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Topic: The Charlwood Fifty
Replies: 52   Last Post: Jun 24, 2013 10:24 PM

 Messages: [ Previous | Next ]
 Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: The Charlwood Fifty
Posted: Jun 6, 2013 1:33 PM

On 6/6/2013 11:23 AM, clicliclic@freenet.de wrote:
>

>
> It would have been nice if Prof. Charlwood could have thrown light on
> problem #49 from his appendix: Did he really want his students to work
> on an elementary evaluation of INT(ASIN(x*SQRT(1-x^2)), x) and fail?

fyi,

In http://www.apmaths.uwo.ca/~arich/CharlwoodIntegrationProblems.pdf

#49 is written as INT(ASIN(x/SQRT(1-x^2)), x)

I just checked the Charlwood?s 2008 paper, and it should be as
you have shown it (i.e. multiplication not division), so there is a
typo in the above pdf file.

> Anyway, here is a real version of the elliptic result:
>
> x*ASIN(x*SQRT(1-x^2))+2*SQRT(1-x^2)*SQRT(x^4-x^2+1)/(2-x^2)+SQRT~
> (x^4/(2-x^2)^2)*SQRT(x^4-x^2+1)/(2*x^2*SQRT((x^4-x^2+1)/(2-x^2)^~
> 2))*(EL_F(ASIN(2*SQRT(1-x^2)/(2-x^2)),SQRT(3)/2)-4*EL_E(ASIN(2*S~
> QRT(1-x^2)/(2-x^2)),SQRT(3)/2))
>
> where the incomplete elliptic integrals are defined as:
>
> EL_F(phi, k) := INT(1/SQRT(1 - k^2*SIN(t)^2), t, 0, phi)
>
> EL_E(phi, k) := INT(SQRT(1 - k^2*SIN(t)^2), t, 0, phi)
>
> Martin.
>

btw, this is what Mathematica gives for this one:

In[1]:= Integrate[ArcSin[x*Sqrt[1 - x^2]], x]

Out[1]= x ArcSin[x Sqrt[1 - x^2]] + (1/Sqrt[
1 - x^2 + x^4])(1 - x^2)^(
3/2) (2 + 2/(-1 + x^2)^2 + 2/(-1 + x^2) + (
2 (-1)^(5/6) Sqrt[(-1 + (-1)^(1/3) + x^2)/(-1 + x^2)] Sqrt[
1 - (-1)^(2/3)/(-1 + x^2)]
EllipticE[I ArcSinh[(-1)^(1/3)/Sqrt[1 - x^2]], (-1)^(2/3)])/
Sqrt[1 - x^2] - ((-1)^(2/3) Sqrt[3 + (3 (-1)^(1/3))/(-1 + x^2)]
Sqrt[1 - (-1)^(2/3)/(-1 + x^2)]
EllipticF[I ArcSinh[(-1)^(1/3)/Sqrt[1 - x^2]], (-1)^(2/3)])/
Sqrt[1 - x^2])

Maple 17 could not seem to be able to do it. returned unevaluated.

--Nasser