
Re: The Charlwood Fifty
Posted:
Jun 17, 2013 6:46 PM


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

