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Inverse Hyperbolic Sine — Definition, Formula & Examples

The inverse hyperbolic sine, written arcsinh⁡(x)\operatorname{arcsinh}(x) or sinh⁡−1(x)\sinh^{-1}(x), is the function that reverses sinh⁡\sinh: if sinh⁡(y)=x\sinh(y) = x, then arcsinh⁡(x)=y\operatorname{arcsinh}(x) = y. It accepts all real numbers as input and appears frequently in integration results.

For any x∈Rx \in \mathbb{R}, the inverse hyperbolic sine is defined by the closed-form expression arcsinh⁡(x)=ln⁡ ⁣(x+x2+1)\operatorname{arcsinh}(x) = \ln\!\left(x + \sqrt{x^2 + 1}\right). Its derivative is ddxarcsinh⁡(x)=1x2+1\dfrac{d}{dx}\operatorname{arcsinh}(x) = \dfrac{1}{\sqrt{x^2+1}}.

Key Formula

arcsinh⁡(x)=ln⁡ ⁣(x+x2+1)\operatorname{arcsinh}(x) = \ln\!\left(x + \sqrt{x^2 + 1}\right)
Where:
  • xx = Any real number
  • ln⁡\ln = Natural logarithm (base $e$)

How It Works

You most often encounter arcsinh⁡\operatorname{arcsinh} as the result of an integral rather than something you set out to compute directly. The integral ∫dxx2+1\displaystyle\int \frac{dx}{\sqrt{x^2+1}} evaluates to arcsinh⁡(x)+C\operatorname{arcsinh}(x)+C, which is equivalent to ln⁡(x+x2+1)+C\ln(x+\sqrt{x^2+1})+C. Recognizing the pattern 1x2+a2\frac{1}{\sqrt{x^2+a^2}} in an integrand signals that an inverse hyperbolic sine (or a trigonometric substitution leading to the same logarithmic form) is the antiderivative. Because arcsinh⁡\operatorname{arcsinh} is defined for all real xx and is a strictly increasing, odd function, its behavior is straightforward — no domain restrictions to worry about.

Worked Example

Problem: Evaluate the definite integral ∫03dxx2+1\displaystyle\int_0^{3} \frac{dx}{\sqrt{x^2+1}}.
Identify the antiderivative: The integrand matches the derivative of arcsinh, so the antiderivative is arcsinh⁡(x)\operatorname{arcsinh}(x).
∫dxx2+1=arcsinh⁡(x)+C\int \frac{dx}{\sqrt{x^2+1}} = \operatorname{arcsinh}(x) + C
Apply the bounds: Evaluate using the logarithmic form at the upper and lower limits.
arcsinh⁡(3)−arcsinh⁡(0)=ln⁡ ⁣(3+10)−ln⁡(0+1)\operatorname{arcsinh}(3) - \operatorname{arcsinh}(0) = \ln\!\left(3+\sqrt{10}\right) - \ln(0+\sqrt{1})
Simplify: Since ln⁡(1)=0\ln(1)=0, only the upper-limit term remains.
ln⁡ ⁣(3+10)≈ln⁡(6.1623)≈1.8184\ln\!\left(3+\sqrt{10}\right) \approx \ln(6.1623) \approx 1.8184
Answer: ∫03dxx2+1=ln⁡ ⁣(3+10)≈1.8184\displaystyle\int_0^{3} \frac{dx}{\sqrt{x^2+1}} = \ln\!\left(3+\sqrt{10}\right) \approx 1.8184

Why It Matters

In Calculus II, integrals involving x2+a2\sqrt{x^2 + a^2} arise regularly in arc-length problems and trigonometric/hyperbolic substitution exercises. Recognizing the arcsinh⁡\operatorname{arcsinh} pattern lets you write the antiderivative immediately instead of completing a full substitution. Engineers also encounter it in catenary curve calculations and signal processing.

Common Mistakes

Mistake: Confusing 1x2+1\frac{1}{\sqrt{x^2+1}} with 1x2−1\frac{1}{\sqrt{x^2-1}} or 11−x2\frac{1}{\sqrt{1-x^2}}, which yield different inverse hyperbolic or trigonometric functions.
Correction: Check the sign inside the radical carefully: +1+1 gives arcsinh⁡\operatorname{arcsinh}, −1-1 (with x>1x>1) gives arccosh⁡\operatorname{arccosh}, and 1−x21-x^2 (with ∣x∣<1|x|<1) gives arcsin⁡\arcsin.