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HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

Continue to have fun with trig.:jump:

Harder version:

Find the area between x-axis and y=f(x) on its maximal domain.

Simpler version:
Show that
 
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fan96

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Re: HSC 2018 MX2 Integration Marathon

Continue to have fun with trig.:jump:

Harder version:

Find the area between x-axis and y=f(x) on its maximal domain.

Simpler version:
Show that
a very nice integral.
 

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stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

This one should be considerably easier than the previous one.:tongue:



The answer is pretty small. (1/32304)
 
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fan96

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Re: HSC 2018 MX2 Integration Marathon

I've reduced the integral to



if someone else wants to finish it from this, but it seems very difficult.

Maybe a different approach might be necessary?
 
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stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

I've reduced the integral to



if someone else wants to finish it from this, but it seems very difficult.

Maybe a different approach might be necessary?
You are almost there. If you put x=tan theta, does it look familiar? You've solved that in the previous one.:devil:
 

fan96

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Re: HSC 2018 MX2 Integration Marathon





By using



and some manipulation, similar to the previous integral, we get









Because the integrand is even,







Integrating by parts,







Because for ,



This integral has been evaluated before to be equal to .

 
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stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

This one is absolutely a beast.
 

stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

This is another beast.:cool2:
This is a skeleton solution.
By substituting u=(x-2)/sqrt(2) and considering f(x)+f(-x), the integral can be re-written as


A tangent substitution will turn it into a format that Wolfram can solve...finally:eek:

https://www.wolframalpha.com/input/?i=integrate+sqrt(1+tan^4+x)+/+(1-tan^2+x)

I know Wolfram used hyperbolic tangent substitution but it is also solvable in MX2 by secant substitution.:tongue:

Alternatively, if you don't mind handling improper integral, you can do some algebraic manipulation to get:

Substituting v=u-1+u and w=u-1-u will lead to two improper (but solvable) integrals because u-1 blows up at 0.
 
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stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

This one may look simple at the first glance but actually trickier than you may have thought.
I'm sure a lot of people will come up with an answer 2.:devil:
 

fan96

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Re: HSC 2018 MX2 Integration Marathon

Hint:

 

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Re: HSC 2018 MX2 Integration Marathon

I think this is my first "Stupid_girl's Integrals" ill get. Maybe idk



Well didnt get a but got a Thats outta be good right?

I had to graph the thing, is it possible without graphing it?
 
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Paradoxica

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Re: HSC 2018 MX2 Integration Marathon

This is a skeleton solution.
By substituting u=(x-2)/sqrt(2) and considering f(x)+f(-x), the integral can be re-written as


A tangent substitution will turn it into a format that Wolfram can solve...finally:eek:

https://www.wolframalpha.com/input/?i=integrate+sqrt(1+tan^4+x)+/+(1-tan^2+x)

I know Wolfram used hyperbolic tangent substitution but it is also solvable in MX2 by secant substitution.:tongue:

Alternatively, if you don't mind handling improper integral, you can do some algebraic manipulation to get:

Substituting v=u-1+u and w=u-1-u will lead to two improper (but solvable) integrals because u-1 blows up at 0.
*soluble
 

Paradoxica

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Re: HSC 2018 MX2 Integration Marathon

I think this is my first "Stupid_girl's Integrals" ill get. Maybe idk



Well didnt get a but got a Thats outta be good right?

I had to graph the thing, is it possible without graphing it?
That's correct, but you don't need to graph the function, you just need to know the sign of the thing under the absolutes at every point in the region.
 

stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

The next task is to find the indefinite integral. Of course the answer is not sin x-cos x+c.


Hint: You may consider the floor function.
 

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