       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*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*Sin[x/2] + 4*Sqrt*Floor[(x+Pi)/(2*Pi)]

David Cantrell

```

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