
Re: Charlwood Fifty test results
Posted:
Jul 6, 2013 3:14 AM


On 7/5/2013 7:20 PM, clicliclic@freenet.de wrote: > > Albert Rich schrieb: >> >> On Friday, July 5, 2013 11:15:41 AM UTC10, clicl...@freenet.de wrote: >> >>> [...] But why is problem 41 untractable? >> >> [...] Problem #41 is >> >> integrate(log(sin(x)) sqrt(1+sin(x)), x) >> >> Rubi starts correctly (I think) using integration by parts that gives >> the log free integral >> >> integrate(cos(x) cot(x)/sqrt(1+sin(x)), x) >> >> Unfortunately Rubi and the other systems tested cannot determine that >> its antiderivative is >> >> 2 cos(x)/sqrt(1+sin(x)) ? 2 arctanh(cos(x)/sqrt(1+sin(x))) >> >> [...] > > The integrand COT(x)*COS(x)/SQRT(1 + SIN(x)) equals (1  SIN(x)^2) > /(SIN(x)*SQRT(1 + SIN(x))) which equals (1  SIN(x))*SQRT(1 + SIN(x)) > /SIN(x). That is, the integrand obtained after integration by parts is a > function of SIN(x) only, and of a type that Rubi should be able to > handle. > > Martin. >
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]]
Nasser

