Expert: Paul Klarreich - 6/28/2007

Question
Previous Question:

Hi Paul! Can you help me with a correct manner of obtaining the reduction
formula for this:
(    du
)----------
( (u^2+a^2)^n
I know the end result, but do not know the correct manner of obtaining it.

Also, is there more than one reduction formula for a particular intergal? I
assume there is but I thought I should ask you.
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The end result of the reduction formula is equal to:
  1                u            (2n-3)      (    du
---------  X --------------  + ---------  X  )--------
(2a^2)(n-1)  (u^2+a^2)^(n-1)   (2a^2)(n-1)  ( (u^2+a^2)n-1

I've nearly busted my brain trying to get the correct answer out of curiosity, but I can't reduce the (u^2+a^2) term without getting an vexating 1/x^2 term in it, which is retrograde to the purpose and it is really bothering me because it is imposing upon my "Calculus Ego" (you know, the ego that all calculus students have that whispers to them that they're absolutely invicible and impervious and, therefore, able to integrate anything the world has to offer. What is that? a tank? dont' worry, I'll intergrate it.) Well, I'm going to try it again, thanks Paul.

Answer
Hi Justin,

There is an old trick that is useful here.  When you have something like  x^2, and you WANT  x^2 + a^2, or something like it, write:

x^2 = x^2 + a^2 - a^2.

That makes things more complex, but you may be able to handle the two parts separately.  I used something like that here.

Here is your integral:

{        dx
|  -----------------
}  (a^2 + x^2)^n

YOU WROTE:
.......................
The end result of the reduction formula is equal to:
   1             u            (2n-3)      (    du
---------   --------------  + ---------    |  --------
(2a^2)(n-1) (u^2+a^2)^(n-1)   (2a^2)(n-1)  ( (u^2+a^2)n-1
..........................
I am changing your ‘u’ to an ‘x’ because the symbol ‘u’ is going to be used later for other things.

WARNING: USE COURIER FONT TO VIEW THIS, BECAUSE THERE ARE FRACTIONS AND OTHER STUFF.

{        dx
|  -------------- =
}  (a^2 + x^2)^n

1   {     a^2 dx
---- |  -------------- =
a^2  }  (a^2 + x^2)^n

FROM NOW ON, I WILL NOT WRITE THE FACTOR OF 1/a^2.  I will put it back at the end.  If I forget, you can put it back.


{  a^2 + x^2 - x^2   
|  ----------------- dx  [THE TRICK]
}  (a^2 + x^2)^n


{  (a^2 + x^2) - x^2   
|  ----------------- dx
}  (a^2 + x^2)^n

{         1              {      x^2   
|  --------------- dx  - | ------------- dx
}  (a^2 + x^2)^n-1       } (a^2 + x^2)^n

I will make an abbreviation here for the first term and write that as:

    {       x^2   
RI - | -------------- dx
    } (a^2 + x^2)^n

RI stands for reduced integral -- with n-1 as the exponent.

OK, now the first of those integrals will show up as part of the reduction formula you want.  The second one has a very useful x^2 in the numerator.  How useful?

1. One piece, an x, will integrate the denominator.  Integrating that will give us (....)^n-1 and the ‘unintegrated’ part of your reduction formula.
2. The other piece, another x, is easy to differentiate.

In other words, do integration by parts on that second term:

{       x^2   
| -------------- dx
} (a^2 + x^2)^n

Choose, for the IBP:

u = x,  and  du = dx

dv =  x(a^2 + x^2)^-n dx = (1/2) 2x(a^2 + x^2)^-n dx

v = (1/2)(a^2 + x^2)^-n+1 OVER (-n+1) or (1-n)
          1
v = -----------------
   2(....)^n-1(1-n)
                x
Now  uv = ----------------
         2(....)^n-1(1-n)

[Sorry, I am getting a bit sloppy with the parenthesizing, but I think you will figure it out.]

 {
- | v du =
 }

   1   {       dx               RI
- ------| ---------------- = - ------
 2(1-n)} (a^2 + x^2)^n-1      2(1-n)

(Remember, RI stands for reduced integral.)

So we are up to:

     {       x^2   
RI  - | -------------- dx   (from before) =
     } (a^2 + x^2)^n

             x              RI
RI  - [ ---------------- - ------ ]
       2(....)^n-1(1-n)   2(1-n)


             x              RI
RI  -  ----------------  + ------
       2(....)^n-1(1-n)   2(1-n)


      RI            x       
RI + ------ -  ----------------
    2(1-n)    2(....)^n-1(1-n)

         1               x       
RI[1  + ------] -  ----------------
       2(1-n)      2(....)^n-1(1-n)

  2 - 2n + 1             x       
RI[----------] -  ----------------
   2(1-n)         2(....)^n-1(1-n)

   3 - 2n             x       
RI[--------] -  ----------------
   2(1-n)       2(....)^n-1(1-n)

That’s where I will leave it.  You can now get your formula by:
1. Flipping the signs on the top and bottom of the first term.
2. Putting the reduced integral in place of RI.
3. Oh, yes, did I forget something?

And, one last comment, to you, Mr. "I can integrate anything":

Try this one:

{
| e^(-x^2) dx
}