
Re: Charlwood Fifty test results
Posted:
Jul 6, 2013 5:10 PM


On Saturday, July 6, 2013 10:28:30 AM UTC10, clicl...@freenet.de wrote:
> Since Rubi manages to return the proper T1based result already, it > shouldn't be hard to make it handle Charlwood's problem 41. It merely > needs to be taught to normalize integrands like COT(x)*COS(x)/ > SQRT(1 + SIN(x)) to their sineonly form. This would be just one > normalization among others performed regularly, and would only require > some twisting or minor extension of existing code, I guess.
Rather than blindly normalizing such integrands to their sineonly form and hoping for the best, it is safer to drive the degree of the cosine factor toward zero using the following hot new rule: If a^2b^2=0 and m>0 is even, then
Int[Cos[c+d*x]^m*Sin[c+d*x]^n*(a+b*Sin[c+d*x])^p, x]
goes to
1/a*Int[Cos[c+d*x]^(m2)*Sin[c+d*x]^n*(a+b*Sin[c+d*x])^(p+1), x]  1/b*Int[Cos[c+d*x]^(m2)*Sin[c+d*x]^(n+1)*(a+b*Sin[c+d*x])^(p+1), x]
This rule is based on the trig identity
Cos[z]^2 = (a+b*Sin[z])/a  Sin[z]*(a+b*Sin[z])/b
which is valid if a^2b^2=0. The new rule will make it possible for the next version of Rubi to integrate a whole new class of expressions including the integrand in question, as well as the much more difficult one
Cos[x]^4/(Sin[x]^3*(a+a*Sin[x])^(3/2))
Albert

