Re: A completely different solution?

Thanks for your time. That idea already occurred to me, but in an Excel
spread*** I calculated the difference between the two expressions for
some values within the domain; and the difference did not at all seem
to be constant over the domain.
So I suspect there is some more fundamental error here...

Frank



Arturo Magidin schreef:
> In article <1118694850.829957.263970@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> speedy <frank.degeeter@xxxxxxxxx> wrote:
> >Hi there.
> >
> >My son asked me to look at an integral he had calculated:
> >integral(dx/sqrt(-x^2-x+2)). He came up with the solution
> >arcsin((2x+1)/3).
>
> Plus a constant.
>
>
>
> >I found the same solution, and checked it with the
> >Wolfram integrator on the net, which confirmed it.
> >
> >However, my son then came up with a solution by his teacher, who
> >tackled the integral in a totally different manner: he replaced the
> >denominator of the integrand by (x-1)*z. From this the integrand =
> >-3z/(z^2+1).
> >And dx = dz * 6z / (z^2+1)^2. Finally, this leads to a solution: -2
> >arctg(((-x-2)/(x-1))^0.5).
>
>
> Plus a constant.
>
> >I could spot no error in these calculations, so I would presume that
> >the solutions are equal
>
>
> Up to a constant.
>
> en tried to prove that using
> >sin(-2x)=-(2tgx)/(1+tg^2x), substituting the value of tgx from the
> >second solution. But I cannot recover that (2x+1)/3 from the initial
> >solution.
> >
> >Where is the error?
>
> The error is assuming that the two solutions are identical. What you
> really have is that the integral is equal to
>
> arcsin((2x+1)/3) + C
>
> and to
>
> -2arctan(sqrt((-x-2)/(x-1))) + D
>
> for some constants C and D. What you need to make sure of is that the
> difference between arcsin((2x+1)/3) and -2arctan(sqrt((-x-2)/(x-1)))
> is constant, not that it is equal to 0.
>
>
> --
> ======================================================================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes")
> ======================================================================
>
> Arturo Magidin
> magidin@xxxxxxxxxxxxxxxxx

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