reserve f,f1,f2,g for PartFunc of REAL,REAL;
reserve A for non empty closed_interval Subset of REAL;
reserve p,r,x,x0 for Real;
reserve n for Element of NAT;
reserve Z for open Subset of REAL;

theorem
  A = [.-sqrt 2/2,sqrt 2/2.] & dom ((arccos)`|].-1,1.[) = dom f2 & (for
  x holds x in ].-1,1.[ & f2.x = -1/sqrt(1-x^2)) & f2|A is continuous implies
  integral(f2,A) = -PI/2
proof
  assume that
A1: A = [.-sqrt 2/2,sqrt 2/2.] and
A2: ( dom ((arccos)`|].-1,1.[) = dom f2 & for x holds x in ].-1,1.[ & f2
  .x = -1/ sqrt(1-x^2) )& f2|A is continuous;
  upper_bound A = sqrt 2/2 & lower_bound A = -sqrt 2/2 by A1,Th37;
  then integral(f2,A) = arccos.(sqrt 2/2)-arccos.(-sqrt 2/2) by A1,A2,Th9,Th63
    .= arccos(sqrt 2/2) - arccos.(-sqrt 2/2) by SIN_COS6:def 4
    .= arccos(sqrt 2/2) - arccos(-sqrt 2/2) by SIN_COS6:def 4
    .=(-2*PI)/4 by Th15,Th16;
  hence thesis;
end;
