
Re: Charlwood Fifty test results
Posted:
Jul 6, 2013 1:25 PM


On 7/6/2013 11:42 AM, clicliclic@freenet.de wrote:
> > If Rubi evaluates the integral > > INT((1  SIN(x))*SQRT(1 + SIN(x))/SIN(x), x) > > to > > 2*SQRT(1  SIN(x))  2*ATANH(SQRT(1  SIN(x))) > > then you have found a bug, since the derivative > > DIF(2*SQRT(1  SIN(x))  2*ATANH(SQRT(1  SIN(x))), x) > > equals > > COS(x)/SQRT(COS(x)^2)*(1  SIN(x))*SQRT(1 + SIN(x))/SIN(x) > > where the piecewise constant COS(x)/SQRT(COS(x)^2) differs from unity. > > Accordingly, a correct evaluation of the integral is > > 2*COS(x)/SQRT(1 + SIN(x))  2*ATANH(COS(x)/SQRT(1 + SIN(x))) >
Ok, please let me backup. Rubi does generate the result you showed
2* T1  2*ATANH( T1 )
Where
T1 = COS(x)/SQRT(1 + SIN(x))
But I took it abone myself to just replace T1 above with T2 = SQRT(1  SIN(x))
since they seemed the same to me(*), so I wrote in this post that Rubi gave
2* T2  2*ATANH( T2 )
Which you complained about. I see now that T1=T2 only from Pi/2..Pi/2.
Sorry about this confusion. I need to review this whole process again to see why it did not work for all x range using this.  (*) In T1, let COS(x) = SQRT(1SIN(X)^2), then
T1 = SQRT( (1SIN(X)^2) / (1 + SIN(x)) )
Let SIN(X) = Y then it is the same as T1 = SQRT ( (1Y^2)/(1+Y) ) which is T1 = SQRT ( 1Y ) But Y = SIN(X), hence T1 = SQRT ( 1 SIN(X) ) which is T2
So this is what happened.
Nasser

