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HSC 2018-2019 MX2 Integration Marathon (2 Viewers)

stupid_girl

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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.
Not sure if anyone attempted to go further from this.

If you are careful with the manipulation, you should have got the final answer.









 

stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

I saw another approach on the internet...however the back substitution may be slightly messier.
 

stupid_girl

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Re: HSC 2018 MX2 Integration Marathon

This is slightly tedious.
 
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Paradoxica

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Re: HSC 2018 MX2 Integration Marathon

I saw another approach on the internet...however the back substitution may be slightly messier.
This only works for x>0 because the resultant primitive does not have a derivative at 0, yet the function to be integrated is clearly defined at 0.
 

stupid_girl

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#83 and #88 are still outstanding and this is a new one.
Feel free to share your attempt.
 

fan96

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First we note that for .

Using

,

, and















 

stupid_girl

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A new one


The answer looks quite ugly and probably can't be simplified further. Taking common denominator and expanding out will make it really messy.
 

stupid_girl

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A new one


The answer looks quite ugly and probably can't be simplified further. Taking common denominator and expanding out will make it really messy.
As usual, reverse quotient rule problems are easy to set but difficult to solve. It becomes a piece of cake IF (a big if) you can spot it.:)
 

stupid_girl

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Using the above substitution, it should be obvious that


The definite integral can be evaluated easily.
 
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stupid_girl

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A few new integrals

If you can solve one of them, then you can probably solve all of them.



 

stupid_girl

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Why is it getting so quiet?:cry:

Let's have a new integral. Show that
 
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HeroWise

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After multiplying cosx and rearranging the quadratic im stuck a little hint please!

Edit:
DAMMIT im not getting the sqrt pi


GOT IT!
 
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HeroWise

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Use Kings property for another integral, add them, and reduce to a smaller fraction.
After that use weierstrauss and after simplifying use partial fractions. Then integrate normally. Sub in the values and u got it Nice question
 

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