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challenging integration question (1 Viewer)

brett86

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this integration question was covered in my tut this morning, its a very interesting question and only requires 3/4u integration techniques

however, solving it does require creativity:



lets see if anyone can beat FinalFantasy to the solution!
 

brett86

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lol, no, there are no complex numbers used in solving the question

however, thats only in the solution ive seen

its possible there may be a solution that does involve using them

EDIT: its a multiplication sign in the middle:

int{(x^2-1)/(x^2+1) x 1/(sqrt(x^4+1))}dx

------------------------------

the "creative" step is made at the start, once uve done this the answer comes out quite easily
 
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brett86

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the first person to solve this will prob be either u who_loves_maths or FinalFantasy
 

brett86

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maybe not, it might take some time to think up the correct ideas

i tried to do it for a while and ended up asking my tutor for the solution

EDIT: ok, ill remove the hint
 

justchillin

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You can solve any of these integrals with trig sub.... im thinking let x^2=tanx...then proceed...
Edit: tidy it up a bit by, on the numberator, saying x^2-1=X^2+1-2...
 
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brett86

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nice work tutor01!

it can also be solved through the substitution:

u<sup>2</sup> = x<sup>2</sup> + <sup>1</sup>/<sub>x<sup>2</sup></sub>
 

who_loves_maths

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maybe the "creativity" brett was talking about was the re-arrangement of the integrand?

ie. (x^2 -1)/((x^2 + 1)*Sqrt(x^4 + 1)) = [(x^2 -1)/(x^3 + x)]*[Sqrt((x^2)/(x^4 +1))]

then make the substitution, u = (x^2)/(x^4 +1)

which then reduces the integral to:

...

I = {integral of} (-0.5)*[1/((1 + 2u)(Sqrt(u)))]

from here it becomes very simple: one can then make a substitution of, say, v = Sqrt(2u), reducing the integral further to:

I = {integral of} (-1/Sqrt2)*[1/(1 + v^2)]

hence: I = (-1/Sqrt2)*ArcTan(v) + c = (-1/Sqrt2)*ArcTan(Sqrt(2u)) + c

and finally converting to: I = (-1/Sqrt2)*ArcTan(Sqrt((2x^2)/(x^4 +1))) + c


Edit: haha, you got your post in there just before me brett, obviously this ^ isn't the "creativity" you talked about :p
 
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brett86

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tutor01 said:
Anyone want to put up a reconciliation of the two solutions - that is demonstrate that my answer and brett's answer differ only by a constant?


hence, the answers differ by the constant <sup>π</sup>/<sub>2√2</sub>
 

FinalFantasy

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sorry didn't c this post until now, was at some stupid camp:(
that question is from the UNSW maths calculus document lol
question 16C :p
 

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