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could i teach myself 4unit from the textbook? (1 Viewer)

shaon0

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melonkitten said:
absoulutely no chance.

perhaps complex numbers, or the integration chapter..

but not the whole course
i would disagree....there are some ppl that can learn MX2 themselves and get >95 in the HSC without much work
 

minijumbuk

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Yeah, you can try. I read my friend's 4u notes on complex numbers just for fun, and I understood them (I do 3u)

The further you get ahead of everyone in the HSC, the better. It's best to finish the course early, and leave plenty of time for revision.
 

youngminii

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Yeah I'm also going to teach myself 4unit
Also a bit of every other subject
=D Gl
 

alcalder

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Since most of Ext 2 Maths is work you need to do yourself and most of it is about the process and not the theory, if you know the theory then no one can teach you the thinking process.

Learning on your own may be take a lot more time from your other subjects, however, so I would be careful. It does help to have a mentor/teacher to run things by and explain the harder concepts from time to time.
 

shaon0

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could someone help me on the following?

integration~ S (limits~ pi/3 and pi/6) (secx)^2.sinx dx

this is from new senior maths 4unit page108 question 52
thanks :)
 
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shaon0 said:
could someone help me on the following?

integration~ S (limits~ pi/3 and pi/6) (secx)^2.sinx dx

this is from new senior maths 4unit page108 question 52
thanks :)
just remember secx is 1/cosx
so you'll actually be integrating sinx/cos^2x i think
 

Js^-1

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∫ {Lower limit = pi/6, Upper Limit = pi/3} sec<sup>2</sup>x . sinx dx
=∫ {pi/6, pi/3} sinx / cos<sup>2</sup>x dx

Let u = cos x
du = -sin x dx

At x = pi/3 u = 1/2
x = pi/6 u = (√ 3)/2

= - ∫ {1/2, (√ 3)/2} u<sup>-</sup><sup>2</sup> du
= ∫ {(√ 3)/2, 1/2} u<sup>-</sup><sup>2</sup> du

= - [1/u] {(√ 3)/2, 1/2}

= 2 - 2/(√ 3)
= (6-(√ 3))/3
 

tommykins

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shaon0 said:
could someone help me on the following?

integration~ S (limits~ pi/3 and pi/6) (secx)^2.sinx dx

this is from new senior maths 4unit page108 question 52
thanks :)
(secx)^2.sinx = sinx/cos²x = secx.tanx

Thus the evaluation of the integral becomes [secx]pi/3->pi/6
 
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shaon0

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Js^-1 said:
∫ {Lower limit = pi/6, Upper Limit = pi/3} sec<sup>2</sup>x . sinx dx
=∫ {pi/6, pi/3} sinx / cos<sup>2</sup>x dx

Let u = cos x
du = -sin x dx

At x = pi/3 u = 1/2
x = pi/6 u = (√ 3)/2

= - ∫ {1/2, (√ 3)/2} u<sup>-</sup><sup>2</sup> du
= ∫ {(√ 3)/2, 1/2} u<sup>-</sup><sup>2</sup> du

= - [1/u] {(√ 3)/2, 1/2}

= 2 - 2/(√ 3)
= (6-(√ 3))/3
thanks for your help
 

melonkitten

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shaon0 said:
i would disagree....there are some ppl that can learn MX2 themselves and get >95 in the HSC without much work
err, if you say so, faggot.
 

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