jmc

Let $x$ be the measure of the each of the equal sides. Since the perimeter is 10 units, the side lengths measure $x$, $x$, and $10-2x$ units. Since the third side length must be positive, we have $10-2x>0$ which implies $x<5$. By the triangle inequality, the sum of the two equal sides must exceed the third side. Solving $x+x>10-2x$ gives $x>2.5$. There are $\boxed{2}$ integers strictly between 2.5 and 5.