Recent content by blitzzzz

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).

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 :

I am still stuck. Using your method, i got sin@^2 (cos@+1)^2/16cos@. it doesn't look simplified at all

Hi, Can anyone help me in solving this equation? It is also shown in the attachment. [(1-COS@)/((1+COS@)(2-COS@)^2) d@