*To*: mathgroup@smc.vnet.net*Subject*: [mg10327] Re: Assuming n is even*From*: hello@there.com (Rod Pinna)*Date*: Thu, 8 Jan 1998 23:40:50 -0500*Organization*: UWA*References*: <68l3e7$mf6@smc.vnet.net> <68q770$202@smc.vnet.net>

Paul Abbott <paul@physics.uwa.edu.au> wrote: >Good to see a posting from the University of Western Australia! It is a >FAQ but the answer is, briefly, no. A recent and related question was: Just trying to keep busy over the break.... Thanks for the responses >In my opinion, the best way to is using pattern-matching and replacement >rules (see The Mathematica Journal 2(4): 31). E.g., for n integral, we >have > > {Cos[(n_)*Pi] -> (-1)^n, Sin[(n_)*Pi] -> 0}; > >Please post your integral so that perhaps readers can make other >suggestions. The above integral is pretty close to what I'm looking at actually. Say I have w1=A1*Sin[n*t]*(Cos[(1/2)*m*\[Pi]*x/L]-1) v1=A3*Cos[n*t]*Sin[m*\[Pi]*x/L] Then with Et1=(1/a)*(D[v1,t]+w1) The integral is \!\(V12 = \[Integral]\_0\%L \(\[Integral]\_0\%\(2*\[Pi]\)Et1\^2\ \[DifferentialD]t \[DifferentialD]x\)\) (Apologies for the rather horrid formatting above) i.e. Integrate[Integrate[Et^2,{x,0,L}],{t,0,2*pi}] And a few integrals of that type. Some of the results given then have cosine terms which are equivalent. I've used ReplaceAll to replace one with the other. If there isn't a better way, that should be ok. Thanks for the help. Rod Pinna (rpinnaX@XcivilX.uwa.edu.au Remove the X for email)