Inverse Hyperbolic Functions
Inverse Hyperbolic Functions
The inverse hyperbolic functions — arsinh, arcosh, and artanh — arise naturally in integration and are directly expressible as logarithms. This logarithmic form is what makes them particularly useful.
Definitions and Domains
| Function | Definition (y = ...) | Means | Domain | Range | ||
|---|---|---|---|---|---|---|
| arsinh x | y such that sinh y = x | — | All reals | All reals | ||
| arcosh x | y such that cosh y = x, y ≥ 0 | — | x ≥ 1 | y ≥ 0 | ||
| artanh x | y such that tanh y = x | — | \ | x\ | < 1 | All reals |
Note: The prefix is ar (area), not arc. The notation arsinh, arcosh, artanh is standard in UK A-Level. You may also see sinh⁻¹, cosh⁻¹, tanh⁻¹.
arcosh requires the restriction y ≥ 0 because cosh is not one-to-one on all of ℝ — we take the positive branch only.
Logarithmic Forms (Must Know)
These are derived by setting y = arsinh x, so sinh y = x, converting to exponentials, and solving the resulting quadratic.
| Function | Logarithmic form | ||
|---|---|---|---|
| arsinh x | ln(x + √(x² + 1)) | ||
| arcosh x | ln(x + √(x² − 1)), x ≥ 1 | ||
| artanh x | (1/2) ln((1 + x)/(1 − x)), \ | x\ | < 1 |
Derivation of arsinh x = ln(x + √(x² + 1))
Let y = arsinh x, so sinh y = x:
- (e^y − e^(−y))/2 = x
- e^y − e^(−y) = 2x
- Let u = e^y: u − 1/u = 2x → u² − 2xu − 1 = 0
- u = (2x ± √(4x² + 4))/2 = x ± √(x² + 1)
- Since u = e^y > 0, and √(x² + 1) > |x|, we must take the + sign:
- u = x + √(x² + 1)
- y = ln(x + √(x² + 1))
The derivation for arcosh is similar but with x² − 1 under the root.
Derivatives
| Function | Derivative | ||
|---|---|---|---|
| arsinh x | 1/√(x² + 1) | ||
| arcosh x | 1/√(x² − 1), x > 1 | ||
| artanh x | 1/(1 − x²), \ | x\ | < 1 |
These can be verified by differentiating the logarithmic forms or by implicit differentiation.
Proof for arsinh x: Let y = arsinh x, so sinh y = x.
Differentiating implicitly: cosh y · dy/dx = 1
dy/dx = 1/cosh y = 1/√(1 + sinh²y) = 1/√(1 + x²)
(using cosh²y − sinh²y = 1 so cosh y = √(1 + sinh²y) = √(1 + x²), taking the positive root since cosh > 0).
Integration Results
These derivatives give the following standard integrals (essential for Further Maths):
| Integral | Result | ||
|---|---|---|---|
| ∫ 1/√(x² + a²) dx | arsinh(x/a) + C = ln(x + √(x² + a²)) + C | ||
| ∫ 1/√(x² − a²) dx | arcosh(x/a) + C = ln(x + √(x² − a²)) + C, x > a | ||
| ∫ 1/(a² − x²) dx | (1/a) artanh(x/a) + C, \ | x\ | < a |
These extend the standard results ∫1/√(a² − x²) dx = arcsin(x/a) + C from core maths.
Integration by Completing the Square
Many integrals involving quadratics under a square root reduce to the standard forms above after completing the square.
Example: ∫ 1/√(x² + 6x + 13) dx
Step 1: Complete the square: x² + 6x + 13 = (x + 3)² + 4
Step 2: Substitute u = x + 3:
∫ 1/√(u² + 4) du = arsinh(u/2) + C = arsinh((x+3)/2) + C
Or equivalently: ln(x + 3 + √(x² + 6x + 13)) + C
Example: ∫ 1/√(4x² − 8x + 3) dx
Step 1: Factor out 4: 4(x² − 2x) + 3 = 4(x−1)² − 4 + 3 = 4(x−1)² − 1
Step 2: ∫ 1/√(4(x−1)² − 1) dx = (1/2) ∫ 1/√((x−1)² − 1/4) dx
Step 3: = (1/2) arcosh(2(x−1)) + C = (1/2) arcosh(2x − 2) + C
Graphs of Inverse Hyperbolic Functions
- arsinh x: passes through the origin, odd function, shape similar to ln x for large x but defined for all x
- arcosh x: defined for x ≥ 1, starts at (1, 0) and increases, concave down
- artanh x: defined for −1 < x < 1, passes through origin, vertical asymptotes at x = ±1
Connection to Standard Integrals Table
| Integrand shape | Method | Result involves |
|---|---|---|
| 1/√(a² − x²) | Trig sub x = a sin θ | arcsin |
| 1/√(x² + a²) | Hyp sub x = a sinh t | arsinh |
| 1/√(x² − a²) | Hyp sub x = a cosh t | arcosh |
| 1/(a² − x²) | Partial fractions or hyp | artanh |
| 1/(x² + a²) | Trig sub x = a tan θ | arctan |
Exam Tips
- Memorise the three logarithmic forms — they are frequently needed and quicker than deriving each time
- The derivative of arsinh x involves x² + 1; the derivative of arcosh involves x² − 1 — don't mix them up
- Always complete the square before applying the standard integral forms
- When an answer involves arsinh or arcosh, the question may ask you to express in logarithmic form — be ready to convert
- Remember that arcosh has a restricted domain (x ≥ 1) — check your limits of integration are valid
- artanh has vertical asymptotes at x = ±1 and is only defined between them