To summarize, the problem states that we prove $$\mathcal L'=\mathcal L+\frac{\mathrm d}{\mathrm dt}F(\mathbf q,t)$$ is also a valid Lagrangian, that is, it satisfies the Euler-Lagrange equations for each coordinate (assuming $F$ is at least $\mathcal C^2$).
I've seen a solution online in which they just substitute $\mathcal L'$ in the E-L equations and prove they're identical to $0$ $\forall i$: $$\frac{\partial \mathcal L'}{\partial q_i}-\frac{\mathrm d}{\mathrm dt}\frac{\partial \mathcal L'}{\partial \dot q_i}=\cdots=\underbrace{\frac{\partial \mathcal L}{\partial q_i}-\frac{\mathrm d}{\mathrm dt}\frac{\partial \mathcal L}{\partial \dot q_i}}_0+\underbrace{\frac{\partial}{\partial q_i}\frac{\mathrm d F}{\mathrm dt}-\frac{\mathrm d}{\mathrm dt}\frac{\partial F}{\partial q_i}}_0=0$$ Instead, what my professor did today to solve this problem was simply proving $$\delta S=\delta\int_{t_1}^{t_2}\mathcal L'\,\mathrm dt=0$$ In fact it is pretty straightforward to follow from here: \begin{align} \delta S&=\delta\int_{t_1}^{t_2}\mathcal L'\,\mathrm dt\\ &=\delta\int_{t_1}^{t_2}\mathcal L\,\mathrm dt+\delta\int_{t_1}^{t_2}\frac{\mathrm d}{\mathrm dt}F\,\mathrm dt\\ &=0+\delta[F(\mathbf q(t),t)|_{t_1}^{t_2}\\ &=\delta F(\mathbf q(t_2),t_2)-\delta F(\mathbf q(t_1),t_1)\overset{?}{=}0 \end{align} but the issue arises when trying to prove the last line... Do I just assume the variation of a constant is $0$ and that's it? Or was the penultimate step incorrect?
