$\displaystyle
\int
\frac{1}{10 - kv^2}
\, dv
$
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
Hi Bucephalus
It's because I want to use trigonometry identity : $\displaystyle 1- \sin^2 \theta = \cos^2 \theta$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 :
Factorise $\displaystyle 10 - kv^2$
$\displaystyle 10-kv^2 = \sqrt{10}^2-\sqrt{k}^2v^2$
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.