I A question about variables of integration

[Rick16]

TL;DR

an integral with a strange variable of integration

I have the equation $F^x=m\frac {d}{dt}(\gamma v^x)$, where $\gamma$ is the Lorentz factor, and $x$ is a superscript, not an exponent. In my textbook the solution is given as $\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}$.

What bothers me is, when I separate the variables I get $\frac {F^x}{m}dt=d(\gamma v^x)$. Can I simply consider $d(\gamma v^x)$ the variable of integration without any further considerations? Can I simply make the substitution $\gamma v^x = u$ and then write $\frac {F^x}{m}\int_0^t\,dt=\int_0^u\,du$? The solution would then be $\frac {F^x}{m}t=u=\gamma v^x$ as in the book. Is it really that simple? I feel uneasy about this because of the twofold dependence of $\gamma v^x$ on $v^x$.

 

[PeroK]

I don't think you have to make any substitutions. In general, if:
$$f(t) = \frac{d}{dt}g(t)$$Then, by the fundamental theorem of calculus:
$$\int f(t) dt = \int \frac{d}{dt}g(t) dt = g(t) + C$$Or, in terms of the definite integral:
$$\int_{t_0}^{t_1} f(t) dt = \int_{t_0}^{t_1}\frac{d}{dt}g(t) dt = g(t_1) - g(t_0)$$

 

[fresh_42]

This is the typical example of integration: It looks awkward, but it works. What we have is
\begin{align*}
\dfrac{F^x}{m}&=\dfrac{d}{dt}\left(\gamma v^x\right)\\
\int \dfrac{F^x}{m}\,dt&=\int \dfrac{d}{dt}\left(\gamma v^x\right)\,dt \\
\dfrac{F^x}{m}\cdot t&\stackrel{u=\gamma v^x}{=}\int u'\cdot \dfrac{du}{u'}=\int du\\
\dfrac{F^x}{m}\cdot t&=u=\gamma v^x
\end{align*}

 

[Rick16]

I made the substitution because without it my integral looked like this: $\int d(\gamma v^x)$, which I found was a very odd looking integral. If instead I write it as $\int \frac {d}{dt}(\gamma v^x)\,dt$, it does not look odd anymore and it all makes sense. Problem solved, thank you very much.