Re: Integral question 2

*To*: mathgroup at smc.vnet.net*Subject*: [mg73714] Re: Integral question 2*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Sun, 25 Feb 2007 04:39:02 -0500 (EST)*References*: <eropmh$9cb$1@smc.vnet.net>

More on my previous message. To be fair I check the symbolic result from the other CAS. Not being very concentrated I claimed that the obtained result is like yours by hand. Of course the situation is not like this. int(1/sqrt(v^2 - c*s^2),s); 1/2 c s arctan(--------------) 2 2 1/2 (v - c s ) ---------------------- 1/2 c diff(%,s); 1/2 3/2 2 c c s -------------- + -------------- 2 2 1/2 2 2 3/2 (v - c s ) (v - c s ) ------------------------------- / 2 \ 1/2 | c s | c |1 + ---------| | 2 2| \ v - c s / simplify(%); 1/((v^2-c*s^2)^(1/2)) as it should be. Moreover evalf(1/2^(1/2)*arctan(2^(1/2)*1/(4^2-2*1^2)^(1/2))); 0=2E2555251436 Contrary to what I said in the original reply the other CAS perfomance is better in this case than Mathematica. Best Regards 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 (v=4, 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