text: what is the indefinite integral of y=(1-(x^2))^(1/2)? please give reasoning for your answer. y=(1-x^2)^(1/2)  here ,we use the substitution x=sinp,then dx=cospdp 1-x^2=1-sin^2(p)=cos^2(p) cosp=(1-x^2)^(1/2)  int(1-x^2)^(1/2)dp =int(cosp*cosp)^(1/2)dp {plugging in substitution} =1/2int(1+cos2p)dp {trig identity} =1/2(p+1/2sin2p)  =1/2(p+sinp*cosp) {trig identity} but,cosp=(1-x^2)^(1/2), p=sin^-1(x) {above} =1/2sin^-1(x) +1/2x*(1-x^2)^(1/2) =1/2{arcsinx+x*(1-x^2)^(1/2)} +c  where c is an arbitrary constant  if there is anything you don't follow,give me a call-this is an important substitution  i hope that this helps