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Not really, I just use those sites for stalking purposes, and MSN to get away from reality.

The art of sleeping was practiced today.

Now who would you be? :) My timetable, Friday, October 19: English Adv AOS Monday, October 22: English Adv Modules Monday, October 29: Extension 2 Mathematics Friday, November 2: English Extension 1 Monday, November 5: Physics Wednesday, November 7: SOR1 Thursday, November 8: Legal...

<3 you beanie ;):shy: :o

Which is the quick and easy way for integrating sin2x.

Looking at them again, I partially understand a couple of them, and could even see where a possible solution may come from (or just deluding myself :o), but I wouldn't have a clue how to actually solve them yet.

Nope, I'm out.

:shy:

:rofl: I know ;)

Yeah, a rather complicated solution which you could use a different method to get the same answer in a much easier way :)

Harsh :p

Yeah, rather tedious, but more tedious to type it all up :p :rofl:

Absolutely not. Anyway, to do it by parts, (omitting the constants) Int. [sin2x]dx = xsin2x - Int. [x*2sinxcosx]dx Now, Int. [x*2sinxcosx]dx = Int. [x*sin2x]dx = (-xcos2x)/2 - Int. [-1/2 * cos2x]dx = (-xcos2x)/2 + 1/2 Int. [cos2x]dx = (-xcos2x)/2 + (sin2x)/4 Therefore...

You know I <3 your sig ;) And you :embarassed:

You wouldn't have got sin pi/90, but sin (x*pi/90), and when you multiply x*pi/90 by 180/pi to convert it back to degrees again, yes, you would get sin 2x(degrees).

Multiply the degree by pi/180, then you can continue on from there :uhhuh: What is the answer btw?

Not really, I'm behind in maths.

Integration by parts for this question was fun enough to keep me occupied for a couple of minutes :p

I do that :o

There are worse track records.