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Inverse hyperbolic functions watch
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Last edited by non-centrally acting; 15-07-2007 at 20:54.
- 11-04-2006 17:25
- 11-04-2006 17:48
1) Split it into 2 [2x/sqrt(x2+1)] +1/sqrt(x2+1). Then for the first part let u = x2+1 and the second part let x = sinh x
2) Let x3 = u then let u = cosh y
- 11-04-2006 22:35
u should get x^2dx=u/3 which cancels the x^2 on the top. This leaves you 1/3 * 1/rt(u^2 - 1)
then use the cosh substituion du = sinhy dy
making the intergal 1/3 (sinh y / rt((sinhy)^2)) = int(1/3)
giving you y/3
then substitute back to get it in terms of x
y/3 = (arccosh u)/3 = 1/3arcosh(x^3)
hope thats okLast edited by Retropmot; 11-04-2006 at 22:44.
- 12-04-2006 00:48
Yes, but it aids explaination by turning into an integral you can instantly recognise as a standard result.