 reserve a,x for Real;
 reserve n for Element of NAT;
 reserve A for non empty closed_interval Subset of REAL;
 reserve f,h,f1,f2 for PartFunc of REAL,REAL;
 reserve Z for open Subset of REAL;

theorem
 A c= Z & f=(sin(#)cos)^ & Z c= dom (ln*tan) & Z = dom f & f|A is continuous
 implies integral(f,A)=(ln*tan).(upper_bound A)-(ln*tan).(lower_bound A)
proof
  assume
A1: A c= Z & f=(sin(#)cos)^
    & Z c= dom (ln*tan) & Z = dom f & f|A is continuous;then
A2: f is_integrable_on A & f|A is bounded by INTEGRA5:10,11;
A3: (ln*tan) is_differentiable_on Z by A1,FDIFF_8:18;
A4:for x st x in Z holds f.x=1/(sin.x*cos.x)
   proof
   let x;
   assume x in Z;
   then ((sin(#)cos)^).x =1/((sin(#)cos).x) by A1,RFUNCT_1:def 2
        .=1/(sin.x*cos.x) by VALUED_1:5;
    hence thesis by A1;
    end;
A5: for x being Element of REAL
       st x in dom ((ln*tan)`|Z) holds ((ln*tan)`|Z).x = f.x
    proof
      let x be Element of REAL;
      assume x in dom ((ln*tan)`|Z);then
A6:   x in Z by A3,FDIFF_1:def 7; then
      ((ln*tan)`|Z).x = 1/(sin.x*cos.x) by A1,FDIFF_8:18
        .=f.x by A4,A6;
      hence thesis;
    end;
   dom ((ln*tan)`|Z) = dom f by A1,A3,FDIFF_1:def 7;then
   ((ln*tan)`|Z)=f by A5,PARTFUN1:5;
   hence thesis by A1,A2,FDIFF_8:18,INTEGRA5:13;
end;
