Integration of partial derivative $\frac{dL}{dq}$ with respect to $t$ where $q$ is implicitly a function of $t$ Is $\int_{t1}^{t2} \frac{\partial L}{\partial q}\delta{q} dt$ equal to $\left[\frac{\partial L}{\partial \dot{q}}\delta{q}\right]_{t1}^{t2} $ if      $q$ implicitly depends on $t$  ? If not I misunderstood the derivation, if yes then why?
 A: Fundamental Theorem of Calculus
The fundamental theorem of calculus states there exists a function, an antiderivative, such that the following is true:
$$
\text{for } \int_a^b f(x) dx \text{, there exists } F(x) \text{, such that } f(x) = F'(x) \text{, that is}
$$
$$
\int_a^b f(x) dx = \int_a^b F'(x)dx = F(b) - F(a) 
$$
Some Preliminaries ($G$ as the Antiderivative):
When we have a multivariate function, then the above integral depends on the other variables, for instance:
$$
\int_a^bF(x, v)dx = g(v)
$$
where $g(v)$ is some function of the velocity--given different velocities, the integral will take on different values.
We know that there must exist an antiderivative, $G(x, v)$, for $F(x, v)$ such that $\frac{dG}{dx} = F(x, v)$:
$$
\frac{d}{dx}G(x, v) = \frac{\partial G}{\partial x} + \frac{\partial G}{\partial v}\frac{dv}{dx}
$$
Therefore:
$$
F(x, v) = \frac{\partial G}{\partial x} + \frac{\partial G}{\partial v}\frac{dv}{dx}
$$
Relating $G(x, v)$ and $\frac{\partial F}{\partial v}$:
\begin{align}
F(x, v) =&\ \frac{d}{dx} G(x, v) = \frac{\partial G}{\partial x} + \frac{\partial G}{\partial v}\frac{dv}{dx} \\ 
\frac{\partial F}{\partial v} =&\ \frac{\partial }{\partial v} \frac{d}{dx} \frac{\partial G}{\partial v} \\
=& \frac{\partial }{\partial v}\left(\frac{\partial G}{\partial x} + \frac{\partial G}{\partial v}\frac{dv}{dx}\right)\\
=& \frac{\partial^2 G}{\partial v\partial x} + \frac{\partial^2 G}{\partial v^2}\frac{dv}{dx} + \frac{\partial G}{\partial v}\frac{\partial \frac{dv}{dx}}{\partial v} \\
=& \frac{\partial^2 G}{\partial v\partial x} + \frac{\partial^2 G}{\partial v^2}\frac{dv}{dx}
\end{align}
($\frac{\partial \frac{dv}{dx}}{\partial v} = 0$ because we should have $v = v(x)$ and thus no dependence on $v$--hopefully this is trivial to see)
Relating $\frac{\partial }{\partial v}\int Fdx$ and $\frac{\partial G}{\partial v}$:
The integral ($\frac{\partial G}{\partial v}$ is the antiderivative):
\begin{align}
\frac{\partial}{\partial v}\int\limits_a^b F(x, v) dx =&\ \frac{\partial}{\partial v}\left(G(b, v) - G(a, v)\right) = \frac{\partial G}{\partial v}(b, v) - \frac{\partial G}{\partial v}(a, v) 
\end{align}
Finding integrand for the antiderivative $\frac{\partial G}{\partial v}$:
\begin{align}
\frac{d}{dx}\frac{\partial G}{\partial v} =&\ \frac{\partial^2 G}{\partial x \partial v} + \frac{\partial^2 G}{\partial v^2} \frac{dv}{dx}
\end{align}
Putting this all together we have shown:
\begin{align}
\int_a^b \frac{\partial F}{\partial v}dx =&\ \int_a^b \left(\frac{\partial^2 G}{\partial v\partial x} + \frac{\partial^2 G}{\partial v^2}\frac{dv}{dx}\right)dx \\
\frac{\partial }{\partial v}\int_a^b Fdx =&\ \int_a^b \left(\frac{d}{dx}\frac{\partial G}{\partial v}\right)dx = \int_a^b \left(\frac{\partial^2 G}{\partial x\partial v} + \frac{\partial^2 G}{\partial v^2}\frac{dv}{dx}\right)dx 
\end{align}
Since $\frac{\partial^2 G}{\partial x\partial v} = \frac{\partial^2 G}{\partial v\partial x}$ for "well behaved" $G(x, v)$, these two integrals are equal and thus:
$$
\int_a^b \frac{\partial F}{\partial v}dx = \frac{\partial }{\partial v}\int_a^b Fdx \text{, q.e.d}
$$
