```Date: Jun 6, 2013 1:33 PM
Author: Nasser Abbasi
Subject: Re: The Charlwood Fifty

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 asyou have shown it (i.e. multiplication not division), so there is atypo 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
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