HSC 2013 Maths Marathon (archive) (1 Viewer)

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Shazer2

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Re: HSC 2013 2U Marathon

I'm not sure, all I know is 1/x integrated is ln(x), not sure about 1/ln(x).
 

JJ345

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Re: HSC 2013 2U Marathon

You would integrate it by writing 1/sinx as sinx/sinx^2 ie. sinx/1-cosx^2
Then perform substitution u=cosx , partial fractions etc. (too hard for 2U). Its just that the limits in your question stuffs things up.
 

HeroicPandas

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Re: HSC 2013 2U Marathon





<------change csc and cot into sine and cosine, apply quotient rule to differentiate

In the end, this is just a memory exercise (unless somehow you knew to multiply top and bottom by that)
 

mahmoudali

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Re: HSC 2013 2U Marathon

Too hard lol

But i do remember something...when integrating a cosecx, you multiply top and bottom by (csc^2 - cot^2) or (csc^2 + cot^2) or (csc + cot) or (csc - cot) then you use
so many options :D

You would integrate it by writing 1/sinx as sinx/sinx^2 ie. sinx/1-cosx^2
Then perform substitution u=cosx , partial fractions etc. (too hard for 2U). Its just that the limits in your question stuffs things up.
so say for instance we change the limits to pi/3 to p/2 the area wouldnt be infinite but is there anyway the function can be differentiated using 2U or 3U even?
 

mahmoudali

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Re: HSC 2013 2U Marathon





<------change csc and cot into sine and cosine, apply quotient rule to differentiate

In the end, this is just a memory exercise (unless somehow you knew to multiply top and bottom by that)
ohh theres the answer how gay is that maths just turned into a guessing game
 

-o-

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Re: HSC 2013 2U Marathon

this thread is stressing my out so much dont know how to do half these questions...
 
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