Tim_Lou Posted January 30, 2005 Report Posted January 30, 2005 how do you integrate sqrt (1-x^2), besides expanding as a infinite series? is there such way?
sanctus Posted January 30, 2005 Report Posted January 30, 2005 I guess with the residual thm it might work, but it's not trivial and not knowing if it really works I didn't want to invest all the time in searching....
Tim_Lou Posted January 30, 2005 Author Report Posted January 30, 2005 if its possible (without infinite series), pi could be expressed in a single term...by finding the integral from 0 to 1 of sqrt(1-x^2) and times it by 4.
maddog Posted January 31, 2005 Report Posted January 31, 2005 how do you integrate sqrt (1-x^2), besides expanding as a infinite series? is there such way? Offhand, I wound try either a substitution or use integration by parts. The substitutionmight work first. The second may not work at all. My guess is use a trig identity like 1 = x^2 + y^2 = sin^2(t) + cos^2(t) y^2 = 1 - x^2y = (1 - x^2)^(1/2)dy = [(1 - x^2)^(-3/2)]2x dx Put this in, I think now you can do without the "by parts integration". :) :) Maddog
sanctus Posted January 31, 2005 Report Posted January 31, 2005 Ok, maddog I also thought about substituion but I thought about y= sin(x), that why I din't find nothing. Well seen...
Tim_Lou Posted February 5, 2005 Author Report Posted February 5, 2005 i used sinx and it does work for me...substitute sink for x, i got integral (cosk)^2 dk,using (cosx)^2 = (cos(2x)+1) /2 and it works for me...
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