Re: simplify a trig expression
- To: mathgroup at smc.vnet.net
- Subject: [mg65446] Re: simplify a trig expression
- From: "Scout" <Scout at nodomain.com>
- Date: Sun, 2 Apr 2006 05:00:05 -0400 (EDT)
- References: <e0j3a1$foe$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Murray Eisenberg" <murray at math.umass.edu> news:e0j3a1$foe$1 at smc.vnet.net... >A direct substitution (with paper and pencil) gives that the integral of > Cos[x]/(Sin[x] + 1) is Log[Sin[x] + 1]. This is valid provided Sin[x] > is not -1. > > Mathematica gives: > > Integrate[Cos[x]/(Sin[x] + 1), x] > 2 Log[Cos[x/2] + Sin[x/2]] > > Is there some simple way to coerce the latter Mathematica-supplied > result into the paper-and-pencil answer? > > The closest I could get is: > > Log[TrigExpand[Expand[(Cos[x/2] + Sin[x/2])^2]]] /. > {Sin[x/2] -> Sqrt[(1 - Cos[x])/2], > Cos[x/2] -> Sqrt[(1 + Cos[x])/2]} > Log[1 + Sqrt[1 - Cos[x]]*Sqrt[1 + Cos[x]]] > > Am I not seeing some easier TrigExpand or TrigReduce method? > > -- > Murray Eisenberg murray at math.umass.edu > Mathematics & Statistics Dept. > Lederle Graduate Research Tower phone 413 549-1020 (H) > University of Massachusetts 413 545-2859 (W) > 710 North Pleasant Street fax 413 545-1801 > Amherst, MA 01003-9305 > Hi Murray, one could think to use Simplify or FullSimplify but it doesn't work! Let's say In[1]:= f1[x_] := 2 Log[Cos[x/2] + Sin[x/2]]; f2[x_] := Log[(Cos[x/2] + Sin[x/2])^2]; g[x_] := Log[1 + Sin[x]]; Note that f2(x) derives from f1(x) applying the logarithm property : n*Log[x]==Log[x^n] , valid in this case. Let's show that the functions are equal: In[2]:= FullSimplify[ f1[x] == g[x], {k \[Element] Integers && x != 1/2 (3 + 4 k)/ Pi}] Out[2]= 2 Log[Cos[x/2] + Sin[x/2]] == Log[1 + Sin[x]] nothing happened, instead: In[3]:= FullSimplify[ f2[x] == g[x], {k \[Element] Integers && x != 1/2 (3 + 4 k)/ Pi}] Out[3]= True So, the problem is that the function FullSimplify[] (Simplify[]) doesn't match the previous Log property! To force the Simplify[] function to use the Log property, write down our transformation function: In[4]:= tf[a_ Log[x_]] := FullSimplify[Log[x^a]]; Thus, In[5]:= Simplify[f1[x], TransformationFunctions -> tf] Out[5]= Log[1 + Sin[x]]. Hope this helps. Regards, ~Scout~