Re: ArcCos[x] with x > 1

• To: mathgroup at smc.vnet.net
• Subject: [mg49311] Re: ArcCos[x] with x > 1
• Date: Tue, 13 Jul 2004 04:32:46 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```On 7/12/04 at 2:11 AM, carlw at u.washington.edu (Carl K. Woll) wrote:

>"Paul Abbott" <paul at physics.uwa.edu.au> wrote in message
>news:cco3io\$4ig\$1 at smc.vnet.net...

>>In fact, Mathematica 5.0 gets the indefinite integral wrong:

>>Simplify[Integrate[r ArcCos[1/(2r)] Sqrt[4r^2 + 1], r], 1/Sqrt[2] > r
>>> 1/2]

>>returns

>>(1/12) (2r^2 + (4r^2 + 1)^(3/2) ArcCos[1/(2r)] - Log[4r^2 +
>>Sqrt[16r^4 - 1]] - Sqrt[16r^4 - 1])

>Why do you state that the indefinite integral is incorrect? In your
>case, I plotted the difference between the derivative of the
>indefinite integral and the integrand and the result was zero, so
>that the indefinite integral is indeed correct, at least for the
>range 1/2 < r < 1/Sqrt[2].

Hmm... When I try to reproduce this result I definitely do not get a plot showing the difference to be zero. And when I try a particular point between 1/2 and 1/Sqrt[2] say 3/5

In[1]:=
g = r*ArcCos[1/(2*r)]*Sqrt[4*r^2 + 1];
f = Simplify[Integrate[g, r],
1/Srt[2] > r > 1/2];
d = D[f, r];
FullSimplify[g - d /. r -> 3/5]

Out[4]=
(1/180)*(-36 + Sqrt[671])

which is clearly not zero since Sqrt[671] is about 26
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