Integration problem

  • Thread starter Skuzzy
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  • #1
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Homework Statement



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

..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:

[tex]\int \frac{g'(x)}{g(x)}=ln \left|(g(x)) \right|[/tex]

but I am still missing something to make this work.

All help appreciated.
 

Answers and Replies

  • #2
68
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[tex]\left\{\begin{array}{rcl}u&=&mg\sin \alpha -kv\\

{\rm d}v&=&-\frac{{\rm d}u}{k}\end{array}\right.[/tex]

[tex]\Rightarrow \int \frac{m}{mg\sin \alpha -kv}{\rm d}v=\int -\frac{m}{k}\frac{{\rm d}u}{u}[/tex]
 
  • #3
35,125
6,872
[tex]\left\{\begin{array}{rcl}u&=&mg\sin \alpha -kv\\

{\rm d}v&=&-\frac{{\rm d}u}{k}\end{array}\right.[/tex]

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

You had dv.
 
  • #4
68
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Make that [tex]u&=&mg\sin \alpha -kv[/tex]
and
[tex]}du&=&-\frac{{\rm d}u}{k}[/tex]

You had dv.

I'm sorry, what?
 
  • #5
11
0
Make that [tex]u&=&mg\sin \alpha -kv[/tex]
and
[tex]}du&=&-\frac{{\rm d}u}{k}[/tex]

You had dv.


Now I'm more confused... How can [tex]}du&=&-\frac{{\rm d}u}{k}[/tex] ?
 
  • #6
525
7
Ignore Mark's statement. Donaldos' approach is correct.
 
  • #7
35,125
6,872
I'm sorry, what?
Apparently you had two errors and I caught only one.
If
[tex]u&=&mg\sin \alpha -kv[/tex]
then
du = -kdv
 
  • #8
525
7
Apparently you had two errors and I caught only one.
If
[tex]u&=&mg\sin \alpha -kv[/tex]
then
du = -kdv

Look again at his post. His answer is the same as what you have ;)
 
  • #9
35,125
6,872
OK, I got it now. With variables that look so much alike, it would have been nice to have the intervening step.
 

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