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Forum index » Science and Technology » Math » Undergraduate
Inverse hyperbolic substitutions in integration
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TedMaul
science forum beginner


Joined: 14 Jun 2005
Posts: 4

PostPosted: Tue Jun 14, 2005 9:46 pm    Post subject: Inverse hyperbolic substitutions in integration Reply with quote
hi all,
I've got an expression to integrate which is as follows:

1/((x^2)Sqrt[1-x^2])

Now I've managed to generate an answer (which I confirmed as correct in Mathematica), but I've been specifically asked to solve using hyperbolic trig, which is proving something of a problem. I've so far been unable to put this into any form which will facilitate the above (thanks largely to the annoying x^2 term). I feel like I'm missing something really obvious. Any help would be very much appreciated. Thanks in advance.
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Arturo Magidin
science forum Guru


Joined: 25 Mar 2005
Posts: 1838

PostPosted: Tue Jun 14, 2005 10:04 pm    Post subject: Re: Inverse hyperbolic substitutions in integration Reply with quote
In article <16251688.1118785624546.JavaMail.jakarta@nitrogen.mathforum.org>,
TedMaul <ben@dris.demon.co.uk> wrote:
Quote:
hi all,
I've got an expression to integrate which is as follows:

1/((x^2)Sqrt[1-x^2])

Now I've managed to generate an answer (which I confirmed as correct
in Mathematica), but I've been specifically asked to solve using
hyperbolic trig, which is proving something of a problem. I've so far
been unable to put this into any form which will facilitate the above
(thanks largely to the annoying x^2 term). I feel like I'm missing
something really obvious. Any help would be very much
appreciated. Thanks in advance.

Hmmm... Well, we can set x = tanh(z). Then

1 - x^2 = 1 - tanh^2(z) = (cosh^2 z - sinh^2 z)/cosh^2 z
= 1/cosh^2 z = sech^2(z). Then

1/(x^2*sqrt[1-x^2]) = 1/(tanh^2(z)sech(z)

Since dx = d(tanh x) = sech^2(z)dz this gives

dx/[x^2*sqrt[1-x^2] = sech^2(z)dz / (tanh^2(z)sech (z))
= sech(z)dz/tanh^2(z)
= cosh(z)dz/sinh^2(z)

which you should spot very easily how to integrate.




--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@math.berkeley.edu
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Virgil
science forum Guru


Joined: 24 Mar 2005
Posts: 5536

PostPosted: Tue Jun 14, 2005 10:25 pm    Post subject: Re: Inverse hyperbolic substitutions in integration Reply with quote
In article
<16251688.1118785624546.JavaMail.jakarta@nitrogen.mathforum.org>,
TedMaul <ben@dris.demon.co.uk> wrote:

Quote:
hi all,
I've got an expression to integrate which is as follows:

1/((x^2)Sqrt[1-x^2])

Now I've managed to generate an answer (which I confirmed as correct in
Mathematica), but I've been specifically asked to solve using hyperbolic
trig, which is proving something of a problem. I've so far been unable to put
this into any form which will facilitate the above (thanks largely to the
annoying x^2 term). I feel like I'm missing something really obvious. Any
help would be very much appreciated. Thanks in advance.

The only hyperbolic identity which springs to mind is

1- tanh(t)^2 = sech(y)^2

suggesting the substitution

x = tanh(t), dx = sech(t)^2 dt.

The resulting integral is manageable, with a bit of imagination.
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TedMaul
science forum beginner


Joined: 14 Jun 2005
Posts: 4

PostPosted: Tue Jun 14, 2005 10:41 pm    Post subject: Re: Inverse hyperbolic substitutions in integration Reply with quote
Thanks for the suggestion. I'd just been trying the substitution x=sech[u], which seems to work.

For anyone who's interested the answer comes to:

-sinh[ln((1+Sqrt[1-x^2])/x)]

I'll take your advice and try with tanh[u] though. Thanks for the help.
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