How do I integrate this equation? (Rational functions)

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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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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.
 
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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
 
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]:
 
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).