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

1. Nov 11, 2009

### wshfulthinker

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. Nov 12, 2009

### Staff: Mentor

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

What is the exact wording of the problem?

3. Nov 12, 2009

### wshfulthinker

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

4. Nov 12, 2009

### Staff: Mentor

5. Nov 12, 2009

### wshfulthinker

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