How do I integrate this equation? (Rational functions)

  • Thread starter Thread starter blitzzzz
  • Start date Start date
  • Tags Tags
    Integration
Click For Summary

Homework Help Overview

The discussion revolves around the integration of a rational function expressed in terms of trigonometric functions, specifically involving the variable θ. The original poster seeks assistance in solving the integral of the function (1-COS@)/((1+COS@)(2-COS@)^2) d@.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the t-substitution method (t = tan(θ/2)) to transform the integral into a rational function. There are varying attempts to simplify the resulting expressions and integrate them. Some participants express confusion about the substitution process and the resulting forms.

Discussion Status

Multiple approaches to the problem are being explored, with participants sharing their attempts and expressing uncertainty about the simplifications and integration steps. Some guidance has been offered regarding substitution methods, but there is no clear consensus on the best path forward.

Contextual Notes

Participants note the complexity of the integral and the challenges faced with the substitution and simplification processes. There is mention of integration techniques such as integration by parts, but confusion remains regarding the application of these methods.

blitzzzz
Messages
4
Reaction score
0
Hi, Can anyone help me in solving this equation? It is also shown in the attachment.

[(1-COS@)/((1+COS@)(2-COS@)^2) d@
 

Attachments

Last edited:
Physics news on Phys.org
In such a situations, the t-formula's always help (but it's long).

I mean, do a substitution

[tex]t=\tan(\theta/2)[/tex]

then

[tex]\cos\theta = \frac{1-t^2}{1+t^2}[/tex]

[tex]\sin\theta = \frac{2t}{1+t^2}[/tex]

[tex]\tan\theta = \frac{2t}{1-t^2}[/tex]

This gives you a (not so nice) rational function.
 
Last edited:
micromass said:
In such a situations, the t-formula's always help (but it's long).

I mean, do a substitution

[tex]t=\tan(\theta/2)[/tex]

then

[tex]\cos\theta = \frac{1-t^2}{1+t^2}[/tex]

[tex]\cos\theta = \frac{2t}{1+t^2}[/tex]

[tex]\tan\theta = \frac{2t}{1-t^2}[/tex]

This gives you a (not so nice) rational function.


I am still stuck. Using your method, i got sin@^2 (cos@+1)^2/16cos@. it doesn't look simplified at all
 
How the hell did you arrive there? This should give you a rational function. Can you show what you did?
 
Just replace all occurences of [tex]cos(\theta)[/tex] with

[tex]\frac{1-t^2}{1+t^2}[/tex]

and change dx with

[tex]\frac{2dt}{1+t^2}[/tex]
 
micromass said:
Just replace all occurences of [tex]cos(\theta)[/tex] with

[tex]\frac{1-t^2}{1+t^2}[/tex]

and change dx with

[tex]\frac{2dt}{1+t^2}[/tex]

this is wot i got after substituting the [tex]cos(\theta)[/tex]:

t^2(1+t^2)^2/(1-t^2)^2

and i got sin@^2 (cos@+1)^2/16cos@ after substituting back the [tex] t=\tan(\theta/2) [/tex] and playing ard with it

Btw i do not understand change dx, cos i have no dx, only [tex]d\theta[/tex]:
 
Ummm, why did you substite it back? Just substitute [tex]t=\tan(\theta)[/tex] and then solve the integral for t. No need to substitute it back...
 
micromass said:
Ummm, why did you substite it back? Just substitute [tex]t=\tan(\theta)[/tex] and then solve the integral for t. No need to substitute it back...

I'm still lost.. this is what i have after replacing all the cos@:

[tex] \frac{(2t^2+2t^4)dt}{(1-t^2)^2} [/tex]
This is what i have to integrate right?
I tried to use integration by parts uv-|vdu. it gets very long and i got stuck when i have to integrate ln(1-t^2).
 
Rational functions are always integrated in the same way:

- perform long division
- split in partial fractions

See math.furman.edu/~dcs/book/c6pdf/sec64.pdf for more information.
If you haven't seen how to integrate rational functions, then thiss integral is impossible...
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K