Tim_Lou Posted January 30, 2005 Report Share Posted January 30, 2005 how do you integrate sqrt (1-x^2), besides expanding as a infinite series? is there such way? Link to comment Share on other sites More sharing options...
sanctus Posted January 30, 2005 Report Share 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.... Link to comment Share on other sites More sharing options...
Tim_Lou Posted January 30, 2005 Author Report Share 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. Link to comment Share on other sites More sharing options...
maddog Posted January 31, 2005 Report Share 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 Link to comment Share on other sites More sharing options...
sanctus Posted January 31, 2005 Report Share 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... Link to comment Share on other sites More sharing options...
Tim_Lou Posted February 5, 2005 Author Report Share 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... Link to comment Share on other sites More sharing options...
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