area of 10 − (9/2) ln 3 ≈ 5.056. With definite integrals you can either change the limits to θ-values or convert back to x and use the original limits.
Example 5: ∫ √(4 − x²) / x dx
With x = 2 sin θ, the integral becomes ∫ (2 csc θ − 2 sin θ) dθ = 2 ln|csc θ − cot θ| + 2 cos θ + C. Converting back gives 2 ln|(2 − √(4 − x²))/x| + √(4 − x²) + C.
Tips
- Complete the square first if you see something like √(x² + 2x + 5) = √((x + 1)² + 4).
- Always draw the triangle: label the sides from the substitution, then read off the other trig ratios.
- Check your answer by differentiating it.
Verify results with the integral calculator and the derivative calculator, or check angles with the inverse trig calculator.
Sources: OpenStax Calculus Volume 2, 3.3 Trigonometric Substitution; OpenStax Calculus Volume 2, Table of Integrals.
Every worked example on this page was checked with Calcist's own engine and an independent reference.