Re: Integral question 3

*To*: mathgroup at smc.vnet.net*Subject*: [mg73716] Re: Integral question 3*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Sun, 25 Feb 2007 04:40:07 -0500 (EST)*References*: <eropmh$9cb$1@smc.vnet.net>

Hello again. A trick to obtain your "hand" result: In[58]:= Integrate[1/Sqrt[EulerGamma^2 - c*s^2], s] % /. EulerGamma -> v % /. {v -> 4, c -> 2, s -> 1} N[%] Out[58]= ArcSin[(Sqrt[c]*s)/EulerGamma]/Sqrt[c] Out[59]= ArcSin[(Sqrt[c]*s)/v]/Sqrt[c] Out[60]= ArcSin[1/(2*Sqrt[2])]/Sqrt[2] Out[61]= 0=2E2555251438123125 Dimitris =CF/=C7 Tulga Ersal =DD=E3=F1=E1=F8=E5: > Dear Mathematica users, > > Let's consider the integral > > Integrate[1/Sqrt[v^2 - c*s^2], s] > > where v and c are positive reals. When I calculate the integral by hand, I get > > ArcSin[(Sqrt[c]*s)/v]/Sqrt[c] > > However, if I evaluate the integral in Mathematica, I get > > (I*Log[(-2*I)*Sqrt[c]*s + 2*Sqrt[-(c*s^2) + v^2]])/Sqrt[c] > > which not only does not look like what I have found by hand, but > also, for v=4, c=2, s=1, for example, gives > > 0.255525 + 1.47039 i > > as opposed to just 0.255525. > > If you evaluate the integral in Mathematica with the said values (v4, c=2) > > Integrate[1/Sqrt[4^2 - 2*s^2], s] > > you get the answer > > ArcSin[s/(2*Sqrt[2])]/Sqrt[2] > > which agrees with what I have found by hand. > > My question is: How can I get what I found by hand using Mathematica > without having to assign values to v and c before evaluating the > integral? I tried adding the assumptions v>0, c>0, but it didn't help. > > I'd appreciate your help. > > Thanks, > Tulga