Explanation of trick used to prove the work energy theorem in 1-D The proof of the Work energy theorem for 1-D starts with $$F(x)=\frac{dv(t)}{dt}$$ Then, they integrate with respect to the position x in both sides of the equation(even though v doesn't depend on x!):
$$\int_{x_a}^{x_b} F(x) dx = \int_{x_a}^{x_b} \frac{dv(t)}{dt} dx $$
And use differentials to substitute dx by $$dx=\dfrac{dx}{dt} dt= v \ dt$$
and the integral on the right becomes $$\int_{t_a}^{t_b} \frac{dv(t)}{dt} v \ dt$$
Of course, now it's possible to integrate that and we get the theorem.
Mathematically I didn't understand the substitution(I never got the whole differentials thing that physicists do)
So I was reflecting on this, and I think they assumed that $v(t)$ could somehow be written as $v(x)$.For example, if $x(t)=t^3$ and $v(t)=3t^2$ , then you could write $t$ as a function of $x$, $t=\sqrt[3]{x}$, and then replace that value on $v(t)$, getting $v(x)=3x^{\frac{2}{3}} $
After that we realize that by the chain rule we have
$$\frac{dv(x(t))}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx} \cdot v(x(t))=v'v  $$ where the ' is the derivative with respect to x.
Then $$ \int_{x_a}^{x_b} \frac{dv(t)}{dt} dx=\int_{x_a}^{x_b} v'v=\frac{v^2(x_b)}{2}-\frac{v^2(x_a)}{2}$$
I need to know whether my guess of what they did(i.e, write $v$ as a function of $x$) is correct or not.
In case I'm in the right, how can they assume that $v$ can be writen as a function of $x$? As far as I know, physicists work most of the time with infinite differentiable functions, but a lot of them can be not injective(like sine and cosine), and so it's impossible to write $t$ as a function of $x$.
 A: As you noted,
\begin{align}
F(x)&=\frac{dv(t)}{dt}
\end{align}
is already very sloppy, so anything which follows should of course be taken with skepticism. Note by the way that you forgot the mass $m$ on the right. Newton's second law $F=ma$ stated more precisely is the following:

Consider a particle of mass $m$ moving in $\Bbb{R}^n$, and a vector field $F:\Bbb{R}^n\to\Bbb{R}^n$. Let $\gamma:I\to\Bbb{R}^n$  (where $I\subset\Bbb{R}$ is an open interval) denote the trajectory taken by the path under the influence of the net force field $F$, parametrized by time. Then, $\gamma$ satisfies the following differential equation: for each $t\in I$
\begin{align}
F(\gamma(t))&=m\ddot{\gamma}(t).
\end{align}
Or as equality of functions, $F\circ \gamma=m\ddot{\gamma}$.

Next, coming to the work done, note that strictly speaking the work done by a force is defined along a curve, and is defined as follows: for any vector field $G:\Bbb{R}^n\to\Bbb{R}^n$ and any curve $\sigma:[a,b]\to\Bbb{R}^n$,
\begin{align}
\text{Work done by $G$ along $\sigma$}&:=\int_{\sigma}\sum_{i=1}^nG_i\,dx^i\tag{$*$}
\end{align}
As of now, the symbol on the right has yet no meaning. We have to give it a definition! And its definition is
\begin{align}
\int_{\sigma}\sum_{i=1}^nG_i\,dx^i&:=\int_{a}^b\sum_{i=1}^nG_i(\sigma(t))\,\dot{\sigma}^i(t)\,dt\equiv\int_a^b\sum_{i=1}^n(G_i\circ \sigma)\,\dot{\sigma}^i,
\end{align}
where on the right, I have used the standard notation for a Riemann integral of a function $[a,b]\to\Bbb{R}$. If I really want to I can write that last expression as
$\int_a^b\sum_{i=1}^nG_i(\sigma(\zeta))\,\dot{\sigma}^i(\zeta)\,d\zeta$, it still means the same thing. Now, it is traditional notation to write the RHS of $(*)$ as $\int_{\sigma}\mathbf{G}\cdot d\mathbf{x}$, and the definition can be written as $\int_a^b\langle G(\sigma(t)),\dot{\sigma}(t)\rangle\,dt$, where $\langle\cdot,\cdot\rangle$ is the standard inner product on $\Bbb{R}^n$. Physicists typically explain the definition by saying "just divide and multiply by $dt$ to get $\int \mathbf{G}\cdot d\mathbf{x}=\int_a^b\mathbf{G}\cdot \frac{d\mathbf{x}}{dt}\,dt=\int_a^b\mathbf{G}\cdot v\,dt$", or sometimes they may just say "substitute $d\mathbf{x}=\mathbf{v}\,dt$", but really this equality is by definition of a line integral. Nothing more. The multiplication and division by $dt$ is definitely a good mnemonic for remembering the definition though.
In view of this, if we apply the definitions along with Newton's second law, we get that the net work done by $\mathbf{F}$ along the particle trajectory $\gamma:[a,b]\to\Bbb{R}^n$ is:
\begin{align}
\text{Work done by $\mathbf{F}$ along $\gamma$}&:=\int_{\gamma}\mathbf{F}\cdot d\mathbf{x}\\
&:=\int_a^b\langle(\mathbf{F}\circ \gamma)(t),\dot{\gamma}(t)\rangle\,dt\\
&=\int_a^b\langle m\ddot{\gamma}(t),\dot{\gamma}(t)\rangle\,dt\\
&=\int_a^b\frac{d}{dt}\left[\frac{m}{2}\|\dot{\gamma}(t)\|^2\right]\,dt\\
&=\frac{m\|\dot{\gamma}(b)\|^2}{2}-
\frac{m\|\dot{\gamma}(a)\|^2}{2},
\end{align}
i.e the net work done is the change in kinetic energy. The second equal sign (which seems to be your concern, is actually just a definition).

In this particular case, it's not like we're trying to invert the functions to rewrite velocity as a function of position, because the position need not be an injective function (eg uniform circular motion in the plane, or an ideal simple harmonic motion on the line).
But your guess is a good one; sometimes that is indeed what they mean. In those cases, they're assuming that the motion of the particle is such that it can be inverted (at least locally, by the inverse function theorem in 1D, this is usually always possible by assuming the velocity is non-zero). For another variant of this sort, take a look at Formalizing a result from vector algebra of motion of curve in a plane. But again, this particular example of work done is simply by the definition of the line integral.
