Quick hyperbolic question

craig · #1 $Old$ May 31st, 2009, 04:10 PM

This is part of a larger integration question about finding the arc length of a curve. Have managed to do the rest of the question but stuck on this last bit when subbing in the integrals.

The part I'm stuck on is:
$\frac{1}{2}\sinh{2u}$ where $u$ is equal to $\sinh^{-1}{12}$ .

I expanded $\sinh^{-1}{12}$ to $\ln{(12 + \sqrt{144 + 1})}$ .

I then have this equation:

$\frac{1}{2}\sinh{2(\ln{(12 + \sqrt{144 + 1})}})$

$\frac{1}{2}\sinh{\ln{(12 + \sqrt{144 + 1})^2}}$

Not too sure where to go from here, any pointers would be greatly appreciated.

Thanks in advance.

Spec · #2 $Old$ May 31st, 2009, 04:33 PM

It's probably easier to use: $\frac{\sinh 2x}{2}=\cosh x \sinh x$ and $\cosh x = \sqrt{1+\sinh^2 x}$

craig · #3 $Old$ June 1st, 2009, 03:33 AM

Thanks, I'd forgot about the double angle formula for sinh.