In the case of a disc rolling without slipping, we have a constraint $\dot{x}=a\dot{\theta}$ where $a$ is the radius of the disc. Note that I have considered $x$ and $\theta$ as the generalized coordinates. By definition, this is a non-holonomic constraint. However, on integrating the constraint, we arrive at $x=a\theta+\phi$ ($\phi$ is a numerical constant of integration), which turns out to be holonomic.
In the method of finding the equations of motion using Lagrangians with a Lagrangian multiplier, we have $\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q_i}}\right)-\frac{\partial L}{\partial q_i} + \lambda \frac{\partial f}{\partial q_i} = 0 $ where $f$ represents the constraint and $q_i$ is the $i$th generalised coordinate. In the case above, $q_i=\{x,\theta\}$. Now, had we used $f=\dot{x}-a\dot{\theta}$ as the constraint, the last term of the modified Euler Lagrange would have been zero. However, if we use the integrated version of the same constraint, we get non zero terms ($\lambda$ and $a\lambda$ for $x$ and $\theta$ respectively). Surprisingly, the latter is correct according to Goldstein. What am I missing here? (I'm specifically referring to the example of a hoop rolling down the incline in chapter two of Goldstein)
This brings me to the more general question: In the method of Lagrangian multipliers, how should I write the constraint relation? To illustrate what I mean, take the following constraint presented in words: the particle moves in a circle of radius $a$. If I denote the position of the particle by $r$ (generalised coordinate), then the constraint dictates $r-a=0$. Alternatively, I can write the same as $r^3-a^3=0$, whose partial derivate with respect to $r$ is not the same as in the $r-a=0$ case. What's going on here?