HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

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Sy123

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Re: MX2 Integration Marathon





(you may assume the limit converges)
 
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Sy123

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Re: MX2 Integration Marathon



 

Sy123

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Sy123

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Re: MX2 Integration Marathon

Alternatively we do IBP with u=x^2, dv = xsqrt(1-x^2) dx

==



 
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You should probably give a definition of inverse cotangent since it isn't explicitly defined in the syllabus.
 

Sy123

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Re: MX2 Integration Marathon

Try doing this one using no trig substitutions, and only the reverse chain rule.

I have a thing for solving complex problems using basic tools.
Good idea



Which can be done using reverse chain rule

What makes integration one of my favorite topics is the vast amount of ways to get to the same answer, I think we've all shown up to 4 ways to the answer so far
 

Sy123

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braintic

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Re: MX2 Integration Marathon

Please STOP using this site to promote your own stupid website.
 

integral95

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Re: MX2 Integration Marathon

for 1<x<2

1/2<(1-1/x)<1

0<(1-1/x)^n<1

Divide the whole inequality by n

Integrate the whole thing with respect to x with limits 0 to 1

0<(integral)<1/n
Then apply the limit as n approaches infinity to get the result
 

Sy123

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Re: MX2 Integration Marathon

Great question :) I noticed you love the ones that use u=x+1/x as a substitution haha

Yeah I do quite like those substitutions, it seems so elegant.
Like those random ones people think of like x=(1-u)/(1+u) or whatever

 
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