reserve r,x,x0,a,b for Real;
reserve n,m for Element of NAT;
reserve A for non empty closed_interval Subset of REAL;
reserve Z for open Subset of REAL;

theorem
  A = [.0,2*PI.] implies integral(( #Z n*sin)(#)cos,A) = 0
proof
  assume A=[.0,2*PI.];
  then upper_bound A=2*PI & lower_bound A=0 by INTEGRA8:37;
  then
  integral(( #Z n*sin)(#)cos,A)=((1/(n+1))(#)( #Z (n+1)*sin)).(2*PI) -((1/
  (n+1))(#)( #Z (n+1)*sin)).0 by Th19
    .= 1/(n+1)*( #Z (n+1)*sin).(2*PI) -((1/(n+1))(#)( #Z (n+1)*sin)).0 by
VALUED_1:6
    .= 1/(n+1)*( #Z (n+1)*sin).dp -1/(n+1)*( #Z (n+1)*sin).0 by VALUED_1:6
    .= 1/(n+1)*(( #Z (n+1)).(sin.(2*PI))) -1/(n+1)*( #Z (n+1)*sin).0 by
FUNCT_1:13,SIN_COS:24
    .= 1/(n+1)*(( #Z (n+1)).(sin.(2*PI))) -1/(n+1)*(( #Z (n+1)).(sin.0)) by
FUNCT_1:13,SIN_COS:24,Lm6
    .= 0 by SIN_COS:30,76;
  hence thesis;
end;
