Catenary Realism: Calculating Weight of Electric Wire

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Homework Statement


when a cable with non-zero mass is connected to a pole at both ends, the shape it assumes is called a catenary.
it can be shown that for an electrical wire whose linear mass density is .9 kg/m strung between poles 30m apart(and making a 22 degree angle at each end) the mathematical equation is
y=(38.1m)[cosh(x/38.1m)-1]
a) an electircal wire of linear mass density .9 kg/m is strung, between poles 30m apart, from west to east on earth. initially the current flowing is zero and therefore the wires shape is that of a catenary. what is the weight of the wire?HINT: the length of the wire(which is not 30m) is found by integrating dl over the catenary. Use the fact the
dl=dx[sqrt(1+(dy/dx)^2) in order to have an integration over the x-axis.

It shows a picture with an equation for the wire. y=(38.1m)[cosh(x/38.1m)-1]


Homework Equations


equation for the wire ...y=(38.1m)[cosh(x/38.1m)-1]
length of the wire... dl=dx[sqrt(1+(dy/dx)^2)
He also gave us all of the equations and proofs of hyperbolic functions.

The Attempt at a Solution


I was not sure what to do since i have never done a problem like this. I was neither shown in class how to do anything close to this.
I started with integrating the length of the wire function. I am having problems with the integration though. I am not sure what to do after this or if its even right what I am doing.
 
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Well, what sort of problems are you having with the integration of dl?
 
Should i plug the equation of the wire into the length of the wire equation? Then i would derive the equation for the wire and then integrate the dl...
 
I can change the sqrt[1+(dy/dx)^2] to cosh(dy/dx), correct?