I'm temtpted to take off extra points for not seeing it.
Try to relax a little. No sense missing the simple algebra.
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?
You're showing signs of stress.
1) 1.5 is not particularly helpful. 1.5/6 = 3/12 = 1/4
2)
3)
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.