Integration help

  • Thread starter blitzzzz
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  • #1
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Hi, Can anyone help me in solving this equation? It is also shown in the attachment.

[(1-COS@)/((1+COS@)(2-COS@)^2) d@
 

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Answers and Replies

  • #2
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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:
  • #3
4
0
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 doesnt look simplified at all
 
  • #4
22,129
3,297
How the hell did you arrive there? This should give you a rational function. Can you show what you did?
 
  • #5
22,129
3,297
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]
 
  • #6
4
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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]:
 
  • #7
22,129
3,297
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...
 
  • #8
4
0
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).
 
  • #9
22,129
3,297
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...
 

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