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

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Nasser Abbasi

Posts: 5,698
Registered: 2/7/05
Re: Charlwood Fifty test results
Posted: Jul 6, 2013 3:14 AM
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On 7/5/2013 7:20 PM, clicliclic@freenet.de wrote:
>
> Albert Rich schrieb:

>>
>> On Friday, July 5, 2013 11:15:41 AM UTC-10, 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




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