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 = [.-PI/2,PI/2.] implies integral(( #Z n*cos)(#)sin,A) = 0
proof
  assume A=[.-PI/2,PI/2.];
  then upper_bound A=PI/2 & lower_bound A=-PI/2 by INTEGRA8:37;
  then
  integral(( #Z n*cos)(#)sin,A)=((-1/(n+1))(#)( #Z (n+1)*cos)).(PI/2) -((-
  1/(n+1))(#)( #Z (n+1)*cos)).(-PI/2) by Th22
    .= (-1/(n+1))*( #Z (n+1)*cos).(PI/2) -((-1/(n+1))(#)( #Z (n+1)*cos)).(-
  PI/2) by VALUED_1:6
    .= (-1/(n+1))*( #Z (n+1)*cos).(PI/2) -(-1/(n+1))*( #Z (n+1)*cos).(-PI/2)
  by VALUED_1:6
    .= (-1/(n+1))*(( #Z (n+1)).(cos.pd)) -(-1/(n+1))*( #Z (n+1)*cos).mpd
   by FUNCT_1:13,SIN_COS:24
    .= (-1/(n+1))*(( #Z (n+1)).(cos.pd)) -(-1/(n+1))*(( #Z (n+1)).(cos.(
  -PI/2))) by FUNCT_1:13,SIN_COS:24
    .= (-1/(n+1))*(( #Z (n+1)).(cos.(PI/2))) -(-1/(n+1))*(( #Z (n+1)).(cos.(
  PI/2))) by SIN_COS:30
    .= 0;
  hence thesis;
end;
