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 Th63:
  A c=].-1,1.[ & 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
  ) = arccos.(upper_bound A)-arccos.(lower_bound A)
proof
  assume that
A1: A c=].-1,1.[ and
A2: dom ((arccos)`|].-1,1.[) = dom f2 and
A3: for x holds x in ].-1,1.[ & f2.x = -1/sqrt(1-x^2) and
A4: f2|A is continuous;
A5: A c= dom f2 by A1,A2,FDIFF_1:def 7,SIN_COS6:106;
A6: dom (arccos`|].-1,1.[ ) = ].-1,1.[ by FDIFF_1:def 7,SIN_COS6:106;
  for x being Element of REAL
    st x in dom ((arccos)`|].-1,1.[) holds ((arccos)`|].-1,1.[).x = f2 .x
  proof
    let x be Element of REAL;
    assume
A7: x in dom (arccos`|].-1,1.[);
    then
A8: -1 < x & x < 1 by A6,XXREAL_1:4;
    ((arccos)`|].-1,1.[).x =diff(arccos,x) by A6,A7,FDIFF_1:def 7,SIN_COS6:106
      .= -1 / sqrt(1-x^2) by A8,SIN_COS6:106
      .= f2.x by A3;
    hence thesis;
  end;
  then
A9: ((arccos)`|].-1,1.[) = f2 by A2,PARTFUN1:5;
  f2 is_integrable_on A by A6,A1,A2,A4,INTEGRA5:11;
  hence thesis by A1,A4,A5,A9,INTEGRA5:10,13,SIN_COS6:106;
end;
