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*cos)(#)sin,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*cos)(#)sin,A)=((-1/(n+1))(#)( #Z (n+1)*cos)).(2*PI) -((-
  1/(n+1))(#)( #Z (n+1)*cos)).0 by Th22
    .= (-1/(n+1))*( #Z (n+1)*cos).(2*PI) -((-1/(n+1))(#)( #Z (n+1)*cos)).0
  by VALUED_1:6
    .= (-1/(n+1))*( #Z (n+1)*cos).(2*PI) -(-1/(n+1))*( #Z (n+1)*cos).0 by
VALUED_1:6
    .= (-1/(n+1))*(( #Z (n+1)).(cos.dp)) -(-1/(n+1))*( #Z (n+1)*cos).0
  by FUNCT_1:13,SIN_COS:24
    .= (-1/(n+1))*(( #Z (n+1)).(cos.(2*PI))) -(-1/(n+1))*(( #Z (n+1)).(cos.0
  )) by FUNCT_1:13,SIN_COS:24,Lm6
    .= 0 by SIN_COS:30,76;
  hence thesis;
end;
