classical mechanics - Question about Principle of least action - Landau

I have this conceptual question: In Landau's book of classical mechanics, about the principle of least action, it's written:

$$\left. \delta S =\frac{\partial L}{\partial v} \delta q \right\rvert_{t_1}^{t_2} + \ \int\limits_{t_1}^{t_2} \ dt \left(\frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}\right) \delta q \ =0 ,$$

where $q=q(t)$ is the position function, $v=v(t)$ velocity function, $S$ the action, and L the Lagrangian of the system.

There is the condition $\delta q(t_1)=\delta q(t_2)=0$. So the first term is zero, and then it says " there remains an integral which must vanish for all values of $\delta q$. This can be so only if the integrand is zero identically.

Well I can't understand why

$$\left(\frac{\partial L}{\partial q}\ - \frac d {dt}\frac {\partial L}{\partial v}\right) =0 \;.$$

Answer

Consider the following integral $$\int_{t_1}^{t_2}F(t)\delta q(t) dt=0,$$ for every possible $\delta q$. Then choose a function $\delta q$ which has a large value but is different from zero only in an infinitesimal neighborhood of a point $t_0\in[t_1,t_2]$. Then, $$0=\int_{t_1}^{t_2}F(t)\delta q(t) dt=\int_{t_0-\epsilon}^{t_0+\epsilon}F(t)\delta q(t) dt=F(t_0)\int_{t_0-\epsilon}^{t_0+\epsilon}\delta q(t) dt.$$ In the last equal sign we took $F$ out of the integral because the function is approximately constant in the infinitesimal interval $2\epsilon$. Now, the last integral above is different from zero, so $F(t_0)$ has to vanish. Repeat this argument for all $t\in[t_1,t_2]$ and you obtain that $F(t)$ is identically zero. This is basically the fundamental lemma of variational calculus.