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 >