I try to integrate
$$\int \frac {dv}{\frac {-c}{m}v^2 - g \sin \theta}$$
I did substituted $u = \frac{c}{m}$ and $w = g \sin \theta$ to get
$$-\int \frac {dv}{uv^2 + w}$$
I'm wondering if I have to do a second substitution. To be honest, I don't know if I can do that or how to do that. Furthermore, maybe I have to rearrange to get $\frac1{1+x^2}$





You've already rewritten the integral to make it easier to work with. The next step, as you've considered, is to try to get it into the form $\int \frac{dx}{1+x^2}$ so you can get an antiderivative in terms of arctangent. To do that: \begin{align*} -\int \frac{dv}{uv^{2} + w} &= -\frac{1}{w}\int\frac{dv}{\frac{u}{w}v^2 + 1}\\ &=-\frac{1}{w}\int \frac{dv}{\left(\sqrt{\frac{u}{w}}v\right)^{2}+1}. \end{align*}
If we use the substitution $x = \sqrt{\frac{u}{w}}v$ and $dx = \sqrt{\frac{u}{w}}\,dv$ we get
\begin{align*} -\frac{1}{w}\int \frac{dv}{\left(\sqrt{\frac{u}{w}}v\right)^{2}+1} &= -\sqrt{\frac{w}{u}}\frac{1}{w}\int\frac{dx}{x^{2} + 1}\\ &=-\frac{1}{\sqrt{uw}}\arctan(x) + C\\ &= -\frac{1}{\sqrt{uw}}\arctan\left(\sqrt{\frac{u}{w}}v\right). \end{align*}
At this point you just need to undo your original substitutions.