[SOLVED] Another Arc Lengths problem

**mollymcf2009** · February 26th 2009, 05:44 PM

Find the arc length:

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

I got for my integral:

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

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

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

**mollymcf2009** · February 26th 2009, 06:04 PM

Ok, I figured out that I can just integrate this as is:

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

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

I got:

$\frac{8 + cosh(9)}{3}<br />$
Which is wrong. Any help is GREATLY appreciated!!

**mr fantastic** · February 27th 2009, 01:40 AM

Originally Posted by mollymcf2009

Find the arc length:

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

I got for my integral:

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

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

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

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

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

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

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