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**HallsofIvy** If pdx+ qdy is locally exact, then it is, as you say, an exact differential. That means that there exist some function F(x,y) such that dF= pdx+ qdy, at least inside $\displaystyle \Omega$.

But then $\displaystyle \oint pdx+ qdy= \oint dF$. Let t be any parameter for the curve and we have $\displaystyle \oint pdx+ qdy= \int_a^b \frac{dF}{dt}dt= F(x(b),y(b))- F(x(a),y(a))$ where a and b are the beginning and ending values for t. But since this is a closed curve, t= a and t= b give the same point so x(b)= x(a), y(b)= y(a) and F(x(b),y(b))- F(x(a),y(a))= 0.