Inverse Hyperbolic Functions

The inverse hyperbolic functions arise when solving equations like sinh x = k. Unlike inverse trig functions, they can be expressed as natural logarithms, making them very useful in integration.

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1 Definitions and Logarithmic Forms

arsinh x = ln(x + sqrt(x^2 + 1)) -- Domain: all reals, Range: all reals

arcosh x = ln(x + sqrt(x^2 - 1)) -- Domain: [1, infinity), Range: [0, infinity)

artanh x = (1/2) ln((1 + x)/(1 - x)) -- Domain: (-1, 1), Range: all reals

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2 Deriving arsinh x

Let y = arsinh x, so sinh y = x.

(e^y - e^(-y))/2 = x

Multiply by e^y: e^(2y) - 2x e^y - 1 = 0

Quadratic in e^y: e^y = x + sqrt(x^2 + 1) (taking the positive root since e^y > 0)

arsinh x = ln(x + sqrt(x^2 + 1))

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3 Deriving arcosh x

Let y = arcosh x, so cosh y = x (with y >= 0).

e^(2y) - 2x e^y + 1 = 0

e^y = x + sqrt(x^2 - 1) (taking the larger root for y >= 0)

arcosh x = ln(x + sqrt(x^2 - 1))

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4 Deriving artanh x

Let y = artanh x, so tanh y = x.

(e^(2y) - 1)/(e^(2y) + 1) = x

e^(2y) = (1 + x)/(1 - x)

artanh x = (1/2) ln((1 + x)/(1 - x))

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5 Worked Example 1

Find arsinh(3/4).

arsinh(3/4) = ln(3/4 + sqrt(9/16 + 1)) = ln(3/4 + 5/4) = ln 2

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6 Worked Example 2

Find arcosh(5/3).

arcosh(5/3) = ln(5/3 + sqrt(25/9 - 1)) = ln(5/3 + 4/3) = ln 3

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7 Worked Example 3

Find artanh(1/3).

artanh(1/3) = (1/2) ln((4/3)/(2/3)) = (1/2) ln 2 = ln(2)/2

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8 Graphs

y = arsinh x: Passes through origin, odd function, increasing, shape like a flattened cubic.

y = arcosh x: Starts at (1, 0), increasing, defined only for x >= 1.

y = artanh x: Passes through origin, odd function, vertical asymptotes at x = -1 and x = 1.

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9 Standard Integrals Involving Inverse Hyperbolic Functions

integral of 1/sqrt(x^2 + a^2) dx = arsinh(x/a) + C

integral of 1/sqrt(x^2 - a^2) dx = arcosh(x/a) + C (for x > a)

integral of 1/(a^2 - x^2) dx = (1/a) artanh(x/a) + C (for |x| < a)

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