Integration Problem: Can't 'See' How It's Done

[Skuzzy]

Homework Statement

I was given that \int \frac{m}{mgsin\alpha-kv} dv = -\frac{m}{k}ln(mgsin\alpha-kv) + C

..but I can't 'see' how this was done.

Homework Equations

The Attempt at a Solution

My first thought is that this is somehow related to the fact that:

\int \frac{g'(x)}{g(x)}=ln \left|(g(x)) \right|

but I am still missing something to make this work.

All help appreciated.

 

[Donaldos]

\left\{\begin{array}{rcl}u&amp;=&amp;mg\sin \alpha -kv\\<br /> <br /> {\rm d}v&amp;=&amp;-\frac{{\rm d}u}{k}\end{array}\right.

\Rightarrow \int \frac{m}{mg\sin \alpha -kv}{\rm d}v=\int -\frac{m}{k}\frac{{\rm d}u}{u}

 

[Mark44]

\left\{\begin{array}{rcl}u&amp;=&amp;mg\sin \alpha -kv\\<br /> <br /> {\rm d}v&amp;=&amp;-\frac{{\rm d}u}{k}\end{array}\right.

\Rightarrow \int \frac{m}{mg\sin \alpha -kv}{\rm d}v=\int -\frac{m}{k}\frac{{\rm d}u}{u}

Make that u&amp;=&amp;mg\sin \alpha -kv
and
}du&amp;=&amp;-\frac{{\rm d}u}{k}

You had dv.

 

[Donaldos]

Make that u&amp;=&amp;mg\sin \alpha -kv
and
}du&amp;=&amp;-\frac{{\rm d}u}{k}

You had dv.

I'm sorry, what?

 

[Skuzzy]

Make that u&amp;=&amp;mg\sin \alpha -kv
and
}du&amp;=&amp;-\frac{{\rm d}u}{k}

You had dv.

Now I'm more confused... How can }du&amp;=&amp;-\frac{{\rm d}u}{k} ?

 

[xepma]

Ignore Mark's statement. Donaldos' approach is correct.

 

[Mark44]

I'm sorry, what?

Apparently you had two errors and I caught only one.
If
u&amp;=&amp;mg\sin \alpha -kv
then
du = -kdv

 

[xepma]

Apparently you had two errors and I caught only one.
If
u&amp;=&amp;mg\sin \alpha -kv
then
du = -kdv

Look again at his post. His answer is the same as what you have ;)

 

[Mark44]

OK, I got it now. With variables that look so much alike, it would have been nice to have the intervening step.