G P
Posted on Friday, 13 February, 2004 - 03:27 pm:
how do you integrate cosech x using the identity sinhx=1/2(e^x-e^-x) and the substitution v=e^x?
Francis Woodhouse
Posted on Friday, 13 February, 2004 - 04:07 pm:
LaTeX Image

Our integral is, therefore:

LaTeX Image

Let v = ex . Then, we have x = ln v, and so dx = dv / v. So, the integral now becomes:

LaTeX Image

Now, use a table of derivatives to look at the derivative of tanh-1 x. Can you see where to go from here?
Michael McLoughlin
Posted on Friday, 13 February, 2004 - 06:19 pm:
If you were interested, you can also do it this way:

LaTeX Image

Taking u=cosh(x) gives:

LaTeX Image
Marcos
Posted on Friday, 13 February, 2004 - 08:06 pm:
Michael,

tanh-1(coshx) is undefined for all x.

I think what you meant to write was:

òdu/(u2-1)=1/2 ln[(u-1)/(u+1)]+C = ln[(ex-1)/(ex+1)]+C

Marcos

Marcos
Posted on Friday, 13 February, 2004 - 08:10 pm:
P.S. Francis, I think it's simpler to just use partial fractions for integrating 1/(x2 - 1) rather than messing around with the derivatives of tanh x.

Marcos
Francis Woodhouse
Posted on Friday, 13 February, 2004 - 08:16 pm:
In that case, we have LaTeX Image. I agree with you, Marcos; that's much simpler to integrate.
Kerwin Hui
Posted on Friday, 13 February, 2004 - 08:17 pm:
In fact, tanh-1 (cosh x) is not real unless x=0. Further, the answer is not right - you seems to be thinking tanh2 x-1=sech2 x which is not true (it is -sech2 x).

Kerwin
Marcos
Posted on Friday, 13 February, 2004 - 08:23 pm:
Kerwin.
We're in agreement: In my terminology, "undefined" = "not real" (!) (in this case it's okay since we are dealing with real functions). Also, when x = 0, the expression is still undefined (as it shoots to infinity).

Marcos
Kerwin Hui
Posted on Friday, 13 February, 2004 - 08:30 pm:
Yes, "not real"="strictly complex".