HSC 2012-14 MX2 Integration Marathon (archive) (8 Viewers)

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Sy123

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Re: MX2 Integration Marathon

Let's generalize that just a little bit



(its doable I tried it)
 

dunjaaa

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Re: MX2 Integration Marathon

chucking a carrot I see
I got I(n)={[k(n-m+1)]/[m+k(n+1)]}I(n-m)
 

integral95

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Re: MX2 Integration Marathon

Hey guys in dat UNSW ASOC integration bee, there was this weird question (carrot would approve)


 

Carrotsticks

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Re: MX2 Integration Marathon

Hey guys in dat UNSW ASOC integration bee, there was this weird question (carrot would approve)


I thought it was a rather different one though lol.

The actual integration is very straightforward, it's just finding the closed form that not too many people are familiar with.
 

dunjaaa

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Re: MX2 Integration Marathon

change into half angles -> int{[x+2sin(x/2)cos(x/2)]/2cos^2(x/2) dx}
=int{1/2 [xsec^2(x/2)]+tan(x/2) dx}
=[xtan(x/2)] between x=0 and x=pi/4 by reverse product rule
=pi/4tan(pi/8)
 

Sy123

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Re: MX2 Integration Marathon

change into half angles -> int{[x+2sin(x/2)cos(x/2)]/2cos^2(x/2) dx}
=int{1/2 [xsec^2(x/2)]+tan(x/2) dx}
=[xtan(x/2)] between x=0 and x=pi/4 by reverse product rule
=pi/4tan(pi/8)
nice work

 
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