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

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 clicliclic@freenet.de Posts: 1,245 Registered: 4/26/08
Re: Charlwood Fifty test results
Posted: Jul 6, 2013 12:42 PM

"Nasser M. Abbasi" schrieb:
>
> Just trying things....
>
> Starting with your final expression above
>
> I1 = (1 - SIN(x))*SQRT(1 + SIN(x)) /SIN(x)
>
> Rub 4.1 uses rule 2072 to integrate it, whichhas quit few
> transformations in it.
>
> When doing substitution y=sin(x) in the above first, and obtaining new
> integrand as function of y only
>
> I2 = (1 - y) Sqrt[1 + y]/(y Sqrt[1 - y^2])
>
> and integrating this instead of I1, Rubi used rule 17, which seems to
> be much simpler.
>
> Both give the same result (after letting y->sin(x) for I2 case),
> which is
>
> -2 ArcTanh[Sqrt[1 - Sin[x]]] + 2 Sqrt[1 - Sin[x]]
>

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)))

Martin.