testing ibpdv and ibpu

05122015, 07:59 PM
(This post was last modified: 05122015 08:05 PM by salvomic.)
Post: #5




RE: testing ibpdv and ibpu
(05122015 07:45 PM)Aries Wrote: Hey Salvo, thank you, Aries! I'm reading it and trying \[ \int_0^1{\sqrt{1x^{2}}}dx \] assume(u>0); a:=subst('integrate(sqrt(1x^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(1x^2)+c (changing also the integral bounds) ... However it's interesting this approach. Salvo ∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C  DM42, DM41X  WP34s Prime Soft. Lib 

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