Re: ArcCos[x] with x > 1

*To*: mathgroup at smc.vnet.net*Subject*: [mg49311] Re: ArcCos[x] with x > 1*From*: Bill Rowe <readnewsciv at earthlink.net>*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 -- To reply via email subtract one hundred and four