Proving that $\int \delta \dot{x} dt = \delta x$ Everytime I've seen this I've assumed it was true. It seems plausible. But I would like to rigorously prove it. I think this is correct, but I would like another opinion because my mathematics is very rusty and I don't have much confidence in what I'm doing.
$ \int  \delta \dot{x}  dt  = \delta x $ is used when deriving the Euler Lagrange Equations for a Lagrangian in Physics. 
***$\delta$ is defined below.
[Note:This formula is used in the derivation because integration by parts is applied to change $ \delta \dot{x} $ to $ \delta x $ and then you group the like terms such that  $a \delta x$ + $b \delta x $ becomes $(a+b)\delta x$.]
So this is what I've done.
I take $\dot{x}$ to be a function. Then $\int \delta \dot{x} = \int \frac{\mathrm{d} \dot{x} }{\mathrm{d} t} dt = g(t)$
So
$\int \delta \dot{x}$ = $\delta x$
Have I left anything out?  
Edit: 
***The $\delta$ of a function is the change in the function for a small change due to a small change in the argument of the function. That is achieved by applying Taylors theorem to first order. (We could apply it to higher orders but, in this case, we are interested in the maximum and minimum of the function and not the concavity of the function. We don't need that higher order information.)
By Taylors Theorem: 
$g(t+ dt)= g(t) + \frac{\mathrm{d} g(t) }{\mathrm{d} t} dt$
Therefore, the change in $g(t)$ is:
$ \delta g(t)= g(t+ dt) - g(t)= \frac{\mathrm{d} g(t) }{\mathrm{d} t} dt$
 A: Well, I'm not sure that your proof is correct. It looks to me like some explanations I used to give myself when I started to play around with Hamilton's principle, before realizing which was the correct way to go through those kinds of calculations. I'll give you the strictly correct answer to your question.
Recall that you're varying the time variable and the position function as follows:
$$\begin{cases}
t\to\tilde{t}=t+\delta t=t+\epsilon T(t)\\
x(t)\to\tilde{x}(\tilde{t})=x(t)+\delta x(t)
\end{cases}$$
where $T$ is a function of time. These are the variations you make in the general case, by which you get the conservation of energy and momentum other than the Euler-Lagrange equation. If you're only interested in the Euler-Lagrange equation, you take $T$ to be zero. In this case you can skip the derivation that follows and go straight to the answer in the last few lines.
You must calculate the variation $\delta x$ and the variation in the velocity, $\delta\dot{x}$. As for $\delta x$, you have:
$$\delta x=\tilde{x}(\tilde{t})-x(t)=\tilde{x}(t+\delta t)-x(t)\approx\tilde{x}(t)+\dot{\tilde{x}}(t)\,\delta t-x(t)$$
Please notice that $\delta t$ is a function of time, so you're working in the hypothesis that you can variate $x$ taking the variation in time to be a constant, which is justified by the fact that the coordinate is being varied "after" the variation in time has occurred, i.e. the coordinate doesn't see the time-dependence of the time variation as it is considered to be varied punctually in time. Now, $\tilde{x}(t)$ is composed by the old coordinate function $x$ and the variation of the coordinate function $\bar{\delta}x$, i.e.
$$\tilde{x}(t)=x(t)+\bar{\delta}x(t)=x(t)+\eta\,X(x)$$
Then you have
$$\delta x\approx\bar{\delta}x(t)+\frac{d}{dt}\,(x(t)+\bar{\delta}x(t))\,\delta t\approx\bar{\delta}x(t)+\dot{x}(t)\,\delta t$$
as the last term was of the second order. You found
$$\delta x(t)=\bar{\delta}x(t)+\dot{x}(t)\,\delta t$$
Now comes the calculation you're looking for. You must calculate the variation in the velocity. The varied velocity is
$$\frac{d \tilde{x}}{d\tilde{t}}(\tilde{t})=\dot{x}(t)+\delta{\dot{x}}(t)$$
and the variation must be read as $\delta(\dot{x})$, which, in general, is different from
$$\frac{d}{dt}\delta\tilde{x}$$
as you're deriving with respect to the new time variable. Let us calculate the variation $\delta{\dot{x}}(t)$,
$$\delta\dot{x}(t)=\frac{d\tilde{x}}{d\tilde{t}}(\tilde{t})-\dot{x}(t)$$
You notice that
$$\frac{dx}{dt}=\frac{dx}{d\tilde{t}}\frac{d\tilde{t}}{dt}=(1+\delta\dot{t})\,\frac{dx}{d\tilde{t}}$$
where the variation of the time variable is equal to the derivative of the variation of time; so you have
$$\delta\dot{x}(t)\approx\frac{d}{d\tilde{t}}[\tilde{x}(\tilde{t})-x(t)]-\frac{dx}{d\tilde{t}}\delta\dot{t}(t)=\frac{d}{d\tilde{t}}[\tilde{x}(\tilde{t})-x(\tilde{t}-\delta t)]-\frac{dx}{d\tilde{t}}\delta\dot{t}(t)\approx\frac{d}{d\tilde{t}}[\tilde{x}(\tilde{t})-x(\tilde{t})+\frac{dx}{d\tilde{t}}\delta t(\tilde{t}))]-\frac{dx}{d\tilde{t}}\delta\dot{t}(t)=\frac{d}{d\tilde{t}}\,\bar{\delta}x(\tilde{t})+\frac{d^{2}x}{d\tilde{t}^{2}}\,\delta t(\tilde{t})+\frac{dx}{d\tilde{t}}\frac{d\delta t}{d\tilde{t}}(\tilde{t})-\frac{dx}{d\tilde{t}}\delta\dot{t}(t)$$
Getting rid of all the second-order-and-above terms, you're left with
$$\delta\dot{x}=\bar{\delta}\dot{x}+\ddot{x}\,\delta t$$
If you compare $\delta\dot{x}$ and $\frac{d}{dt}\delta x$ you'll notice that they are NOT equivalent, as
$$\begin{cases}
\delta\dot{x}=\bar{\delta}\dot{x}+\ddot{x}\,\delta t\\
\frac{d}{dt}\delta x=\bar{\delta}\dot{x}+\dot{x}\,\delta\dot{t}+\ddot{x}\,\delta t
\end{cases}$$
Then, when you integrate
$$\int\delta\dot{x}\ dt$$
you must ask yourself whether you're integrating $\delta\dot{x}$ or $\frac{d}{dt}\delta x$, depending on where the integrand came from. As for the latter,
$$\int\frac{d}{dt}\delta x\ dt=\delta x$$
obviously. As for the former,
$$\int\delta\dot{x}\ dt\neq\delta x$$
Notice, however, that the two integrals are equivalent in the case where $\delta t$ is zero, i.e. if you only need to get Euler-Lagrange's equations, or when the variation $\delta t$ is constant.
