Integrals That Give arsinh and arcosh

A plus inside the root turns the answer over.

Differentiating arsinh x

Let y = arsinh x, so that x = sinh y. Differentiating with respect to y gives dx/dy = cosh y, and turning that over gives dy/dx = 1 / cosh y.

To write that in terms of x, use the identity cosh²y = 1 + sinh²y. Since cosh y is always positive, cosh y = √(1 + sinh²y) = √(1 + x²). So arsinh x differentiates to 1/√(x² + 1).

arcosh x works the same way. With x = cosh y and y ≥ 0, dx/dy = sinh y, and sinh y = √(cosh²y − 1) = √(x² − 1). So arcosh x differentiates to 1/√(x² − 1), for x > 1.

x² + y² = 1area = t/2 = 0.75x² − y² = 1area = t/2 = 0.75(cosh t, sinh t) = (2.35, 2.13)t = 1.5

the hyperbolic sector and the circular sector both have area t/2, so t is an angle for the hyperbola: (cosh t, sinh t) is to x² − y² = 1 what (cos t, sin t) is to x² + y² = 1

Sweep t until both sectors have area ½

The point (cosh t, sinh t) runs along the hyperbola x² − y² = 1, as (cos t, sin t) runs round the circle x² + y² = 1. At t = 1.5 it is at (2.35, 2.13). Every point of the hyperbola has cosh²t − sinh²t = 1, the identity that turns √(1 + sinh²y) into cosh y. Drag t to 1, where both shaded sectors have area ½.

Reading the derivatives backwards

An integral is a derivative read backwards. Since arsinh x differentiates to 1/√(x² + 1), ∫ dx/√(x² + 1) = arsinh x + c. Since arcosh x differentiates to 1/√(x² − 1), ∫ dx/√(x² − 1) = arcosh x + c, for x > 1.

They sit beside ∫ dx/√(1 − x²) = arcsin x + c, which comes from differentiating arcsin x the same way. The three roots differ only in their signs.

A general a

For ∫ dx/√(x² + a²), with a > 0, substitute x = a sinh u. Then dx = a cosh u du, and the root becomes √(a²sinh²u + a²) = a√(sinh²u + 1) = a cosh u. The a cosh u from dx cancels the a cosh u from the root, and the integral is ∫ du = u + c.

Since x = a sinh u, u = arsinh(x/a). So ∫ dx/√(x² + a²) = arsinh(x/a) + c. In the same way, x = a cosh u gives ∫ dx/√(x² − a²) = arcosh(x/a) + c, for x > a.

Check by differentiating. The chain rule gives 1/a times 1/√(x²/a² + 1), and √(x²/a² + 1) = √(x² + a²) / a, so the a cancels and the derivative is 1/√(x² + a²). For ∫ dx/√(x² + 16), a = 4 and the answer is arsinh(x/4) + c.

Compare ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + c, where a factor of 1/a stands in front. Here there is none: the root of a² is a, and that a cancels the a in dx = a cosh u du.

xy

The area under y = 1/√(x² + 1) from 0 to 4/3 is arsinh(4/3). The integral of 1/√(x² + 9) from 0 to 4 is the same number, arsinh(4/3), because the a = 3 only scales the input.

Which family

The sign inside the root decides. In √(9 − x²) the x² is taken away, and x = 3 sin θ clears the root, since 9 − 9 sin²θ = 9 cos²θ. So ∫ dx/√(9 − x²) = arcsin(x/3) + c.

In √(9 + x²) the x² is added, and x = 3 sinh u clears the root, since 9 + 9 sinh²u = 9 cosh²u. So ∫ dx/√(9 + x²) = arsinh(x/3) + c.

In √(x² − 9) the 9 is taken away from the x², and x = 3 cosh u clears the root, since 9 cosh²u − 9 = 9 sinh²u. So ∫ dx/√(x² − 9) = arcosh(x/3) + c, for x > 3.

A quadratic in another shape can be brought to one of these by completing the square. x² + 4x + 13 = (x + 2)² + 9, a square plus 9, so ∫ dx/√(x² + 4x + 13) = arsinh((x + 2)/3) + c.

