Good afternoon,

I am working on my last surface area problem, and am kind of hung up on the following:

Wait... could that be re-written as?:

That doesn't really help either...

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- July 15th 2010, 09:30 AMMechEngIntegrate cosh(x)^2
Good afternoon,

I am working on my last surface area problem, and am kind of hung up on the following:

Wait... could that be re-written as?:

That doesn't really help either... - July 15th 2010, 09:39 AMTed
Is it ?

If so, this integral is unelementary.

Do you mean ? - July 15th 2010, 09:40 AMchisigma
The easiest way is the following identity...

(1)

Kind regards

- July 15th 2010, 09:44 AMMechEng
It is in fact which I have written as

Where is the identity derived from? - July 15th 2010, 09:45 AMTed
Its a well-known identity.

You can prove it by using the definitions of the Heperbolic Functions.

******Edit:******

Another way to do it:

and it is easy to integrate. - July 15th 2010, 10:15 AMMechEng
Would that be the double angle formula for

Pardon my ignorance, but are all of the trig identities applicable to the hyperbolic functions? - July 15th 2010, 10:23 AMTed
No. Not all of them.

Also, sometimes there is a small difference like the signs.

for example:

but

- July 15th 2010, 10:46 AMMechEng
Just to check my work...

What is the surface area of for when revolved about the x-axis?

I get:

Is this a reasonable looking answer? - July 15th 2010, 12:07 PMroninpro
Actually, yes, with some slight modifications. Have a look at Hyperbolic function - Wikipedia, the free encyclopedia, and look for "Osborn's Rule".