$\displaystyle

\int

\frac{1}{10 - kv^2}

\, dv

$

- Aug 18th 2009, 05:21 AMBucephalus[SOLVED] Can someone give me a start integrating this please?
$\displaystyle

\int

\frac{1}{10 - kv^2}

\, dv

$

- Aug 18th 2009, 05:35 AMsongoku
Hi Bucephalus

If k is a constant, you can try :

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

or

2. Partial fraction by first factoring the denominator - Aug 18th 2009, 06:45 AMBucephalusMore understanding please
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"

$\displaystyle

\frac{1}{10} - \frac{1}{(kv)v}

$

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

bucephalus - Aug 18th 2009, 07:02 AMmr fantastic
- Aug 18th 2009, 09:20 AMsongoku
Hi Bucephalus

Quote:

I want to understand how you got that expression in your first helper

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

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

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

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

For the second one, follow Mr. F suggestion :

Quote:

Factorise $\displaystyle 10 - kv^2$

- Aug 19th 2009, 01:33 AMBucephalusFactorising
so factorise this.

$\displaystyle 10 - kv^2$

The only way I can think of to factorise this is:

$\displaystyle 10(1-\frac{k}{10}v^2)$

Is this what you mean? - Aug 19th 2009, 03:09 AMpickslides
$\displaystyle 10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2$

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

You get

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

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

Find A and B and the integral will be alot easier in this form. - Aug 19th 2009, 03:18 AMBucephalusThanks Pickslides
That's an awesome response.

Thanks for your help.