# Homework Help: Integral help

1. Apr 3, 2010

### theneedtoknow

Is this integral doable at all ?

x^4 dx / (x^2+a^2)^(3/2)

I have tried several u-substitutions but none of them seem to make the form of the integral any simpler

2. Apr 3, 2010

### Dick

I looks doable to me. Did you try a trig substitution like x=a*tan(t)?

3. Apr 4, 2010

### theneedtoknow

Thanks :)
so this is how far I get:

if x=atan(u)
then dx=asec^2(u)du
my numerator becomes x^4dx = a^5tan^4(u)sec^2(u)du

since (x^2+a^2)^1/2 = asec(u)
then my denominator will be a^3sec^3(u)

my old limits of integration were x=0 to x=R
new ones become u=0 to u=tan^-1(R/a)

so my new integral is:

integral from 0 to tan^-1(R/a) of a^2tan^4(u)cos(u)du

unfortunately this is where I get suck again. :S I tried some kind of tan^2=sec^2-1 substitution but I couldn't figure out how to do the resulting integral either

4. Apr 4, 2010

### Dick

Now try v=sin(u) and remember tan(u)=sin(u)/cos(u). I said it was doable. I didn't say it was easy.

5. Apr 4, 2010

### theneedtoknow

So , if I let v=sinu, then dv=cosudu ---> du=dv/cosu >> tan^4ucosu = v^4dv /cos^4(u) = v^4 dv /(1-v^2)^2

Do i now use long division to reduce the power of the denominator by one, and then solve the remainder with partial fractions?

6. Apr 4, 2010

Right.

7. Apr 4, 2010

### theneedtoknow

Wow thank you so much!

That was one hell of an integral..!

8. Apr 4, 2010

### Count Iblis

x = a sinh(t) is a lot simpler. With some effort you can see the answer using mental computations, without using paper and pencil.