RE: Strange results
- To: mathgroup at smc.vnet.net
- Subject: [mg15429] RE: [mg15389] Strange results
- From: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
- Date: Wed, 13 Jan 1999 20:57:40 -0500
- Sender: owner-wri-mathgroup at wolfram.com
Ing. Alessandro Toscano wrote: The following in/out does not make sense to me: In[2]:= p=Integrate[Cos[x]^n,x] Out[2]= \!\(\(-\(\(Cos[x]\^\(1 + n\)\ Hypergeometric2F1[\(1 + n\)\/2, 1\/2, \(3 + n\)\/2, Cos[x]\^2]\ Sin[x]\)\/\(\((1 + n)\)\ \ at Sin[x]\^2\)\)\)\) In[4]:= p//.n->0//PowerExpand Out[4]= -ArcSin[Cos[x]] _____________ REPLY: If the answer from Integrate differs from the one you expect by an arbitrary constant there isn't a problem. Now if we add Pi/2 to the answer above we get something very close to what you expected. (see below) In[5]:= FullSimplify[Pi/2-ArcSin[Cos[x]]] Out[5]= ArcCos[Cos[x]] This of course simplifies to (x) for 0<=x<=Pi. For other values of (x) this is off by a multiple of 2*Pi. I doubt it's possible to express Integrate[Cos[x]^n, x] in closed form without such pit falls. _________________ In[7]:= p//.n->1//Simplify//PowerExpand Out[7]= -1+Sin[x] Isn't it true that (Integrate[Cos[x]^0,x] == x? Isn't it true that (Integrate[Cos[x]^1,x] == Sin[x]? __________________ REPLY: This differs from what you expected by the constant (-1). This is allowed for the same reason. The indefinite integral isn't unique. Cheers, Ted Ersek