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 = [.2*n*PI,(2*n+1)*PI.] implies integral(sin+cos,A) = 2
proof
  assume A = [.2*n*PI,(2*n+1)*PI.];
  then upper_bound A=(2*n+1)*PI & lower_bound A=(2*n*PI) by Th37;
  then
  integral(sin+cos,A) = (-cos).((2*n+1)*PI) - (-cos).(2*n*PI) + sin.((2*n+
  1)*PI) - sin.(2*n*PI) by Th69
    .= -(cos.((2*n+1)*PI)) - (-cos).(2*n*PI) + sin.((2*n+1)*PI) - sin.(2*n*
  PI) by VALUED_1:8
    .= -(cos.((2*n+1)*PI)) - -(cos(0+2*n*PI)) + sin.((2*n+1)*PI) - sin.(0+2*
  n*PI) by VALUED_1:8
    .= -(cos.((2*n+1)*PI)) - -(cos(0)) + sin.((2*n+1)*PI) - sin.(0+2*n*PI)
  by Th3
    .= -(cos.((2*n+1)*PI)) + (cos(0)) + sin.((2*n+1)*PI) - sin.(0+2*n*PI)
    .= -(cos.((2*n+1)*PI)) + (cos(0+2*PI)) + sin.((2*n+1)*PI) - sin.(0+2*n*
  PI) by SIN_COS:79
    .= -(cos.((2*n+1)*PI)) + 1 + sin.((2*n+1)*PI) - sin(0+2*n*PI) by SIN_COS:77
    .= -(cos.((2*n+1)*PI)) + 1 + sin.((2*n+1)*PI) - sin(0) by Th1
    .= -(cos.((2*n+1)*PI)) + 1 + sin.((2*n+1)*PI) - sin(0+2*PI) by SIN_COS:79
    .= -(cos(0+(2*n+1)*PI)) + 1 + sin(0+(2*n+1)*PI) by SIN_COS:77
    .= -(cos(0+(2*n+1)*PI)) + 1 + -sin(0) by Th2
    .= - -cos(0) + 1 + -sin(0) by Th4
    .= cos(0) + 1 - sin(0)
    .= cos(0+2*PI) + 1 - sin(0) by SIN_COS:79
    .= cos(0+2*PI) + 1 - sin(0+2*PI) by SIN_COS:79
    .= 2 by SIN_COS:77;
  hence thesis;
end;
