Re: Simplifying Log to ArcCos Expressions

• To: mathgroup at smc.vnet.net
• Subject: [mg56888] Re: [mg56866] Simplifying Log to ArcCos Expressions
• From: DrBob <drbob at bigfoot.com>
• Date: Mon, 9 May 2005 01:46:10 -0400 (EDT)
• References: <200505080610.CAA02252@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```How about substituting k/r == Cos[t] BEFORE integrating?

expr1 = (k/r^2)*(1/Sqrt[1 - k^2/r^2])
D[k/Cos@t, t]expr1 /. r -> k/Cos@t // Simplify // PowerExpand
Integrate[%, t] + C
% /. t -> ArcCos[k/r]

k/(Sqrt[1 - k^2/r^2]*r^2)
1
C + t
C + ArcCos[k/r]

Sin[t] would work equally well.

Bobby

On Sun, 8 May 2005 02:10:16 -0400 (EDT), David Park <djmp at earthlink.net> wrote:

> Dear MathGroup,
>
> I want to integrate the following expression and get a simple answer.
>
> expr1 = (k/r^2)*(1/Sqrt[1 - k^2/r^2])
>
> The answer is actually quite simple: ArcCos[k/r] + constant. But what a lot of work for me to get it! Perhaps someone can show a simpler path. (I'm working with Version 5.0.1.)
>
> expr2 = Integrate[expr1, r]
> -((Sqrt[k^2 - r^2]*Log[(2*(k + Sqrt[k^2 - r^2]))/
>       r])/(Sqrt[1 - k^2/r^2]*r))
>
> Then I have to do all the following simplification steps...
>
> expr2[[{2, 3, 4}]]
> Numerator[%]/(Denominator[%] /. Sqrt[a_]*(b_) :>
>     Sqrt[Distribute[a*b^2]])
> % /. (a_)^(1/2)/(b_)^2^(-1) -> (a/b)^(1/2)
> Simplify[%, r >= k]
> Expand[%*FunctionExpand[expr2[[{1, 5}]]]]
> expr3 = %[[2]]
>
> expr3
> MapAt[Distribute, %, {{2, 1}}]
> % /. Sqrt[a_]/(b_) :> Sqrt[Distribute[a/b^2]]
> % /. r -> k/z
> % /. Log[(z_) + Sqrt[(z_)^2 - 1]] -> I*ArcCos[z]
> % /. z -> k/r
>
> Thanks in advance for a more direct path.
>
> David Park
>
>
>
>
>

--
DrBob at bigfoot.com

```

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