# [SOLVED] Another Arc Lengths problem

• Feb 26th 2009, 05:44 PM
mollymcf2009
[SOLVED] Another Arc Lengths problem
Find the arc length:

$\displaystyle y = 3 + \frac{1}{3} cosh(3x)$ for $\displaystyle 0 \leq x \leq 3$

I got for my integral:

$\displaystyle \int\limits^{3}_{0} \sqrt{1 + sinh^2(3x)} dx$

$\displaystyle = \int\limits^{3}_{0}1 + sinh(3x) dx$

What do I do from here? Should I substitute w/ double angle?
• Feb 26th 2009, 06:04 PM
mollymcf2009
Ok, I figured out that I can just integrate this as is:

$\displaystyle \int\limits^{3}_{0} 1 + sinh(3x) dx$

$\displaystyle = x + \frac{1}{3} cosh (3x) |^{3}_{0}$ **BUT THIS IS THE ORIGINAL PROBLEM?!?!?!?

I got:

$\displaystyle \frac{8 + cosh(9)}{3}$
Which is wrong. Any help is GREATLY appreciated!!
• Feb 27th 2009, 01:40 AM
mr fantastic
Quote:

Originally Posted by mollymcf2009
Find the arc length:

$\displaystyle y = 3 + \frac{1}{3} cosh(3x)$ for $\displaystyle 0 \leq x \leq 3$

I got for my integral:

$\displaystyle \int\limits^{3}_{0} \sqrt{1 + sinh^2(3x)} dx$

$\displaystyle = \int\limits^{3}_{0}1 + sinh(3x) dx$

What do I do from here? Should I substitute w/ double angle?

Doesn't $\displaystyle \cosh^2 A - \sinh^2 A = 1 ?$

In which case, won't $\displaystyle 1 + \sinh^2 (3x) = \cosh^2 (3x) ?$

In which case won't your integral become $\displaystyle \int \limits^{3}_{0} \cosh (3x) \, dx ?$