# Integration help

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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• equation.bmp
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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.

Last edited:
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 doesnt 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 $$cos(\theta)$$ with

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

and change dx with

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

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 $$t=\tan(\theta/2)$$ and playing ard with it

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

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...

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@:

$$\frac{(2t^2+2t^4)dt}{(1-t^2)^2}$$
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