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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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