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Re: Integration using mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg72646] Re: Integration using mathematica
• From: dh <dh at metrohm.ch>
• Date: Sat, 13 Jan 2007 04:54:09 -0500 (EST)
• Organization: hispeed.ch
• References: <eo7mt2$1ps$1@smc.vnet.net>

Hi,
your expression has syntax errors. I assume you mean:
0.013 *Exp [15.16(ArcTan[0.002 x]]^2 *ArcTan[0.002 x]
For an example I set all numerical constant to 1, you can put them back 
yourself.
t1=Exp [ 2 ArcTan[x] ] *ArcTan[x]
this calls for substitution of ArcTan[x] by a new variable y. dx is 
thereby transformed into: (1+x^2 )dy= (1+Tan[y]^2) dy. And the integral:
t2=Integrate[ Exp[2 y] y (1 + Tan[y]^2), y]//Simplify
Finally we replace x back:
t3=t2/.y->ArcTan[x]//Simplify
You may make a quick check by verifying that D[t3,x]-t1 gives zero if 
you replace x by a numerical value.
Daniel



Negede wrote:
> I have difficulty in integrating the following equation on Mathematica.
> Does any one of you know how to do it on Mathematica? Any trick to
> aply? or another software to use? Please help me. The function is as
> follows:
> 
> 0.013 *Exp [15.16(ArcTan(0.002 x)])^2 *(ArcTan(0.002 x)
> 
> Thanks in advance
>