# Integration problem

## 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.

## 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.

$$\left\{\begin{array}{rcl}u&=&mg\sin \alpha -kv\\ {\rm d}v&=&-\frac{{\rm d}u}{k}\end{array}\right.$$

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

Mark44
Mentor
$$\left\{\begin{array}{rcl}u&=&mg\sin \alpha -kv\\ {\rm d}v&=&-\frac{{\rm d}u}{k}\end{array}\right.$$

$$\Rightarrow \int \frac{m}{mg\sin \alpha -kv}{\rm d}v=\int -\frac{m}{k}\frac{{\rm d}u}{u}$$
Make that $$u&=&mg\sin \alpha -kv$$
and
$$}du&=&-\frac{{\rm d}u}{k}$$

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

I'm sorry, what?

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

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

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

Mark44
Mentor
I'm sorry, what?
Apparently you had two errors and I caught only one.
If
$$u&=&mg\sin \alpha -kv$$
then
du = -kdv

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

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

Mark44
Mentor
OK, I got it now. With variables that look so much alike, it would have been nice to have the intervening step.