theorem
  A = [.0,PI/2.] implies integral(cos,A) = 1
proof
  assume A=[.0,PI/2.];
  then upper_bound A=PI/2 & lower_bound A=0 by Th37;
  then integral(cos,A) = 1 - sin.0 by Th39,SIN_COS:77
    .= 1 - cos.(PI/2 - 0) by SIN_COS:78
    .= 1 - 0 by SIN_COS:77;
  hence thesis;
end;
