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Integrals and Inverse Hyperbolic Functions

Calculus 2 · Chapter 3: Hyperbolic FunctionsSection 3.4 · lesson 15 of 5412 min · free

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In this lesson

Running every derivative backwards - and the catenary that started it.

  • Read the six derivative rules backwards as antiderivatives
  • Express each inverse hyperbolic function as a logarithm
  • Use them to resolve \sqrt{x^{2}+a^{2}} and \sqrt{x^{2}-a^{2}} integrals
  • Compute the length of a hanging cable in closed form

Transcript

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MAT-265 - Calculus 2 · Section 3.4

Integrals and Inverse Hyperbolic Functions

A structural engineer designing a pedestrian bridge needs three things. The exact shape of the supporting cable. The length of material to order. And the tension at the anchor points. All three flow from a single function — the hyperbolic cosine — and from the calculus that surrounds it. Working out the cable's length, in particular, leads to an integral that no ordinary trig substitution resolves. But the inverse hyperbolic functions handle it cleanly. The previous section established the derivatives of all six hyperbolic functions. And every derivative formula, read backward, is an antiderivative formula. So the differentiation rules immediately hand over a whole family of integrals — for free.

Reversing the Rules

Every Derivative Formula Is an Antiderivative Formula

Let's collect them. Reversing each differentiation rule produces a basic integral. The integral of cosh x is sinh x plus C. The integral of sinh x is cosh x plus C. The integral of sech squared x is tanh x. The integral of csch squared x is minus coth x. The integral of sech x tanh x is minus sech x. And the integral of csch x coth x is minus csch x. These mirror the circular integration formulas with predictable sign differences. And they combine with substitution exactly as their circular cousins do. Let's do two quick ones. Evaluate the integral of x cosh of x squared. The inner function x squared suggests u equals x squared, so d u is two x d x — meaning x d x is one half d u. The integral becomes one half the integral of cosh u, which is one half sinh of x squared, plus C. Second. Evaluate the integral of tanh x. Rewrite tanh as sinh over cosh, and substitute u equals cosh x, so d u is sinh x d x. You get the integral of d u over u, which is log absolute u. So log of cosh x, plus C. And notice — no absolute value is needed here, because cosh x is positive for every x. Compare that to the circular result: the integral of tangent x is minus log absolute cosine x. Again, the hyperbolic version has the sign and the absolute value smoothed out.

The Inverses

Every One Is a Logarithm

Now the second family. The inverse hyperbolic functions. Sinh is one-to-one across its entire domain — it's strictly increasing — so it has an inverse defined for all real numbers. Cosh is not one-to-one. Being even, it fails the horizontal line test. So its inverse is defined only on the principal branch, x at least one, with output y at least zero. And tanh is one-to-one, with inverse defined on the open interval from minus one to one. Now here's a striking feature that sets these apart from inverse circular functions. Each one can be written explicitly as a logarithm. And that's no accident. The hyperbolic functions are built from exponentials — so their inverses are built from logarithms. Let's derive one. Start from x equals sinh y, which is one half of e to the y minus e to the minus y. Multiply through by two e to the y to clear the negative exponential. You get e to the two y, minus two x e to the y, minus one, equals zero. And that is a quadratic in e to the y. Apply the quadratic formula with w equals e to the y, and you get w equals x, plus or minus root x squared plus one. Now — which root? Since w is e to the y, it must be positive. And root x squared plus one is always bigger than the absolute value of x. So only the plus root is valid. Take the natural log, and you have it. Inverse sinh of x equals log of x plus root x squared plus one. For all real x. The same approach gives the other two formulas.

Figure 3.4.1

Graphs of the Principal Inverses

Look at the three graphs, because the domains tell the story. On the left, inverse sinh. Defined for all x, increasing smoothly through the origin. It's the mirror image of sinh across the line y equals x, and since sinh was defined everywhere and unbounded, its inverse is too. In the middle, inverse cosh. It exists only for x at least one, starting at the point one comma zero. That's because cosh has range one to infinity — so its inverse can only accept inputs from that range. On the right, inverse tanh. Defined on minus one to one, with vertical asymptotes at x equals plus and minus one. And notice the elegance there. Tanh had horizontal asymptotes at y equals plus and minus one. Its inverse has vertical asymptotes at x equals plus and minus one. Reflecting across y equals x turns one into the other.

The Payoff

Integrals the Radicals Would Not Give Up

Now the real payoff. Differentiating the logarithmic formulas — or using implicit differentiation on x equals sinh y — yields compact derivative rules. The derivative of inverse sinh x is one over root x squared plus one. The derivative of inverse cosh x is one over root x squared minus one, for x greater than one. And the derivative of inverse tanh x is one over one minus x squared, for absolute x less than one. Now read those backwards as integration formulas. The integral of d x over root x squared plus one is inverse sinh x. The integral of d x over root x squared minus one is inverse cosh x. And the integral of d x over one minus x squared is inverse tanh x. More general versions, obtained by substitution, handle a positive constant a. The integral of d x over root x squared plus a squared is inverse sinh of x over a. These resolve integrals whose radicands — root x squared plus a squared, and root x squared minus a squared — resist ordinary methods. Now here's something worth thinking about. Ordinary trig substitution can also evaluate the integral of d x over root x squared plus nine, using x equals three tangent theta. But that work is longer, and it ends in a logarithm anyway. So why do the inverse hyperbolic formulas feel so much cleaner here? Because root x squared plus a squared is the natural companion of the fundamental identity, rearranged as one plus sinh squared equals cosh squared. The radical and the identity were built for each other. So the substitution leaves no leftover trigonometric cleanup at all.

Worked Examples 3 and 4

Indefinite, Then Definite

Let's do one of each. Indefinite, then definite. Evaluate the integral of d x over root x squared plus nine. Here a squared is nine, so a is three. Apply the formula directly and you get inverse sinh of x over three, plus C. Using the logarithmic form instead, this equals log of x plus root x squared plus nine, plus a different constant. The two answers differ only by log three, which the constant absorbs. Either form is acceptable. Now the definite version. The integral from zero to four of the same thing. Using the antiderivative just found, evaluate inverse sinh of x over three, from zero to four. That's inverse sinh of four thirds, minus inverse sinh of zero. And here the logarithmic form earns its keep, because you need actual numbers. Inverse sinh of zero is log of zero plus root one, which is log one. Which is zero. Inverse sinh of four thirds is log of four thirds, plus root of sixteen ninths plus one. That inner root is root twenty-five ninths, which is five thirds. So you get log of four thirds plus five thirds. Which is log of nine thirds. Which is log three. The answer is log three. About one point zero nine eight six.

Application

The Catenary, and the Length of a Cable

Let's close with the curve that motivated this whole chapter. A perfectly flexible cable of uniform weight, suspended from two points and hanging freely, settles into a shape governed by a balance. Gravity pulling down. Tension along the cable. Solving the resulting differential equation produces the catenary. y equals a cosh of x over a. Here x is horizontal distance from the lowest point, and a is a positive constant set by the cable's weight per unit length and its horizontal tension. The lowest point sits at zero comma a, since cosh zero is one. A catenary is often mistaken for a parabola. Look at the picture — near the bottom they're genuinely similar. But the catenary rises more steeply toward the supports. And that distinction matters in engineering. A freely hanging cable is a catenary. But a cable supporting a uniformly distributed horizontal load — like a suspension bridge deck — is closer to a parabola. Now let's find the cable's length. And watch what happens. Arc length is the integral of root one plus y prime squared. Differentiate: y prime is a sinh of x over a, times one over a. Which is just sinh of x over a. So one plus y prime squared is one plus sinh squared of x over a. And here the fundamental identity collapses the radical beautifully. One plus sinh squared equals cosh squared. So the square root is simply cosh of x over a. Positive, because cosh always is. The messy arc-length integral has become the integral of a single cosh. Integrate, use the oddness of sinh, and the length is two a sinh of b over a. A closed-form answer an engineer can evaluate in one step, once a and the half-span b are known. That collapse — from a messy radical to a single cosh — is exactly why hyperbolic functions are the right language for this problem.

Key Takeaways

What to Carry Forward

Five things to carry forward. Every hyperbolic derivative, read backwards, is an antiderivative. The six basic integrals mirror the circular ones with predictable signs. The integral of tanh x is log of cosh x, with no absolute value — because cosh is always positive. Each inverse hyperbolic function is a logarithm. Inverse sinh is log of x plus root x squared plus one. Inverse cosh is log of x plus root x squared minus one, for x at least one. And inverse tanh is one half log of one plus x over one minus x. The inverses unlock integrals with root x squared plus a squared, and root x squared minus a squared, more cleanly than trig substitution — because the radical and the identity were made for each other. And a hanging cable is a cosh of x over a, with length two a sinh of b over a. The arc-length radical collapses to a single cosh. That closes Chapter Three. Next up is Chapter Four — Applications of Integration, beginning with areas between curves.

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FAQ

Questions students ask us

About Integrals and Inverse Hyperbolic Functions

How do you write inverse hyperbolic functions as logarithms?
Each inverse hyperbolic function equals a natural logarithm, because the hyperbolic functions are built from exponentials. Specifically, \sinh^{-1}x = \ln(x+\sqrt{x^2+1}) for all real x, \cosh^{-1}x = \ln(x+\sqrt{x^2-1}) for x \ge 1, and \tanh^{-1}x = \frac{1}{2}\ln\frac{1+x}{1-x} for |x|<1. To derive the first, set x=\sinh y, multiply by 2e^y, and solve the quadratic e^{2y}-2xe^y-1=0 for e^y, keeping only the positive root.
How do you integrate 1 over the square root of x^2 + a^2?
Use \int \frac{dx}{\sqrt{x^2+a^2}} = \sinh^{-1}(x/a)+C, which equals \ln(x+\sqrt{x^2+a^2})+C with a different constant. For example, \int_0^4 \frac{dx}{\sqrt{x^2+9}} = \sinh^{-1}(4/3)-\sinh^{-1}(0). The logarithmic form gives \ln(4/3+5/3) - \ln 1 = \ln 3 \approx 1.0986. Similarly, \int \frac{dx}{\sqrt{x^2-a^2}} = \cosh^{-1}(x/a)+C for x>a. This approach is shorter than trig substitution with x = 3\tan\theta.
What is the arc length of a hanging cable (catenary)?
A cable hanging as y = a\cosh(x/a) from x=-b to x=b has length 2a\sinh(b/a). Since y' = \sinh(x/a), the arc-length integrand becomes \sqrt{1+\sinh^2(x/a)} = \cosh(x/a) by the identity 1+\sinh^2 = \cosh^2. Integrating gives a\sinh(x/a). Because sinh is odd, evaluating from -b to b doubles a\sinh(b/a). Here b is the half-span, and a depends on the cable's weight per unit length and its horizontal tension.
What is the integral of tanh x, and does it need an absolute value?
\int \tanh x\,dx = \ln(\cosh x)+C, and no absolute value is needed because \cosh x>0 for every x. Write \tanh x = \sinh x/\cosh x and substitute u=\cosh x, du=\sinh x\,dx, which gives \int du/u. A common slip is copying the circular pattern too closely. The formula \int\tan x\,dx = -\ln|\cos x|+C has a minus sign and an absolute value, but the hyperbolic version has neither.

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