# Integration Help

• May 27th 2006, 05:44 PM
eulinator
Integration Help
Please for some help with this integration problem.
$
\int \frac{dx}{\sqrt{(a x^4 + b x^2 + c)}}
$

i.e.
integral ((a x^4 + b x^2 + c)^–½) dx
• May 27th 2006, 06:24 PM
ThePerfectHacker
Are you certain it is not,
$
\int \frac{xdx}{\sqrt{(a x^4 + b x^2 + c)}}
$
??
• May 27th 2006, 06:31 PM
chancey
I havn't done any intergration yet, but what I remember from calculus is to rationalise the denominator

$\int dx \frac{\sqrt {ax^4 + bx^2 + c}}{ax^4 + bx^2 + c}$

Don't know if that helps...
• May 27th 2006, 07:28 PM
eulinator
Thanks for the rationalize approach. I will look into that.

The problem is based on predicting the rotation of a solid body after any given time. As far as I know there is no yet found solution. The integration is done numerically on a computer. I am desparately trying to get a theoretical formula. Alas, I am not that great with integration; I am at GCE A-level.

The equation is actually correctly stated.
• May 27th 2006, 07:32 PM
ThePerfectHacker
Maybe there reason why it is done numerically is because it cannot be precisely computed!

There are cases like that so do not be supprised.
• May 27th 2006, 07:48 PM
eulinator
Quote:

Originally Posted by ThePerfectHacker
Maybe there reason why it is done numerically is because it cannot be precisely computed!

There are cases like that so do not be supprised.

Ok, I will continue with the computer method.

I actually broke down the numerical method into this itegration and a matrix inverse. I expect that this approach have been attempted before, and hoped against.

• May 27th 2006, 10:22 PM
eulinator
new equation form
I have a broken down the original equation using partial fractions

$
\frac{1}{\sqrt{(a x^4 + b x^2 + c)}}
$

Using the roots of a quadratic equation, let
$m = \frac{-b+\sqrt{b^2-4ac}}{2a}$ and $n = \frac{-b-\sqrt{b^2-4ac}}{2a}$

Therefore becomes
$
\frac{1}{\sqrt{(x^2 + m)(x^2 + n)}}
$

Trying pratial fractions,
$
\frac{1}{\sqrt{(x^2 + m)(x^2 + n)}} = \frac{A}{\sqrt{(x^2 + m)}} + \frac{B}{\sqrt{(x^2 + n)}}
$

Therefore
$
A\sqrt{(x^2 + n)} + B\sqrt{(x^2 + m)} = 1
$

When $x = i \sqrt{m}$then $1 = A \sqrt{((i \sqrt{m})^2+n)}$and $A=\frac{1}{\sqrt{-m+n}}=\frac{1}{\sqrt{n-m}}$
Likewise
$B=\frac{1}{\sqrt{m-n}}$

So the original becomes
$
\frac
{1}
{
(\sqrt{n-m})(\sqrt{x^2+m})
} +
\frac
{1}
{
(\sqrt{m-n})(\sqrt{x^2+n})
}
$

Expanding to avoid imaginary values
$
\frac
{1}
{\sqrt{nx^2+nm-mx^2-m^2} } +
\frac
{1}
{\sqrt{mx^2+mn-nx^2-n^2} }
$

The new form is
$
\frac{1}
{\sqrt{(n-m)x^2+(nm-m^2)} } +
\frac{1}
{\sqrt{(m-n)x^2+(mn-n^2)} }
$

How can this proceed further? I noticed that it does not fall exactly in the arcsin value for the integral
• May 28th 2006, 03:11 AM
topsquark
Don't quote me on this.

I don't have my book with me at the moment, but the integral in question (I think!) appears in my Mechanics book. If I remember correctly, the integral can be reduced to an elliptic integral. They have no closed form.

-Dan
• May 28th 2006, 10:07 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
...elliptic integral...

I had a good feeling somebody was going to mention that!
But as far as my post #5 is correct. That is why they used numerical methods.
• May 28th 2006, 04:19 PM
eulinator
The equation was not correct :( due to the ± issue of the square root. I will try to re-formulate the equation keeping the sign in mind. Consequently, the result I gave in post one would be rather unsolvable.

I also looked at Elliptic Integral and Jacobian Elliptic Functions. I could not follow Closed Form.

You are right: This integral form is done in numerical methods.
• Jun 3rd 2006, 12:09 AM
chancey
I don't know if your still working on it, but maybe this may help The Integral Calculator
• Jun 6th 2006, 11:57 PM
eulinator
Quote:

Originally Posted by chancey
I don't know if your still working on it, but maybe this may help The Integral Calculator

Wow, this is great.

Thanks.
• Jun 15th 2006, 12:39 PM
eulinator
I think that I finally understand the closed form and the ellipsoid concept

Firstly, the first equation is not of a 'function' therefore the area under the graph would not be closed, and to explain it more vividly: the eqution is that of an ellipse!

EDIT: I think that the arctan integral is only valid for a range of x<1

Well, I did the equation for this over, if any is interested in helping. I am putting it in the Advance Topics section.