As a logarithm

Solving x = sinh y for y gives arsinh x = ln(x + √(x² + 1)), and in the same way arcosh x = ln(x + √(x² − 1)) for x ≥ 1. So arsinh(x/a) = ln(x/a + √(x²/a² + 1)) = ln(x + √(x² + a²)) − ln a.

The − ln a is a constant, so it joins the c. ∫ dx/√(x² + a²) = ln(x + √(x² + a²)) + c is an equally correct answer, and so is ∫ dx/√(x² − a²) = ln(x + √(x² − a²)) + c, for x > a.

A definite integral comes out as a logarithm either way. The integral of 1/√(x² + 9) from 0 to 4 is arsinh(4/3) − arsinh 0. Since √(16 + 9) = 5, the logarithm form gives ln(4 + 5) − ln(0 + 3) = ln 9 − ln 3 = ln 3, about 1.099.

The integral of 1/√(x² − 144) from 13 to 20 is arcosh(20/12) − arcosh(13/12). Here √(400 − 144) = 16 and √(169 − 144) = 5, so it is ln(20 + 16) − ln(13 + 5) = ln 36 − ln 18 = ln 2, about 0.693. The two − ln 12 terms cancel.

The usual mistakes

Using arsinh for a minus inside. √(9 − x²) gives arcsin(x/3); arsinh needs the x² added.

Dropping the a. ∫ dx/√(x² + 16) is arsinh(x/4) + c; arsinh x + c differentiates to 1/√(x² + 1), the integrand with 1 in place of 16.

Putting the a in front. 4 arsinh(x/4) + c differentiates to 4/√(x² + 16), four times the integrand.

Mixing up the two hyperbolic roots. √(x² + 1) gives arsinh; √(x² − 1) gives arcosh, and only for x > 1.

Two beads on a wire

In the application below, a bead slides along a spinning wire and the time it takes is the integral of one over its speed. One bead’s speed has a root of r² − 0.16 and the other’s a root of r² + 0.16, so one time is an arcosh and the other an arsinh, each written as a logarithm.

Worked example: A Bead Sliding Out Along a Spinning Wire: Two Ways of Starting, and How Long Each Takes to Reach the End

Question A bead is threaded on a straight wire 1.2 meters long, which a motor spins in a horizontal plane about one end at a steady 2 radians per second. The bead slides along the wire without friction. When the bead is r meters from the axis, its speed v meters per second along the wire satisfies v2 − 4r2 = k, where the constant k is fixed by how the bead starts. (a) The bead is held 0.4 meters from the axis, at rest on the wire, and let go. How long does it take to reach the far end of the wire? (b) On an identical wire spun the same way, a second bead starts at the axis and is flicked outward along the wire at 0.8 meters per second, at the instant the first bead is let go. How long does it take to reach its far end, and which bead gets there first? Give each time exactly as a logarithm, then to three significant figures.

  1. 1.(a) At the start v = 0 and r = 0.4, so k = 0 − 4 × 0.16 = −0.64. Then v2 = 4r2 − 0.64 = 4(r2 − 0.16), so v = 2√r2 − 0.16, and the time to go from r = 0.4 to r = 1.2 is ∫0.41.2dr2√r2 − 0.16 seconds.

    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mv2= 4r2− 0.64, from rest at r = 0.4
    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mv2= 4r2− 0.64, from rest at r = 0.4
    (a) From rest at r = 0.4, k = −0.64, so v = 2√r2 − 0.16 and the time is ∫0.41.2dr2√r2 − 0.16: the area under 1v.

  2. 2.A square minus a constant under the root: substitute r = 0.4cosh u. Then dr = 0.4sinh u du and √r2 − 0.16 = 0.4√cosh2 u − 1 = 0.4sinh u, so the integrand becomes 0.4sinh u2 × 0.4sinh udu = 12du. Done once and for all, this is ∫dr√r2 − a2 = arcoshra + C.

    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2
    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2
    Substitute r = 0.4cosh u: then √r2 − 0.16 = 0.4sinh u and dr = 0.4sinh u du, so the integrand is 12du.

  3. 3.(a) The limits travel with the substitution: r = 0.4 gives cosh u = 1, so u = 0, and r = 1.2 gives cosh u = 3, so u = arcosh 3 = ln(3 + √8). The time is 12ln(3 + 2√2) = ln(1 + √2) = 0.881 seconds, since (1 + √2)2 = 3 + 2√2. The integrand is infinite at r = 0.4, where the bead is at rest, but the area under it is finite.

    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 m0.881 sv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2(a) half of arcosh 3 = 0.881 s
    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 m0.881 sv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2(a) half of arcosh 3 = 0.881 s
    (a) The limits become u = 0 and u = arcosh 3, so the time is 12ln(3 + √8) = ln(1 + √2) = 0.881 seconds.

  4. 4.(b) The second bead has v = 0.8 at r = 0, so k = 0.64 − 0 = 0.64 and v = 2√r2 + 0.16. The sign inside the root has turned over, so substitute r = 0.4sinh u: then dr = 0.4cosh u du and √r2 + 0.16 = 0.4√sinh2 u + 1 = 0.4cosh u, and again the integrand is 12du. Done once and for all, this is ∫dr√r2 + a2 = arsinhra + C.

    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mflicked from the axis0.881 sv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2(a) half of arcosh 3 = 0.881 sv2= 4r2+ 0.64, flicked at 0.8 m/sr = 0.4 sinh u makes dr/v = du/2
    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mflicked from the axis0.881 sv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2(a) half of arcosh 3 = 0.881 sv2= 4r2+ 0.64, flicked at 0.8 m/sr = 0.4 sinh u makes dr/v = du/2
    (b) Flicked from the axis at 0.8 m/s, k = 0.64, so v = 2√r2 + 0.16; the plus sign calls for r = 0.4sinh u, and again the integrand is 12du.

  5. 5.(b) The limits: r = 0 gives u = 0, and r = 1.2 gives sinh u = 3, so u = arsinh 3 = ln(3 + √10). The time is 12ln(3 + √10) = 0.909 seconds, so the bead let go from rest reaches its end first, by 0.028 seconds: the flicked bead starts faster but has 1.2 meters to cover instead of 0.8. Check: since the time is u2, the positions are r = 0.4cosh 2n and r = 0.4sinh 2n after n seconds, and 0.4cosh(2 × 0.881) = 1.20 and 0.4sinh(2 × 0.909) = 1.20.

    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mflicked from the axis0.881 s0.909 sv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2(a) half of arcosh 3 = 0.881 sv2= 4r2+ 0.64, flicked at 0.8 m/sr = 0.4 sinh u makes dr/v = du/2(b) half of arsinh 3 = 0.909 s, 0.028 s later
    012300.40.81.2r, meters from the axis1/v, seconds per meter012300.40.81.2r, meters from the axis1/v, seconds per meterfrom rest at 0.4 mflicked from the axis0.881 s0.909 sv2= 4r2− 0.64, from rest at r = 0.4r = 0.4 cosh u makes dr/v = du/2(a) half of arcosh 3 = 0.881 sv2= 4r2+ 0.64, flicked at 0.8 m/sr = 0.4 sinh u makes dr/v = du/2(b) half of arsinh 3 = 0.909 s, 0.028 s later
    (b) The time is 12arsinh 3 = 12ln(3 + √10) = 0.909 seconds, so the bead let go from rest arrives first, by 0.028 seconds.

Answer: (a) ln(1 + √2) = 0.881 seconds; (b) 12ln(3 + √10) = 0.909 seconds, so the bead let go from rest gets there first, by 0.028 seconds

Common mistakes

  • Using arsinh for both beads. The sign inside the root decides: r2 − 0.16 needs r = 0.4cosh u and gives arcosh, while r2 + 0.16 needs r = 0.4sinh u and gives arsinh. Putting arsinh into part (a) gives 12(arsinh 3 − arsinh 1) = 0.469 seconds, about half the true time.
  • Leaving out the 2 in front of the root. The speed is twice the root, so the time is half the integral of one over the root; dropping the 2 doubles both answers, to 1.76 and 1.82 seconds.

More techniques of integration problems, worked step by step →

Practice Integrals That Give arsinh and arcosh in the app