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 = [.x+2*n*PI,x+(2*n+1)*PI.] implies integral(sin(#)cos,A) = 0
proof
  assume A = [.x+2*n*PI,x+(2*n+1)*PI.];
  then upper_bound A=x+(2*n+1)*PI & lower_bound A=x+2*n*PI by Th37;
  then
  integral(sin(#)cos,A) = 1/2*(cos.(x+2*n*PI)*cos.(x+2*n*PI)-cos.(x+(2*n+1
  )*PI)*cos.(x+(2*n+1)*PI)) by Th90
    .= 1/2*(cos(x)*cos(x+2*n*PI)-cos(x+(2*n+1)*PI)*cos(x+(2*n+1)*PI)) by Th3
    .= 1/2*(cos(x)*cos(x)-cos(x+(2*n+1)*PI)*cos(x+(2*n+1)*PI)) by Th3
    .= 1/2*(cos(x)*cos(x)- (-cos(x))*cos(x+(2*n+1)*PI)) by Th4
    .= 1/2*(cos(x)*cos(x)- (-cos(x))*(-cos(x))) by Th4
    .= 0;
  hence thesis;
end;
