HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

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dunjaaa

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Re: MX2 Integration Marathon

Screen shot 2014-09-09 at 7.53.13 PM.png for n>0
 
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Ikki

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Re: MX2 Integration Marathon

By inspection if we expand the bottom the power of x will be 5/6. Hence let u=x^1/6
Then the integral is reduced to 6u^2/(1+u^2), factor out constants, +1-1 and you get a 6 as a constant and a tan inverse integration (of 1=pi/4)...
=6*(1-pi/4)
=6-2pi/3

Nice shifty U-sub haha
 
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Re: MX2 Integration Marathon

By inspection if we expand the bottom the power of x will be 5/6. Hence let u=x^1/6
Then the integral is reduced to 6u^2/(1+u^2), factor out constants, +1-1 and you get a 6 as a constant and a tan inverse integration (of 1=pi/4)...
=6*(1-pi/4)
=6-2pi/3

Nice shifty U-sub haha
Nice !
I did two substitutions lol. Nevertheless got the same answer.
 

dunjaaa

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Re: MX2 Integration Marathon

Nice work! Here's what I did: Factor x^n from the (1+x^n), then let u=1+1/x^n and it falls out nicely. Nice generalisation aye :D
 
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Ikki

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Re: MX2 Integration Marathon

Hehe made an easy q, What is this to 2 dp:
 
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Ikki

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Re: MX2 Integration Marathon

Here's an actual challenging Q: (Well not for you guys ofcourse :p)
 

dunjaaa

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Re: MX2 Integration Marathon

I remember doing this question in a trial paper :p. I'll post a solution tomorrow if nobody has solved it yet. Hint: Use a trig-substitution
 

dunjaaa

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Re: MX2 Integration Marathon

Screen shot 2014-09-09 at 9.59.51 PM.png :p, what's with the decimals lol
 

Ikki

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Re: MX2 Integration Marathon

Well done guys :)
Whoops @dunjaa switch the boundaries and remember that it's e to the power of the answer :p
 
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