reserve f,f1,f2,g for PartFunc of REAL,REAL;
reserve A for non empty closed_interval Subset of REAL;
reserve p,r,x,x0 for Real;
reserve n for Element of NAT;
reserve Z for open Subset of REAL;

theorem
  A = [.x+2*n*PI,x+(2*n+1)*PI.] implies integral(sin+cos,A) = 2*cos x - 2*sin 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+cos,A) = (-cos).(x+(2*n+1)*PI) - (-cos).(x+2*n*PI) + sin.(x
  +(2*n+1)*PI) - sin.(x+2*n*PI) by Th69
    .= -(cos.(x+(2*n+1)*PI)) - (-cos).(x+2*n*PI) + sin.(x+(2*n+1)*PI) - sin.
  (x+2*n*PI) by VALUED_1:8
    .= -(cos(x+(2*n+1)*PI)) - -(cos(x+2*n*PI)) + sin(x+(2*n+1)*PI) - sin(x+2
  *n*PI) by VALUED_1:8
    .= -(-cos x) - -(cos(x+2*n*PI)) + sin(x+(2*n+1)*PI) - sin(x+2*n*PI) by Th4
    .= -(-cos x) - -cos x + sin(x+(2*n+1)*PI) - sin(x+2*n*PI) by Th3
    .= -(-cos x) - -cos x + -sin x - sin(x+2*n*PI) by Th2
    .= cos x + cos x -sin x - sin x by Th1
    .= 2* cos x - 2*sin x;
  hence thesis;
end;
