
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.

