Derivation of Method of Substitution for Integration Let $f(x)$ be a continuous function on $[a, b]$ and $g(x)$ be a strictly monotonic, continuous, and differentiable function on $[a, b]$. Now, consider the integral $$\int^{b}_{a}f(g(x))dx$$ which can be written as the limit of the following sum (we know the limit of this sum must exist as $f(g(x))$ is a continuous function, and all continuous functions can be integrated): $$\sum^{n}_{v = 1}f(g(\xi_v))\Delta x_v$$ where $\Delta x_v$ refers to the v-th sub-interval of [a, b] and $\xi_v$ refers to an arbitrary point in the v-th sub-interval.
Since $g(x)$ is a strictly monotonic function there is a 1-to-1 mapping between the interval [a, b] on the x-axis, and an interval $\alpha \leq u \leq \beta$ of the values of $u = g(x)$, where $\alpha = g(a)$, $\beta = g(b)$. Thus, let $u_v$ be the corresponding point to $\xi_v$ i.e. $u_v = g(\xi_v)$ and $\Delta u_v$ be the corresponding sub-interval to $\Delta x_v$. Now, we can rewrite the sum as: $$\sum^{n}_{v=1}f(u_v)\frac{\Delta x_v}{\Delta u_v}\Delta u_v$$
Now, by the mean value theorem, we can write $\frac{\Delta x_v}{\Delta u_v}=\phi^{'}(\eta_v)$, where $\eta_v$ is a point in the interior of $\Delta u_v$ and $x = \phi(u)$ denotes the inverse function of $g(x)$. If we now select the value of $\xi_v$ in such way that $\xi_v$ and $\eta_v$ coincide, i.e. $\xi_v = \phi(\eta_v)$, $\eta_v = g(\xi_v)$, then we can re-write the sum as: $$\sum^{n}_{v=1}f(\eta_v)\phi^{'}(\eta_v)\Delta u_v$$
If we take the limit of this sum as $n \to \infty$, we obtain the expression: $$\int^{\beta}_{\alpha}f(u)\frac{dx}{du}du,$$ which is the formula for method of substitution.
This is my interpretation of Courant's proof for deriving the method of substitution, but I just wanted to make sure my interpretation is correct.
 A: Your interpretation is basically correct, but let me point out some fine points that you may have missed.  Courant assumes, not only that $g$ is differentiable, but also that its derivative is continuous and never 0.  Why?  To ensure that its inverse $\phi$ is continuously differentiable.  Simply knowing that $g$ is continuously differentiable is not enough to ensure that $\phi$ is differentiable.  For example, if $g(x) = x^3$, then $g$ is monotone and continuously differentiable, but the inverse is $\phi(u) = \sqrt[3]{u}$, which is not differentiable at 0.
To be sure that the limit of $\sum_{v=1}^n f(g(\xi_v))\,\Delta x_v$ is $\int_a^b f(g(x))\,dx$, we need to know that the maximum of $\Delta x_v$ (the mesh of the partition) approaches 0, and similarly for the Riemann sum used to compute $\int_\alpha^\beta f(u) \frac{dx}{du}\,du$.  Courant has a footnote in which he explains how to ensure this.  He chooses the partitions of the $u$-interval $[\alpha, \beta]$ first so that their meshes approach 0, and then he applies the uniform continuity of $\phi$ to conclude that the meshes of the corresponding partitions of the $x$-interval $[a, b]$ also approach 0.
I find Courant's use of the notation $\xi_v$ a bit confusing.  It sounds at first like he is fixing a choice of $\xi_v$ at the beginning of the proof, but then in the middle of the proof he says "If we now select the values $\xi_v$ in such a way that ..."  So the $\xi_v$ are not chosen until halfway through the proof, and the uses of the notation $\xi_v$ earlier in the proof have to be understood as referring to numbers that will be chosen, but have not yet been chosen at that point in the proof.
Finally, let me point out that, although Courant assumes that $g$ is continuously differentiable and its derivative is never 0, all he really needs is that $\phi$ is continuously differentiable.  Occasionally this makes a difference.  For example, consider this integral:
$$
\int_a^b \frac{dx}{(\sqrt[3]{x})^2+1}
$$
To solve it, we might let $u = \sqrt[3]{x}$, so $x = u^3$ and $dx = 3u^2\,du$.  The substitution gives us
$$
\int_a^b \frac{dx}{(\sqrt[3]{x})^2+1} = \int_{\sqrt[3]{a}}^\sqrt[3]{b} \frac{3u^2\,du}{u^2+1} = \left[3u - 3\tan^{-1}u\right]_{\sqrt[3]{a}}^{\sqrt[3]{b}}.
$$
This is correct, but it doesn't quite fit Courant's theorem.  In this case we have $g(x) = \sqrt[3]{x}$, which is not differentiable at 0.  But $\phi(u) = u^3$, which is continuously differentiable everywhere, and that's all that is required in Courant's proof.
For a different explanation of this kind of substitution, see my book Calculus: A Rigorous First Course, Section 8.4.
