Simplifying Integrands: How to Factor Out Constants in the Denominator?

  • Thread starter Thread starter ohms law
  • Start date Start date
  • Tags Tags
    Simplifying
Click For Summary

Homework Help Overview

The discussion revolves around simplifying integrands, specifically focusing on the algebraic manipulation of constants in the denominator of integrals. The subject area is calculus, particularly integration techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the process of factoring out constants from the denominator of an integral. Questions arise regarding the manipulation of terms, particularly the transformation of \(\frac{x^2}{a^2}\) and its implications for simplification.

Discussion Status

The discussion is active, with participants sharing insights and examples to clarify the factoring process. Some guidance has been offered regarding the algebraic steps involved, though there remains some uncertainty about specific transformations.

Contextual Notes

Participants note that the problem may initially appear more algebraic than calculus-related, highlighting the interplay between different mathematical concepts. There is an acknowledgment of the need for clearer understanding of the manipulation of terms.

ohms law
Messages
70
Reaction score
0
This equality is given in an example, in my textbook:

[itex]\int\frac{1}{a^2+x^2}dx=\frac{1}{a^2}\int\frac{1}{1+(\frac{x}{a})^2}dx[/itex]

So, my question is simply... how?
This is really more of an algebra question, but it'd really help me to see some more detail here.
 
Physics news on Phys.org
[itex]\frac{1}{a^2+x^2} = \frac{1}{a^2(1+ \frac{x^2}{a^2})} = \frac{1}{a^2}\frac{1}{1+ (\frac{x}{a})^2}[/itex]
 
I see that... sort of. It's a form of factoring, obviously.
I don't quite get the [itex]\frac{x^2}{a^2}[/itex] bit, though...
Well, wait: [itex]a^2\times1=a^2[/itex] obviously, and [itex]a^2\times\frac{x^2}{a^2}=x^2[/itex] right? Neat trick, there. Completely obvious too, now that I think about it, but... I guess that I just didn't realize that could be done. Weird.
 
Just think of it as factoring out an a2 from the bottom. Here's an easier example:
[tex]\frac{1}{4+x^{2}}=\frac{1}{(4)1+\frac{x^{2}}{4}}=(\frac{1}{4})\frac{1}{1+{\frac{x^{2}}{4}}}[/tex]
 

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K