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
If in one year everyone doing the HSC does extremely well is it possible to get more than 50 99.95's?
Only the top 50 students get 99.95, afaik, so if 100 people get full marks in their exams do they all get 99.95?
I think the best way to resolve the bold part is by going to a tutor where many JR and students from top schools go. Some tutors test their students on regular basis and the students there can be very competitive. To resolve the other things you mentioned it really depends on the individual. I...