Can I Solve x^4 dx / (x^2+a^2)^(3/2)?

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In summary, the conversation discusses a difficult integral and various substitution methods to solve it. The final method suggested is x = a sinh(t), which simplifies the integral significantly.
  • #1
theneedtoknow
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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
 
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  • #2
I looks doable to me. Did you try a trig substitution like x=a*tan(t)?
 
  • #3
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
theneedtoknow said:
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.
 
  • #5
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
theneedtoknow said:
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.
 
  • #7
Wow thank you so much!

That was one hell of an integral..!
 
  • #8
x = a sinh(t) is a lot simpler. With some effort you can see the answer using mental computations, without using paper and pencil.
 

1. Can I solve this equation without knowing the value of "a"?

Yes, you can solve this equation without knowing the exact value of "a". You can use algebraic manipulation and integration techniques to find a general solution in terms of "a".

2. Is there a specific method to solve this type of equation?

Yes, there are specific integration techniques that can be used to solve this type of equation, such as substitution or partial fractions. It is important to carefully analyze the equation and choose the most appropriate method.

3. Can this equation be solved analytically or does it require numerical methods?

This equation can be solved analytically using integration techniques. However, if the equation is complex or the value of "a" is large, numerical methods may be more efficient.

4. How do I know if my solution is correct?

You can check your solution by differentiating it and plugging it back into the original equation. If the result is equivalent, then your solution is correct. Additionally, you can use online tools or consult with a math expert for confirmation.

5. Can this equation be solved for any value of "a"?

Yes, this equation can be solved for any value of "a". However, the value of "a" may affect the complexity of the solution and the integration techniques used.

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