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

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Albert D. Rich

Posts: 217
From: Hawaii Island
Registered: 5/30/09
Re: Charlwood Fifty test results
Posted: Jul 6, 2013 5:10 PM
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On Saturday, July 6, 2013 10:28:30 AM UTC-10, clicl...@freenet.de wrote:

> Since Rubi manages to return the proper T1-based 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 sine-only 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 sine-only 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^2-b^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]^(m-2)*Sin[c+d*x]^n*(a+b*Sin[c+d*x])^(p+1), x] -
1/b*Int[Cos[c+d*x]^(m-2)*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^2-b^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



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