How do I integrate this equation? (Rational functions)

[blitzzzz]

Hi, Can anyone help me in solving this equation? It is also shown in the attachment.

[(1-COS@)/((1+COS@)(2-COS@)^2) d@

 

[micromass]

In such a situations, the t-formula's always help (but it's long).

I mean, do a substitution

t=\tan(\theta/2)

then

\cos\theta = \frac{1-t^2}{1+t^2}

\sin\theta = \frac{2t}{1+t^2}

\tan\theta = \frac{2t}{1-t^2}

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

 

[blitzzzz]

In such a situations, the t-formula's always help (but it's long).

I mean, do a substitution

t=\tan(\theta/2)

then

\cos\theta = \frac{1-t^2}{1+t^2}

\cos\theta = \frac{2t}{1+t^2}

\tan\theta = \frac{2t}{1-t^2}

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]

How the hell did you arrive there? This should give you a rational function. Can you show what you did?

 

[micromass]

Just replace all occurences of cos(\theta) with

\frac{1-t^2}{1+t^2}

and change dx with

\frac{2dt}{1+t^2}

 

[blitzzzz]

Just replace all occurences of cos(\theta) with

\frac{1-t^2}{1+t^2}

and change dx with

\frac{2dt}{1+t^2}

this is wot i got after substituting the cos(\theta):

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

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

Btw i do not understand change dx, cos i have no dx, only d\theta:

 

[micromass]

Ummm, why did you substite it back? Just substitute t=\tan(\theta) and then solve the integral for t. No need to substitute it back...

 

[blitzzzz]

Ummm, why did you substite it back? Just substitute t=\tan(\theta) 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@:

<br /> \frac{(2t^2+2t^4)dt}{(1-t^2)^2} <br />
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).

 

[micromass]

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...