Re: integrat trig radical

*To*: mathgroup at smc.vnet.net*Subject*: [mg39593] Re: integrat trig radical*From*: "David W. Cantrell" <DWCantrell at sigmaxi.org>*Date*: Tue, 25 Feb 2003 02:56:24 -0500 (EST)*References*: <b3ck48$7vs$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Friedrich Laher <mathefritz at schmieder-laher.de> wrote: > is there any mathematical reason for mathematica, not 1st internally > using > Cos[x/2] = Sqrt[(1 + Cos[x])/2] > before > integrating Sqrt[1 + Cos[x]] ? Sure! Cos[x/2] = Sqrt[(1 + Cos[x])/2] is false for some values of x. For example, for Pi < x < 3*Pi, Cos[x/2] is negative while Sqrt[(1 + Cos[x])/2] is positive. > It > even refuses to answer True > to > Integrate[Sqrt[1 + Cos[x]],x] == 2*Sqrt[2]*Sin[x/2] I'm glad it refuses! But your problem raises a more interesting question: Since Sqrt[1 + Cos[x]] is continuous for all x, it must have a continuous antiderivative. Yet the antiderivative provided by Mathematica is not continuous everywhere. So why doesn't Mathematica give us a continuous antiderivative here? Well, it should be noted that there are cases when, although we know that a continuous antiderivative exists, we also know that it cannot be written in closed form in terms of familiar functions. HOWEVER, this problem is not such a case. Here is my answer for a continuous antiderivative for Sqrt[1 + Cos[x]]: (-1)^Floor[(x+Pi)/(2*Pi)]*2*Sqrt[2]*Sin[x/2] + 4*Sqrt[2]*Floor[(x+Pi)/(2*Pi)] David Cantrell