# Integrate cosh(x)^2

• July 15th 2010, 09:30 AM
MechEng
Integrate cosh(x)^2
Good afternoon,

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

$
\int{cosh(x)^2} dx
$

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

$\int{1+sinh(x)^2} dx$

That doesn't really help either...
• July 15th 2010, 09:39 AM
Ted
Is it $cosh(x^2)$ ?
If so, this integral is unelementary.

Do you mean $cosh^2(x)$ ?
• July 15th 2010, 09:40 AM
chisigma
The easiest way is the following identity...

$\displaystyle \cosh^{2} x = \frac{1}{2} + \frac{1}{2}\ \cosh 2x$ (1)

Kind regards

$\chi$ $\sigma$
• July 15th 2010, 09:44 AM
MechEng
It is in fact $cosh^2(x)$ which I have written as $cosh(x)^2$

Where is the identity derived from?
• July 15th 2010, 09:45 AM
Ted
Its a well-known identity.
You can prove it by using the definitions of the Heperbolic Functions.

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

Another way to do it:

$cosh^2(x) = \left( \dfrac{e^{x}+e^{-x}}{2} \right)^2=\dfrac{e^{2x}+e^{-2x}+2}{2}=\frac{1}{2}e^{2x}+\frac{1}{2}e^{-2x}+1$

and it is easy to integrate.
• July 15th 2010, 10:15 AM
MechEng
Would that be the double angle formula for $cos^{2}x=\frac{1+cos2x}{2}$

Pardon my ignorance, but are all of the trig identities applicable to the hyperbolic functions?
• July 15th 2010, 10:23 AM
Ted
No. Not all of them.
Also, sometimes there is a small difference like the signs.

for example:

$cos(x+y)=cos(x)cos(y) - sin(x)sin(y)$

but

$cosh(x+y)=cosh(x)cosh(y) + sinh(x) sinh(y)$
• July 15th 2010, 10:46 AM
MechEng
Just to check my work...

What is the surface area of $f(x)=\frac{e^{x}+e^{-x}}{2}$ for $0 \leq x \leq 2$ when revolved about the x-axis?

I get:

$S=\pi(2+sinh(4))$

Is this a reasonable looking answer?
• July 15th 2010, 12:07 PM
roninpro
Quote:

Originally Posted by MechEng
Would that be the double angle formula for $cos^{2}x=\frac{1+cos2x}{2}$

Pardon my ignorance, but are all of the trig identities applicable to the hyperbolic functions?

Actually, yes, with some slight modifications. Have a look at Hyperbolic function - Wikipedia, the free encyclopedia, and look for "Osborn's Rule".