theorem
  A = [.2*n*PI,(2*n+1)*PI.] implies integral(-sin,A) = -2
proof
  assume A=[.2*n*PI,(2*n+1)*PI.];
  then upper_bound A=(2*n+1)*PI & lower_bound A=2*n*PI by Th37;
  then integral(-sin,A) = cos(0+(2*n+1)*PI) - cos(0+2*n*PI) by Th46
    .= -cos(0) - cos(0+2*n*PI) by Th4
    .= -cos(0) - cos(0) by Th3
    .= -cos(0+2 * PI) - cos(0) by SIN_COS:79
    .= -1 - 1 by SIN_COS:77,79;
  hence thesis;
end;
