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Help with hyperbolic functions: cosh(arcsinh(x/2))=?

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to solve:
    1/4tanh(θ) + c


    2. Relevant equations
    x=2sinh(θ)
    θ = arcsinh(x/2)


    3. The attempt at a solution

    I worked out that since tanh(θ) = sin(θ)/cosh(θ)
    then
    1/4tanh(θ) + c = x/8cosh(θ) + c

    But i don't know how to work out cosh(θ) or cosh(arcsinh(x/2))
    I looked on the internet and i think there seems to be a rule? but i don't understand how to work it out by hand. Thankyou

    P.S the answer is x/(4√(x² + 4)) + c
     
  2. jcsd
  3. Nov 12, 2009 #2

    Mark44

    Staff: Mentor

    This is not an equation, so there is nothing to solve.

    What is the exact wording of the problem?
     
  4. Nov 12, 2009 #3
    Sorry, i was trying to cut out the part i needed help with, this is the whole question:

    Evaluate the following indefinite integral:
    ∫1/(x² + 4)^(3/2).dx

    Here is my working out;

    put x = 2sinh(θ)
    dx/dθ = 2cosh(θ)
    dx = 2cosh(θ).dθ


    (x² + 4)^(3/2) = (4sinh²(θ) + 4)^(3/2)

    Since cosh²(θ) - sinh²(θ) = 1
    => cosh²(θ) = 1 + sinh²(θ)

    so,
    (x² + 4)^(3/2) = (2²cosh²(θ))^(3/2)
    = 8cosh³(θ)

    So ∫1/(x² + 4)^(3/2).dx = ∫(1/8cosh³(θ)). 2cosh(θ).dθ
    = ∫1/4cosh²(θ).dθ
    = (1/4)tanh(θ) + c

    and then that's where i worked out that since tanh(θ) = sin(θ)/cosh(θ) then:
    1/4tanh(θ) + c = x/8cosh(θ) + c

    but i don't know how to work out the cosh(θ) part of it
     
  5. Nov 12, 2009 #4

    Mark44

    Staff: Mentor

  6. Nov 12, 2009 #5
    yeah it says cosh(arcsinh(x)) = √(1 + x²)
    But i want to know how you get that from cosh(arcsinh(x)). I have been trying to find proof for it.


    edit:
    never mind, i have figured it out now!
    let y = arcsinhx
    x = sinh(y)

    use the identity cosh^2(y) - sinh^2(y) = 1
    etc
     
    Last edited: Nov 12, 2009
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