= Solution
For a variation $q\mapsto q+\epsilon\eta$ with $\eta(t_0)=\eta(t_1)=0$, the first variation is
$$
\delta S=\int_{t_0}^{t_1}(L_q\eta+L_{\dot q}\dot\eta)dt=[L_{\dot q}\eta]_{t_0}^{t_1}+\int_{t_0}^{t_1}\left(L_q-\frac d{dt}L_{\dot q}\right)\eta\,dt.
$$
The endpoint term vanishes. Since the interior variation is arbitrary, the <fundamental lemma of the calculus of variations> gives the <Euler-Lagrange equation>
$$
\boxed{\frac d{dt}\frac{\partial L}{\partial\dot q}-\frac{\partial L}{\partial q}=0.}
$$
Along such a trajectory,
$$
\frac d{dt}\left(\dot qL_{\dot q}-L\right)=\dot q\left(\frac d{dt}L_{\dot q}-L_q\right)-L_t=-L_t.
$$
If $L$ has no explicit time dependence, the first <integral> is therefore
$$
\boxed{\dot q\frac{\partial L}{\partial\dot q}-L=\text{constant}.}
$$
Its negative is the equivalent Beltrami form. For the usual time-independent mechanical <Lagrangian> $L=T-V$ with quadratic velocity dependence, this conserved quantity is the total <mechanical energy> $T+V$.
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