- #1
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.
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]
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...
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