Integrate $\int \frac {dv}{\frac {-c}{m}v^2 - g \sin \theta}$ 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 $\frac1{1+x^2}$
 A: You have $$-\int \frac{dv}{w+uv^2}=-\frac{1}{w}\int\frac{dv}{1+(v\sqrt{\frac{u}{w}})^2}$$
Now consider the substitution $v\sqrt{\frac{u}{w}}=\tan(x)$.
A: 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.
A: $$I= \int\dfrac{dv}{-\frac{c}{m}v^2-g\sin\theta} = -\int\dfrac{dv}{\frac{c}{m}v^2+g\sin\theta} = -\int\dfrac{dv}{\left(\sqrt\frac{c}{m}\ v\right)^2+(\sqrt{g\sin\theta})^2} $$
(Assuming $0<\sin\theta<1$)
$$I = -\sqrt{\dfrac{m}{c}}\int\dfrac{\sqrt{\dfrac{c}{m}}dv}{\left(\sqrt\frac{c}{m}\ v\right)^2+(\sqrt{g\sin\theta})^2} = -\sqrt{\dfrac{m}{cg\sin\theta}}\arctan\left(v\sqrt{\dfrac{c}{mg\sin\theta}}\right)+k$$
A: Divide the numerator by $u$ to get $$-\frac{1}{u}\int_{ }^{ }\frac{dv}{v^{2}+\left(\frac{w}{u}\right)}$$
Now you may apply the standard integral, a list of which is given here and may provide some additional help.
Also, I have posted some pictures below of the standard integrals and the common integration techniques. Might help someone. (I know that the question didn't demand this and I will remove this part if this attracts too many downvotes or opposing comments)

 





