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)but then i am stuck after this. Sorry this is a bit messy but can anybody help?

(16x^2-36=36coshu^2 - 36)

with all the subtituation i get to the integral of 1.5sinh(u)/6sinh(u) du