In a closed system of masses, the Lagrangian, \(L\), is some function of the coordinates, \(q\), the velocities, \(\dot{q}\), and time, \(t\).
The principle of least action holds that the motion of the masses evolves such that the action, \(S\), is minimized.
\[ S = \int_{t_1}^{t_2} L(q, \dot{q}, t) \mathrm{d}t \]
If \(q(t)\) is the function that minimizes \(S\), then any other function \(q(t) + \delta q(t)\) increases \(S\). The function \(\delta q\) is a variation of the trajectory \(q\) and all such variation functions must be small everywhere between \(t_1\) and \(t_2\) and satisfy \(\delta q(t_1) = \delta q(t_2) = 0\).
Then for our action to be minimized, the first variation of the integral should be zero.
\[ \delta S = \delta \int_{t_1}^{t^2} L(q, \dot{q}, t) \mathrm{d}t = 0\\ \]
or
\[ \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}} \right) \delta q \mathrm{d}t = 0\\ \]
The action:
\[ S(q, \dot{q}, t) = \int_{t_1}^{t_2} L(q, \dot{q}, t) \mathrm{d}t \]
The first variation with \(q(t) + \epsilon \eta(t)\),
\[ \begin{align} \delta S(q, \dot{q}, t, \eta) &= \frac{\mathrm{d}}{\mathrm{d} \epsilon} S(q + \epsilon \eta, \dot{q} + \epsilon \dot{\eta}, t) \Big|_{\epsilon = 0}\\ &= \int_{t_1}^{t_2} \frac{\mathrm{d}}{\mathrm{d} \epsilon} L(q + \epsilon \eta, \dot{q} + \epsilon \dot{\eta}, t) \Big|_{\epsilon = 0} \mathrm{d}t\\ &= \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} \eta + \frac{\partial L}{\partial \dot{q}} \dot{\eta} \right) \mathrm{d}t\\ &= \int_{t_1}^{t_2} \frac{\partial L}{\partial q} \eta \mathrm{d}t + \int_{t_1}^{t_2} \frac{\partial L}{\partial \dot{q}} \dot{\eta} \mathrm{d}t\\ \end{align} \]
Using integration by parts, i.e. \(\int_{t_1}^{t_2} u(t)\dot{v}(t) \mathrm{d}t = \Big[ u(t) v(t) \Big]_{t_1}^{t_2} - \int_{t_1}^{t_2} \dot{u}(t) v(t) \mathrm{d}t\), substituting with $u = and \(\dot{v} = \dot{\eta}\),
\[ \begin{align} \delta S(q, \dot{q}, t, \eta) &= \int_{t_1}^{t_2} \frac{\partial L}{\partial q} \eta \mathrm{d}t + \Big[ \frac{\partial L}{\partial \dot{q}} \eta \Big]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}} \eta \mathrm{d}t\\ \end{align} \]
Noting that \(\eta(t_1) = \eta(t_2) = 0\) cancels the middle term and factoring out \(\eta\), we require the variation equal zero.
\[ \delta S(q, \dot{q}, t, \eta) = \int_{t_1}^{t_2} \Big( \frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}} \Big) \eta \mathrm{d}t = 0 \]For this to hold for all possible variations we must have the integrand zero everywhere.
\[ \frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}} = 0 \] This is the Euler–Lagrange equation.