# Integral help

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

## Answers and Replies

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

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

Homework Helper
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

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.

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?

Homework Helper
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?

Right.

theneedtoknow
Wow thank you so much!

That was one hell of an integral..!

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.