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Topic: Charlwood Fifty test results
Replies: 16   Last Post: Sep 19, 2013 10:09 PM

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Nasser Abbasi

Posts: 5,647
Registered: 2/7/05
Re: Charlwood Fifty test results
Posted: Jul 6, 2013 1:25 PM
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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(1-SIN(X)^2), then

T1 = SQRT( (1-SIN(X)^2) / (1 + SIN(x)) )

Let
SIN(X) = Y
then it is the same as
T1 = SQRT ( (1-Y^2)/(1+Y) )
which is
T1 = SQRT ( 1-Y )
But Y = SIN(X), hence
T1 = SQRT ( 1- SIN(X) )
which is T2

So this is what happened.

--Nasser




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