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