1. ## hyperbolic functions

Can anybody help me here?

Use hyperbolic functions to find the value of...
The integral of 1/square root of(16x^2 - 36) dx (between the values 12 and 6)
my working so far is....
4x=6cosh(u) 4dx/du=6sinh(u)
(16x^2-36=36coshu^2 - 36)
with all the subtituation i get to the integral of 1.5sinh(u)/6sinh(u) du
but then i am stuck after this. Sorry this is a bit messy but can anybody help?

2. $\frac{sinh(u)}{sinh(u)} = What?$

I'm temtpted to take off extra points for not seeing it.

Try to relax a little. No sense missing the simple algebra.

3. Yeh but in my example it has
integral of 1.5sinh(u)/6sinh(u) du= 1/4 intregral of arccosh(2/3x)+C
i dont understand how this works..

4. You're showing signs of stress.

1) 1.5 is not particularly helpful. 1.5/6 = 3/12 = 1/4

2) $\frac{1}{4}\frac{sinh(u)}{sinh(u)} = \frac{1}{4}$

3) $\int \frac{1}{4} du = \frac{1}{4}u + C$

4) You're not going to make me do ALL the work, are you? Look at your original substitution and move #3 back to terms of 'x'.

5) You seem to have an extra "integral" in that answer.