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testing ibpdv and ibpu
05-12-2015, 07:59 PM (This post was last modified: 05-12-2015 08:05 PM by salvomic.)
Post: #5
RE: testing ibpdv and ibpu
(05-12-2015 07:45 PM)Aries Wrote:  Hey Salvo,
as to the integration by substitution, take a look here:
http://www.hpmuseum.org/forum/thread-226...l#pid20100

Aries

thank you, Aries!

I'm reading it and trying
\[ \int_0^1{\sqrt{1-x^{2}}}dx \]
assume(u>0); a:=subst('integrate(sqrt(1-x^2)),0,1,x)',x=sin(u)); a;
and I get [u, π/4 π/4]
Manually I have solution = (½)arcsin1 = π/4 (only?)
Is it right?
I thought to have this series: ∫cos(u)^2du, then (u/2)+(sin(2u))/4+c and finally (½)(arcsin(x)+x*sqrt(1-x^2)+c (changing also the integral bounds) ...

However it's interesting this approach.

Salvo

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Messages In This Thread
testing ibpdv and ibpu - salvomic - 05-12-2015, 01:57 PM
RE: testing ibpdv and ibpu - parisse - 05-12-2015, 02:44 PM
RE: testing ibpdv and ibpu - salvomic - 05-12-2015, 02:45 PM
RE: testing ibpdv and ibpu - Aries - 05-12-2015, 07:45 PM
RE: testing ibpdv and ibpu - salvomic - 05-12-2015 07:59 PM
RE: testing ibpdv and ibpu - parisse - 05-13-2015, 05:14 AM
RE: testing ibpdv and ibpu - salvomic - 05-13-2015, 07:18 AM
RE: testing ibpdv and ibpu - Arno K - 05-12-2015, 10:18 PM
RE: testing ibpdv and ibpu - Arno K - 05-13-2015, 07:34 AM
RE: testing ibpdv and ibpu - salvomic - 05-13-2015, 07:57 AM
RE: testing ibpdv and ibpu - Arno K - 05-13-2015, 08:16 AM
RE: testing ibpdv and ibpu - salvomic - 05-13-2015, 08:33 AM
RE: testing ibpdv and ibpu - Arno K - 05-13-2015, 08:52 AM
RE: testing ibpdv and ibpu - salvomic - 05-13-2015, 08:57 AM
RE: testing ibpdv and ibpu - Arno K - 05-13-2015, 09:32 AM
RE: testing ibpdv and ibpu - salvomic - 05-13-2015, 09:43 AM
RE: testing ibpdv and ibpu - parisse - 05-13-2015, 02:43 PM
RE: testing ibpdv and ibpu - Arno K - 05-13-2015, 05:22 PM



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