I don't have a recommended textbook, but here's my potted version of Calculus of Variations:
Let's suppose that you have a function that is written as the integral:
[math]S = \int_a^b L\left(y(x), y'(x)\right)\,dx[/math]
This reads as 'the integral (over [math]x[/math]) of some function, [math]L[/math], that is itself a function of some other function [math]y(x)[/math] and its first derivative, [math]y'(x)[/math].
In many cases, you don't know what the function [math]x(t)[/math] is - ally know is the form of the function [math]L\left(y(x), y'(x)\right)[/math]. However, you want to know what function [math]x(t)[/math] minimises - or at least finds an extremum of - the functional [math]S[/math]. This might seem an esoteric mathematical question to ask, but in practice may real-world problems can be cast in this form and, indeed, the bulk of modern physics is an exercise in solving problems of this kind.
Anyway, the Calculus of Variations aims to answer that question and, textbook derivations aside, the answer to the question is give by the Euler-Lagrange equations:
[math]\frac{d}{dx}\,\frac{\partial L}{\partial y'(x)}=\frac{\partial L}{\partial y(x)}[/math]
Any function [math]x(t)[/math] (subject to meting certain boundary condition requirements) that solves the above differential equation also minimises (or at least is an extremum of the functional [math]S[/math]). So, if you know the function [math]L[/math] in terms of [math]y(x)[/math] and its first derivative [math]y'(x)[/math], the procedure is to solve the above Euler Lagrange differential equation for [math]y(x)[/math] - and the result gives you the function that minimises the functional [math]S[/math].
An example - shortest path between two points
A standard example of how to apply this is: find the shortest distance between two points on an x-y graph. Now, we know that the answer is a straight line, of course, but we can use the Calculus of Variations (i.e., the Euler Lagrange equation) to prove its.
Let's suppose that the two points are [math](0,\,0)[/math] and [math](1,\,1)[/math]. And let's take an arbitrary squiggly path [math]y(x)[/math] connecting these two points. Let's take any very small line element on this curve that measures [math]dx[/math] horizontally; and [math]dy[/math] vertically. The length of this line element is [math]\sqrt{dx^2 + dy^2}[/math]. And so the total length of the squiggly line (which we'll call [math]S[/math]!) is given by the expression:
[math]S = \int\sqrt{dx^2 + dy^2}[/math]
And, after a little re-arranging, we get:
[math]S = \int_0^1\sqrt{1 + \frac{dy}{dx}^2}\,dx[/math]
which can also be re-written as:
[math]S = \int_0^1 L\left(y(x), y'(x)\right)\,dx[/math]
where [math]L\left(y(x), y'(x)\right) = \sqrt{1+y'(x)^2} [/math]
So, now, we've cast the problem of finding the shortest path into a standard calculus of variations problem. So, we can now use Calculus of Variations (i.e., the Euler-Lagrange equation) to solve for the squiggly path, [math]y(x)[/math], that gives us the shortest path between the two points. So, the procedure is to write down the E-L equation.
First, we differentiate the function, [math]L[/math] with respect to [math]y'(x)[/math]. (In carrying out this differentiation, we ignore the fact that it is a function and just treat the whole thing [math]y'(x)[/math] as a variable). We get:
[math]\frac{\partial L}{\partial y'(x)} = \frac{y'(x)}{\sqrt{1+y'(x)^2}}[/math]
Next, we differentiate the expression on the right with respect to the variable [math]x[/math]:
[math]\frac{d}{dx}\,\frac{\partial L}{\partial y'(x)}=\frac{d}{dx}\,\frac{y'(x)}{\sqrt{1+y'(x)^2}} = \frac{y''(x)}{(1+y'(x)^2)^{3/2}}[/math]
Finally, we differentiate the function, [math]L[/math] with respect to [math]y(x)[/math]. This one's pretty easy:
[math]\frac{\partial L}{\partial y(x)} = 0[/math]
So, putting it all together, the E-L equation is:
[math] \frac{y''(x)}{(1+y'(x)^2)^{3/2}} = 0[/math]
Or, more succinctly:
[math] y''(x) = 0[/math]
So, any function that satisfies this differential equation (and meets the boundary conditions that it passes through the initial and final points of the curve) also minimises (extremises!) the length of the curve - i.e., is the shortest path between the two points.
The general form of the solution to [math] y''(x) = 0[/math] is:
[math]y = a\,x +b[/math]
And since we require that this curve pass through the initial and final points [math](0,\,0)[/math] and [math](1,\,1)[/math], we can solve for the coefficients [math]a[/math] and [math]b[/math] to find that [math]a=1[/math] and [math]b=0[/math].
In other words, using Calculus of Variations, the shortest path between the points [math](0,\,0)[/math] and [math](1,\,1)[/math] is:
[math]y(x) = x[/math]
Yay!
Application to physics
All of this seems fairly academic were it not for the fact that dynamical systems in classical mechanics are usually variational problems of this kind. The modern formulation is to construct an 'action', [math]S[/math], from a Lagrangian [math]L(x(t), x'(t))[/math], where the Lagrangian is expressed in terms of some trajectory [math]x(t)[/math] and its first derivative [math]x'(t)[/math]. The equations of motion of this physical system are then just the E-L equations (which, of course, arise from the Calculus of Variations). In other words, in classical mechanics, the permitted equations of motion are precisely those that minimise (or at least extremes) the action, [math]S[/math].
(Courtesy of Richard Feynman, there is an extension of this idea to quantum mechanics and quantum field theories, but we won't go there!)
Anyway, for many physical systems, we have rules for constructing the Lagrangian. So, for example, in Newtonian mechanics, the Lagrangian can be written as:
[math]L = T - V[/math]
where [math]T[/math] is the kinetic energy term; and [math]V[/math] is the potential energy term.
So, let's take the one-dimensional standard Kepler gravitational problem of a (small) satellite moving in the gravitational field of some large gravitational body (e.g., the Earth). Let's suppose that the trajectory is [math]x(t)[/math]. Then, the kinetic energy term is:
[math]T = \frac{1}{2}\,m\,x'(t)^2[/math]
where x'(t) is the velocity of the satellite; and [math]m[/math] is the mass of the satellite.
The potential energy term is:
[math]V = -\frac{G\,m\,M}{x(t)}[/math]
where [math]G[/math] is the universal gravitational constant; and [math]M[/math] is the mass of the gravitating body.
The Lagrangian, [math]L[/math] of the system is just:
[math]L = \frac{1}{2}\,m\,x'(t)^2 + \frac{G\,m\,M}{x(t)}[/math]
To find the equations of motion of the small satellite, we then write down and solve the E-L equations. This gives:
[math]m\,x''(t) = -\frac{G\,m\,M}{x(t)^2}[/math]
and cancelling the [math]m[/math] on both sides gives:
[math]x''(t) = -\frac{G\,M}{x(t)^2} = -\frac{\mu}{x(t)^2} [/math]
which is, of course, just the standard form of the equation of motion for a body falling in an inverse square gravitational field.
One can just as easily extend this to a two-dimensional Kepler problem. Here, we can immediately write down the kinetic and potential energy terms to get:
[math]L = \frac{1}{2}\,m\,\left(x'(t)^2 + y'(t)^2\right) + \frac{G\,m\,M}{\sqrt{x(t)^2 + y(t)^2}}[/math]
and from this, we can immediately find the equations of motion by solving the E-L equations.
Some final thought
In physics, the real power of the Calculations of Variations approach is that we can develop some simple rules for constructing the system Lagrangian. Once we have this function, we then have sure-fire approach (solve the E-L equations!) to finding the equations of motion of that system - no matter how complex it is.
