I was told a trick by my professors for certain integrals.
Given that the path of a particle in 1D is $x(t)$, then the following is true
$$ \ddot{x}=\frac{d\dot{x}}{dt}=\frac{d\dot{x}}{dx}\frac{dx}{dt}=\dot{x}\frac{d\dot{x}}{dx} $$
My impression is that this is true for any path, so long as $x$ only depends on $t$.
The reason I ask is for a specific problem in Thornton and Marion. I think the answers I am finding online might be wrong.
Specifically this is 2-1; suppose a force is acting on a single particle. If the force has the form $F=F(x_i,t)=f(x_i)g(t)$, determine if the equation of motion is integrable.
The standard answer I am finding copied and pasted is:
$$ m\ddot{x}_i=f(x_i)g(t)\\ \text{Since it is the case that } \ddot{x}_i = \frac{d\dot{x}}{dt} \text{, then we have}\\ m\frac{d\dot{x}}{f(x_i)}=g(t)dt $$
Which is then just declared not integrable without further clarification.
Suppose that the substitution for $\ddot{x}$ I mentioned is generally true for functions of one variable.Then we can rewrite the EOM as
$$ m\ddot{x}_i=f(x_i)g(t)\\ m\dot{x}\frac{d\dot{x}_i}{dx_i}=f(x_i)g(t)\\ \text{Which can be rewritten as } m\dot{x}d\dot{x}=f(x_i)g(t)dx_i $$
To which I would argue that as long as $g(t)$ is not dependent on $x_i$, then the above should be perfectly integrable. Specifically:
$$ \frac{m}{2}\dot{x}^2=g(t)\int f(x_i)dx_i\\ \text{which can be modified to give }\\ \sqrt{\frac{m}{2}}\dot{x}=\sqrt{g(t)}\sqrt{\int f(x_i)dx_i}\\ \text{which just simply gives } \frac{dx_i}{\sqrt{\int f(x_i)dx_i}}=\sqrt{\frac{2g(t)}{m}}dt $$
Which would appear to still be integrable.
So have people been copying the wrong solution for years, or am I wrong for some reason I do not see?
There are two ways to think about this trick. One takes the original autonomous equation $m\ddot x=f(x)$, multiplies it with $\dot x$ and integrates using the chain rule to get $\frac{m}2\dot x^2=F(x)$, where $F'=f$.
The other point-of-view is that the independent variable gets changed from $t$ to $x$. This demands that the solution under consideration is monotonous, at least in some segment around the initial point. To avoid symbolic confusion, declare a function with a new letter, $\dot x=u(x)$. Then $\ddot x=u'(x)\dot x=u'(x)u(x)$.
In your case, the substitution of the independent variable would also make $t$ a function of $x$, so that the equation becomes $$ mu'(x)u(x)=f(x)g(t(x)) $$ Due to the unknown function $t(x)$, this is not directly integrable or separable.