theorem
  A = [.0,PI.] implies integral(sin(#)cos,A) = 0
proof
  assume A = [.0,PI.];
  then upper_bound A=PI & lower_bound A=0 by Th37;
  then integral(sin(#)cos,A) =1/2*(cos.0*cos.0-cos.PI*cos.PI) by Th90
    .=1/2*(cos.(0+2*PI)*cos.0-cos.PI*cos.PI) by SIN_COS:78
    .=1/2*(1*1-(-1)*(-1)) by SIN_COS:76,78;
  hence thesis;
end;
