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Use subsitution, its a lot easier than you think cos³@ = (1 - sin²@) x cos@ Therefore the: ∫cos³@ d@ =∫ (1 - sin²@) x cos@ d@ Let u = sin@ so du = cos@.d@ Substituting these in, you get the ∫ (1 - u² )du ....... *now this...

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Hang on a minute...so you actually do mean the integration of sin3x, and not sin^3 x Why are you going through all that mumbo jumbo for? The integration of sin3x is just -(cos3x)/3 And if you were to find the integration of sin^3 x, it will be easier to use substitution sin^3(x) = (1 -...

They are just using the formula: sin(p) - sin(q) = 2sin[(p-q)/2] . cos[(p+q)/2] cos(p) - cos(q) = -2sin[(p-q)/2] . sin[(p+q)/2] You can work these formulas out by mucking around with the basic trig identies: e.g. sin (a+b) - sin (a-b) = 2sinbcosa Let a+b=p Let a-b =q b = (p-q)/2 a =...

Thx for this, but this is common practise :) And i dont think you mean the integration of sin3x, but rather the integration of a trig function that is powered to a certain degree...

I dont like history :)

You couldve have also factorised that equation 2x^3 - x^2 - 6x +3 x^2(2x - 1) - 3(2x - 1) (2x - 1)(x^2 - 3) Factor R: -1, 1, -3, 3 Factor S: 1, 2 1/2 is a possible solution...

Let x = 2sec@ dx = 2sec@tan@d@ I =∫(4sec^2 @ - 4)^1/2 . 2sec@tan@d@ 2sec@ =∫2tan@ . tan@ = 2∫sec^2 @ - 1 = 2[tan@ - @] + c Put it back in terms of x, then ur done...

We are supposed to do an essay in our half yearly about 2 gwen harwood poems, making reference to various interpretations and including our own... Lazyman's guide to make essays sound more personal? Just a few linking phrases or something..

From my experience, the most important things to keep note of: 1) Always draw diagrams 2) PS=ePM...i dont know how many times i've missed this, which makes my working out very long 3) Sum and product of roots, often used for the locus of some point

thx man :) I was doubting the validity of this method when i first started trying to do this question

Does that actually work?? Because i tried letting e^x = sinu, e^x dx = cosudu I got up to: I= -∫secu du = -ln (secu + tanu) + c Then i got stuck...

hehe forgot that u^2 - 1 can be factorised Will you also be able to use the formula 1/2a. (ln |(a+x)/(a-x)|) When integrating 1/(a^2 - x^2)

I dont know the symbol inputs... Whats the integration of (e^x)/(e^2x - 1 )

Just remember that you need to also let n=0, not just n=1

^^^^^^^^^^^^ Uhh..wanna get out and post somewhere else dude!!

I did Prize Giving for my Eng Assessment a few weeks ago and now i need 2 Gwen Harwood poems for my half yearly. I would like to keep Prize Giving as one...is there another poem that it can relate to?

woops, missed the mistake of the I

I got 44.6% for 1 b), are u sure you didnt make a typo? I think ur mistake here is that P=9900 For 2 b) , you forgot to convert weeks back to years, but you also didnt put the correct value of "n"... it is supposed to be 24+3 months, which is 2.25 years Same for 3 b) "n" is 3 years + 6 months...

Generally, the derivative of ln(f(x)) is f'(x)/f(x)