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Yep, I already see where I went wrong, namely leaving tan 2x out of there somewhere (which I'm assuming was important). Stuff this, I'm doing it on paper, scanning it and posting the image. Can't read my own typing...

why thank you. I'm only new at it, that's why all the fractions are still in x/y form for instance

\int\sec^3 2x\,dx = \int\sec^2 2x sec 2x\,dx = 1/2\int\(1 - tan2x) sec 2x\,d 2x = -1/2(1/2 sec 2x) + 1/2\int\sec 2x\,dx + C = -1/4sec 2x + 1/2(x sec 2x - \int\ x \,d(sec 2x)) + C = -1/4sec 2x + (x sec 2x)/2 - 1/2(sec 2x (x^2)/2) + C = -1/4sec 2x + (x sec 2x)/2 - x^2 sec 2x = sec...

Respect...

An essay accounting for the decline of living conditions under the rule of Stalin, involved a long, in depth discussion of how if Dr. Phil was there, he would have been able to talk everybody through it

Same way cis(2pi/3) is one of the points on the unit circle z^3 = 1

It's not really a plane of reality. More its a graphical interpretation of complex numbers, which is why you can't have a complex number locus. RULE 1 of Complex Numbers: They cannot be graphed

wouldn't: arg [(1+i)z] = arg[z] + arg[iz] ???

Complex numbers are always so much easier if you use the Argand Diagram, particularly these questions. Here's a starter:

70% is a great mark. In the HSC, a raw mark of around 60/120 will get you a 90+ The exam is designed so you can't get full marks.

I wasn't aware that you could have loci on the imaginary axis, just vectors.

Slightly unrelated, but ohexploitable, is your avatar from Gossip Girl? Also, as we have an absence of questions: If In = (Integral from 1 to 0) x^n tan(inverse)x dx, n = 0, 1, 2, ... Show that In = 1/(n+1) * π/2 - 1/[n(n+1)] - (n-1)/(n+1) * I(n-2), n = 2, 3, 4, ... Sorry, don't know latex

No way, curve sketching is so easy. You just have to remember what each does and you're set. Hating integration, don't mind complex numbers Graphs > Complex Numbers > Conics > Volumes > Harder Ext. 1 > Polynomials > Integration Haven't done mechanics yet