but it works fine with the identity, I get: \frac{1}{2}\int_{0}^{\frac{\pi }{2}}sin6\theta sec\theta -sin4\pi sec\theta d\theta which can be evaluated using part a.
Oh wait, there are 2 parts to this question.
Part 1: If I_{n}=\int sin n\theta sec\theta d\theta find I_{n} which is I_{n}=\frac{2cos(n-1)\theta }{n-1}-I_{n-2} and the second part is hence prove...
so I don't think I can use the substitution.
Yes it is, I'm really hoping that they don't care about extracurricular activities and competitions because I haven't done any lol hope 99+ ATAR is enough.
If I don't get in I'll apply for other US unis like Stanford, Chicago and Harvard. If all fails I'll go to UNSW or ANU and apply at MIT for...
Ok, thanks man. mx2 should be enough preparation then.
I am thinking about applying for MIT for undergrad engineering or physics but they say on there website they only accept 150/4000 international applicants per year hahaha