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