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MathGroup Archive 2000

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Re: Mathematica gives bad integral ??

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24315] Re: Mathematica gives bad integral ??
  • From: Ronald Bruck <bruck at math.usc.edu>
  • Date: Sun, 9 Jul 2000 04:52:27 -0400 (EDT)
  • Organization: Univ of Southern California
  • References: <8k3n38$3pk@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <8k3n38$3pk at smc.vnet.net>, "J.R. Chaffer" <jrchaff at mcn.net> 
wrote:

:Hi,  this newbie gets erroneous results with Mathematica
:4.0 (for students), with the following integral.  Hopefully
:someone can tell me why, and what I may be doing
:wrong.  I have tried "Assumptions -> x e Reals", or
:x > 0, with same results.  Integral in question is,
:
:Integrate[1/Sqrt[1-Sin[2x]]]
:
:The result is somewhat involved, instead of the expected
:result (Schaum, "Calculus" 4E, p. 297),
:
:  integral = - (1/Sqrt[2])Log[Abs[Csc[Pi/4-x]-Cot[Pi/4-x]]]
:
:One expects to get differing forms with any computer
:algebra system, since there are so many equivalent forms
:of algebraic expressions.  However, Mathematica's form
:and the Schaum (correct) form differ by significant
:numerical values, as plotting shows (i.e., not some E-16
:or some such).
:
:Further, and what really seems wrong, is that when one
:differentiates Mathematica's result for the integral, one
:does NOT get the original integrand, or anything even
:close, numerically.
:
:So, I am confused.  Anyone who knows the explanation
:would be welcome to share it.

When **I** use Mathematica 4.01 to find the integral, and plot the 
difference between the purported integral and 1/Sqrt[1-Sin[2x]] on 
[0,2Pi], all I get is numerical noise.

Mathematica returns a rather complicated expression involving complex 
numbers, which nevertheless when FullSimplify'd yields

ArcTanh[(1 + Tan[x/2])/Sqrt[2]](Cos[x]-Sin[x]))/Sqrt[1/2-Sin[x]Cos[x]].

Since 

     Sqrt[1/2-Sin[x]Cos[x]] = Sqrt[1-Sin[2x]]/Sqrt[2]

    = Sqrt[(Cos[x]-Sin[x])^2]/Sqrt[2] = Abs[Cos[x]-Sin[x]]/Sqrt[2],

this is equivalent to

  Sqrt[2] ArcTanh[(1 + Tan[x/2])/Sqrt[2]] Sign[Cos[x] - Sin[x]].

I don't see the problem, because this is correct.  On intervals where 
cos x > sin x,

  Sqrt[2] ArcTanh[(1 + Tan[x/2])/Sqrt[2]]

is an antiderivative, and on intervals where cos x < sin x, the negative 
of that expression is an antiderivative.  (You can't have intervals 
containing x with both behaviors, since that would make the integrand 
undefined.)

I was using Mathematica 4.01.  Is there that big a difference between 
the student and professional version?

-- 
Due to University fiscal constraints, .sigs may not be exceed one
line.


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