Trigonometric Substitution Tool

Master integrals containing √(a²±x²) and √(x²-a²) using step‑by‑step trig substitution. Visualize with the reference triangle.

Standard substitutions:

  • √(a² – x²) → x = a·sinθ , dx = a·cosθ dθ
  • √(x² + a²) → x = a·tanθ , dx = a·sec²θ dθ
  • √(x² – a²) → x = a·secθ , dx = a·secθ·tanθ dθ
Examples: sqrt(a^2 - x^2), 1/(x^2+a^2)^(3/2), sqrt(x^2-a^2)/x
√(a² – x²)
1/√(x²+a²)
√(x² – a²)
x²/√(a²–x²)
1/(x²+a²)
If blank, 'a' remains as symbol.
Performing substitution...

Understanding Trigonometric Substitution

Trigonometric substitution is a powerful technique for integrals involving radicals of the forms √(a²−x²), √(x²+a²), or √(x²−a²). By substituting x with a trigonometric function of θ, the radical simplifies using Pythagorean identities, transforming the integral into a trigonometric integral that is often easier to evaluate.

Three core cases:

Form Substitution Differential Identity used Range of θ (typical)
√(a²−x²) x = a sinθ dx = a cosθ dθ 1−sin²θ = cos²θ −π/2 ≤ θ ≤ π/2
√(x²+a²) x = a tanθ dx = a sec²θ dθ 1+tan²θ = sec²θ −π/2 < θ < π/2
√(x²−a²) x = a secθ dx = a secθ tanθ dθ sec²θ−1 = tan²θ 0 ≤ θ < π/2 or π/2 < θ ≤ π

The ranges ensure the square root can be written without absolute values (e.g., cosθ ≥ 0 for sin substitution).

Why These Substitutions Work

Each substitution is chosen to turn the expression under the square root into a perfect square:

  • √(a²−x²): With x = a sinθ, we have a²−x² = a²(1−sin²θ) = a² cos²θ → √ = a|cosθ| = a cosθ (for θ in [−π/2, π/2]).
  • √(x²+a²): With x = a tanθ, x²+a² = a²(tan²θ+1) = a² sec²θ → √ = a|secθ| = a secθ (for θ in (−π/2, π/2) where secθ > 0).
  • √(x²−a²): With x = a secθ, x²−a² = a²(sec²θ−1) = a² tan²θ → √ = a|tanθ|. For θ in (0, π/2) tanθ ≥ 0.

Step‑by‑Step Process

1
Identify the form – Look for √(a²−x²), √(x²+a²), or √(x²−a²). If necessary, complete the square first.
2
Choose the substitution – Use the table above to replace x and dx.
3
Simplify the integrand – Apply the corresponding identity to eliminate the square root. The result will be a trigonometric integral.
4
Integrate with respect to θ – Use standard trigonometric integral formulas.
5
Back‑substitute using the reference triangle – Draw the right triangle that defines the substitution; then express all trigonometric functions of θ in terms of x and a.

Detailed Examples

Let x = a sinθ, dx = a cosθ dθ. Then √(a²−x²) = √(a²−a² sin²θ) = a√(cos²θ) = a cosθ (for −π/2 ≤ θ ≤ π/2).

Integral becomes ∫ a cosθ · a cosθ dθ = a² ∫ cos²θ dθ = a² ∫ (1+cos2θ)/2 dθ = a²(θ/2 + sin2θ/4) + C.

Now back‑substitute: sinθ = x/a → θ = arcsin(x/a). From the triangle, cosθ = √(a²−x²)/a, and sin2θ = 2 sinθ cosθ = 2(x/a)(√(a²−x²)/a).

Thus final result: ∫ √(a²−x²) dx = (a²/2) arcsin(x/a) + (x/2)√(a²−x²) + C.

Let x = a tanθ, dx = a sec²θ dθ. Then √(x²+a²) = √(a² tan²θ + a²) = a√(sec²θ) = a secθ (for −π/2 < θ < π/2).

Integral becomes ∫ (a sec²θ dθ) / (a secθ) = ∫ secθ dθ = ln|secθ + tanθ| + C.

From the triangle: tanθ = x/a, secθ = √(x²+a²)/a. So back‑substitution gives ∫ dx/√(x²+a²) = ln| x/a + √(x²+a²)/a | + C = ln| x + √(x²+a²) | − ln|a| + C.

The constant −ln|a| can be absorbed into C, so final answer: ln| x + √(x²+a²) | + C.

Let x = a secθ, dx = a secθ tanθ dθ. Then √(x²−a²) = √(a² sec²θ − a²) = a√(tan²θ) = a tanθ (for 0 < θ < π/2).

The integrand becomes (a tanθ) / (a secθ) · a secθ tanθ dθ = a tan²θ dθ = a (sec²θ − 1) dθ.

Integral: a ∫ (sec²θ − 1) dθ = a (tanθ − θ) + C.

From the triangle: tanθ = √(x²−a²)/a, and secθ = x/a ⇒ θ = arcsec(x/a). So final result: ∫ √(x²−a²)/x dx = √(x²−a²) − a arcsec(x/a) + C.

Handling Definite Integrals

For definite integrals, you can either:

  • Change the limits to θ – using the substitution equation x = a·trig(θ).
  • Integrate in θ and then back‑substitute – evaluate the resulting expression at the original x‑limits.

Remember to adjust the range of θ so that the substitution is one‑to‑one on the interval of integration.

When to Use Trigonometric vs. Hyperbolic Substitution

For integrals involving √(x²+a²) or √(x²−a²), hyperbolic substitutions (x = a sinh t or x = a cosh t) can sometimes be more efficient because the hyperbolic identities are simpler (cosh²t−sinh²t = 1). However, trigonometric substitution is more traditional and often preferred when the final answer is expected in terms of inverse trigonometric functions. Both methods are valid.

Calculator features: symbolic manipulation with nerdamer.js, dynamic reference triangle, step display. Works for many algebraic combinations. The tool shows the substitution and simplification steps; the actual integration with respect to θ can be performed using our Integral Calculator.

Frequently Asked Questions

When the integrand contains √(a²−x²), √(x²+a²), or √(x²−a²). It also helps for quadratic forms like (x²+a²)ⁿ with half‑integer exponents.

Complete the square first. For example, √(2x−x²) can be rewritten as √(1−(x−1)²) and then substitute u = x−1.

You can either change the limits to θ using the substitution equation, or integrate in θ and convert back to x before evaluating.

This tool focuses on the substitution step and simplification. For actual integration of the θ‑integral, you can use our Integral Calculator (see related tools).