What is the formula for integrating (a^2 - x^2)^n using integration by parts?

[AndersCarlos]

Homework Statement

Use integration by parts to derive the formula:

\int (a^2 - x^2)^n dx = \frac{x(a^2-x^2)^n}{2n+1} + \frac{2a^2n}{2n+1}\int \frac{(a^2 - x^2)^n}{(a^2 - x^2)} dx + C

Homework Equations

Integration by parts general formula
∫udv = uv - ∫vdu

The Attempt at a Solution

I tried the following:
u = (a^2 - x^2)^n, dv = 1
This gave me another integral, as expected. However, further applications of integration by parts didn't give me the right answer. I am still looking for other terms for 'u' and 'dv'.

 

[SammyS]

Homework Statement

Use integration by parts to derive the formula:

\int (a^2 - x^2)^n dx = \frac{x(a^2-x^2)^n}{2n+1} + \frac{2a^2n}{2n+1}\int \frac{(a^2 - x^2)^n}{(a^2 - x^2)} dx + C

Homework Equations

Integration by parts general formula
∫udv = uv - ∫vdu

The Attempt at a Solution

I tried the following:
u = (a^2 - x^2)^n, dv = 1
This gave me another integral, as expected. However, further applications of integration by parts didn't give me the right answer. I am still looking for other terms for 'u' and 'dv'.

What did you get when you used u = (a^2 - x^2)^n,\ \ dv = 1\,?

There's a bit of a trick involved in getting to the correct solution.

 

[AndersCarlos]

Well, it became this:

x(a^2 - x^2) + \int \frac{2nx(a^2 - x^2)^ndx}{(a^2-x^2)} So I took 2n out of the integral, as it's constant, then, I've tried using: u = (a^2 - x^2)^(n-1) and dv = x, which will result in another integral: ∫nx^3(a^2 - x^2)^(n-2)dx, so this way will just generate (n-k) integrals, while that choosing u = x and dv = (a^2 - x^2)^(n-1), creates: ∫∫(a^2 - x^2)^(n-1)dxdx. (I took out the (uv) part just to show that both my ways weren't correct). Sorry if there is anything wrong. I did these calculations really quick.

 

[SammyS]

Well, it became this:

x(a^2 - x^2) + \int \frac{2nx(a^2 - x^2)^ndx}{(a^2-x^2)} So I took 2n out of the integral, as it's constant, then, I've tried using: u = (a^2 - x^2)^(n-1) and dv = x, which will result in another integral: ∫nx^3(a^2 - x^2)^(n-2)dx, so this way will just generate (n-k) integrals, while that choosing u = x and dv = (a^2 - x^2)^(n-1), creates: ∫∫(a^2 - x^2)^(n-1)dxdx. (I took out the (uv) part just to show that both my ways weren't correct). Sorry if there is anything wrong. I did these calculations really quick.

That should be \displaystyle x(a^2 - x^2)^n+\int \frac{2nx^2(a^2 - x^2)^ndx}{(a^2-x^2)}\,, which can be written: \displaystyle x(a^2 - x^2)^n+2n\int x^2(a^2 - x^2)^{n-1}dx\,.

Added in Edit: (I pressed save message instead of preview.)

Expand the factor of x² in the integral as \ -\{(a^2-x^2)-a^2\}\,.

 

[AndersCarlos]

Sorry, forgot to put the 'n' exponent. I would put the (n-1) exponent, but tex shows it wrongly... x^(n-1)
So, where should I expand the factor? The initial integral or the one after the integration by parts? Sorry, for not understanding, but in what does expanding the factor will help?

 

[SammyS]

Sorry, forgot to put the 'n' exponent. I would put the (n-1) exponent, but tex shows it wrongly... x^(n-1)
So, where should I expand the factor? The initial integral or the one after the integration by parts? Sorry, for not understanding, but in what does expanding the factor will help?

To get tex to show it correctly enclose it in {}, e.g. (a^2-x^2)^{n-1} .

In the integral \displaystyle 2n\int x^2(a^2 - x^2)^{n-1}dx there's a factor x^2.

Write that x^2 as \ \ x^2=-\{(a^2-x^2)-a^2\}\,. Distribute the (a²-x²)ⁿ through that, but don't break up the (a²-x²) . Split into two integrals, one of which is the same as the original.​

Alternatively:

Write the original integral as:
\displaystyle \int (a^2 - x^2)^n dx=\int\left((a^2-x^2)(a^2-x^2)^{n-1}\right)dx

\displaystyle =\int\left(a^2(a^2-x^2)^{n-1}\right)dx-\int\left(x^2(a^2-x^2)^{n-1}\right)dx​

Use integration by parts on the second integral on the last line, letting u=x, and dv will be all the rest.

The integration by parts will include an integral identical to your initial integral. Use algebra to solve for that integral.​

 

[AndersCarlos]

SammyS:

Finally I was able to get it correctly, thank you very much.