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Topic: The Charlwood Fifty
Replies: 52   Last Post: Jun 24, 2013 10:24 PM

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 Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: The Charlwood Fifty
Posted: Jun 17, 2013 6:46 PM
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On 6/17/2013 5:09 PM, clicliclic@freenet.de wrote:
>
> "Nasser M. Abbasi" schrieb:

>>
>> On a related point, would you please help me understand how
>> free version of reduce transformed
>>
>> arcsin(x)*log(x)
>>
>> to
>>
>> arcsin( sin(g0) ) * cos(g0) * log( sin(g0) )
>>
>> by replacing x with sin(g0).
>>
>> i.e Where does cos(g0) term come from in the above transformation?
>>
>> Here is a link to the reduce trace for integral #1, which it
>> could not do btw. And the above was the first step in the process.
>>
>> http://12000.org/my_notes/ten_hard_integrals/reduce_logs/1/HTML/trace_1.html
>>

>
> In the original INT(ASIN(x)*LN(x), x), Reduce makes the variable
> substitution x = SIN(y), dx = COS(y)*dy (I've written y for g0). So the
> factor COS(y) just represents the derivative dx/dy that must be included
> in the transformed integrand.
>

Ah! Yes, ofcourse, but the trace was confusing. If you see, it says
it was transforming arcsin(x)*log(x) and NOT arcsin(x)*log(x)*dx

If dx is included, then it works out. Ok, here it is step
by step again.

given:
I = arcsin(x) log(x) dx

let x=sin(y), then dx/dy=cos(y), or dx=cos(y)*dy
hence I becomes

arcsin(sin(y)) log(sin(y) cos(y) dy

So it was the change from dx to dy which introduced the cos(y)
term, that is what I overlooked since trace did not show dx in
what it started with.

thanks
--Nasser

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