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) = 0
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) =1/2*(cos.(2*n*PI)*cos.(2*n*PI)-cos.((2*n+1)*PI)*
  cos.((2*n+1)*PI)) by Th90
    .=1/2*(cos(0)*cos(0+2*n*PI)-cos.((2*n+1)*PI)*cos.((2*n+1)*PI)) by Th3
    .=1/2*(cos(0)*cos(0)-cos.((2*n+1)*PI)*cos.((2*n+1)*PI)) by Th3
    .=1/2*(cos(0+2*PI)*cos(0)-cos.((2*n+1)*PI)*cos.((2*n+1)*PI)) by SIN_COS:79
    .=1/2*(1*1-cos.((2*n+1)*PI)*cos.((2*n+1)*PI)) by SIN_COS:77,79
    .=1/2*(1*1- (-cos(0))*cos(0+(2*n+1)*PI)) by Th4
    .=1/2*(1*1- (-cos(0))*(-cos(0))) by Th4
    .=1/2*(1*1- cos(0)*cos(0))
    .=1/2*(1*1- cos(0+2*PI)*cos(0)) by SIN_COS:79
    .=1/2*(1*1- cos(0+2*PI)*cos(0+2*PI)) by SIN_COS:79;
  hence thesis by SIN_COS:77;
end;
