# [SOLVED] Can someone give me a start integrating this please?

• August 18th 2009, 05:21 AM
Bucephalus
[SOLVED] Can someone give me a start integrating this please?
$
\int
\frac{1}{10 - kv^2}
\, dv
$

• August 18th 2009, 05:35 AM
songoku
Hi Bucephalus

If k is a constant, you can try :

1. Let : $v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta$

or

2. Partial fraction by first factoring the denominator
• August 18th 2009, 06:45 AM
Bucephalus
Quote:

Originally Posted by songoku
Hi Bucephalus

If k is a constant, you can try :

1. Let : $v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta$

or

2. Partial fraction by first factoring the denominator

Hi Songoku

Two things:
1) I want to understand how you got that expression in your first helper.
2) I have done partial fraction decomposition before, but I don't understand what you're saying "factoring the denominator"
$
\frac{1}{10} - \frac{1}{(kv)v}
$

I'm sorry, I'm really having trouble understanding.
bucephalus
• August 18th 2009, 07:02 AM
mr fantastic
Quote:

Originally Posted by Bucephalus
[snip]
2) I have done partial fraction decomposition before, but I don't understand what you're saying "factoring the denominator"
$
\frac{1}{10} - \frac{1}{(kv)v}
$

I'm sorry, I'm really having trouble understanding.
bucephalus

Factorise $10 - kv^2$.

And if you're studying at the level of integral calculus I would sincerely hope that you know that $\frac{1}{10 - kv^2}$ is certainly NOT equal to $\frac{1}{10} - \frac{1}{(kv)v}$ !!!
• August 18th 2009, 09:20 AM
songoku
Hi Bucephalus
Quote:

I want to understand how you got that expression in your first helper
It's because I want to use trigonometry identity : $1- \sin^2 \theta = \cos^2 \theta$

Using $v=\frac{\sqrt{10}}{\sqrt{k}} \sin \theta$ will change the denominator to :

$10-kv^2=10-10 \sin^2 \theta=10(1-\sin^2 \theta)=10 \cos^2 \theta$

In fact, you also can use : $\frac{\sqrt{10}}{\sqrt{k}} \cos \theta$ because it will be relevant to identity : $1- \cos^2 \theta = \sin^2 \theta$

If the question turns out to be $10+kv^2$ , you can use substitution : $v=\frac{\sqrt{10}}{\sqrt{k}} \tan \theta$ because it is relevant to identity : $1+\; tan^2 \theta = \sec^2 \theta$

For the second one, follow Mr. F suggestion :
Quote:

Factorise $10 - kv^2$
• August 19th 2009, 01:33 AM
Bucephalus
Factorising
so factorise this.
$10 - kv^2$

The only way I can think of to factorise this is:
$10(1-\frac{k}{10}v^2)$

Is this what you mean?
• August 19th 2009, 03:09 AM
pickslides
$10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2$

now using the difference of two squares $a^2-b^2 = (a-b)(a+b)$

You get

$10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2 = \sqrt{10}^2-(\sqrt{k}v)^2 = (\sqrt{10}-\sqrt{k}v)(\sqrt{10}+\sqrt{k}v)$

So now $\frac{1}{10-kv^2} = \frac{A}{\sqrt{10}-\sqrt{k}v}+\frac{B}{\sqrt{10}+\sqrt{k}v}$

Find A and B and the integral will be alot easier in this form.
• August 19th 2009, 03:18 AM
Bucephalus
Thanks Pickslides
That's an awesome response.