Mathmaticians, REALLY hard integration !!

[alphaxyz]

how do u integrate { (x^n) / [ (a^2) - (x^2) ] ^ (1/2) } dx ?
this is to find a formula

 

[alphaxyz]

i've tried x = asin(u)

but only got half far

 

[alphaxyz]

bump

 

[warlord]

x^m/((x^2-a^2)^(1/2))=-(x^(m-1)*(a^2-x^2)^(1/2))/m+(m-1)*a^2/m*(integral of)x^(m-2)/(a^2-x^2)^(1/2)dx

thats from the table of integrals in the Handbook of Chem and Phys, its not perfect, but it works.

good luck

 

[alphaxyz]

yup yup, thx.

but i need tp PROOF it

 

[warlord]

use the divided difference therom, that should get you quite aways.

expecially now that you know the answer

 

[alphaxyz]

divided difference therom??

what's that?

 

[amnesiac]

I used to know this. But after I managed to eke through Calc 3 with a C minus (got a 14% average but had a REALLY kind-hearted prof...and 3 students in the class) I just let all the info kind of reverse-osmose out of my head into the atmosphere. That's what calculators are for.

 

[FrontlineWarrior]

math is suck suck

 

[somethingwitty]

i dont know if anyone here has or has heard of maple, but here's what I got:

>int ((x^n)/(sqrt(a^2-x^2)), x);

I got:

2
(1 + n) x
x hypergeom([1/2, 1/2 + 1/2 n], [3/2 + 1/2 n], ----)
2
a
1/2 -----------------------------------------------------------
2
sqrt(a ) (1/2 + 1/2 n)


good luck figuring that one out! I dont even know what a "hypergeom" is...

<edit> it's even tougher to figure out with formatting errors, one sec:
R1 := csgn(a)*x^(1+n)*hypergeom([1/2, 1/2+1/2*n],[3/2+1/2*n],x^2/(a^2))/(a*(1+n))

ok, that's maple trying to simplify...as i said, good luck. </edit>

 

[MereMortal]

You can try trig substitutions, but you must still utilize integration by parts, so you might as well start there.

So, integration by parts gives integral{u*dv}=u*v - integral{v*du}

Let u=x^(n-1), dv=x*dx/[a^2-x^2]^1/2
==>du=(n-1)*x^(n-2)*dx, v=-[a^2-x^2]^1/2 (straightforward integral)

Then integral(x^n*dx/[a^2-x^2]^1/2}=-x^(n-1)*[a^2-x^2]^1/2 + (n-1)* integral{x^(n-2)*[a^2-x^2]^1/2*dx}

If you need to solve this for high n, you just keep using this method until you reduce the power of x to something manageable.

Edit: somethingwitty, hypergeom is the hypergeometric function. This is useful as it is the solution to certain differential equations.

 

[iamwiz82]

i hate maple, and all it stands for, because it epitomizes Kettering U.'s satanic professor, Judith Wolbert....


Saying her name strikes fear in the hearts of many people.

 

[bcterps]

Oh man, I remember Maple its been a looooong time, it sucks compared to Mathematica though.

--Ben

 

[derek]

mathematic is the best and so is MSU

 

[Triumph]

7.