testing ibpdv and ibpu

[salvomic]

(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