HSC 2016 MX2 Integration Marathon (archive) (3 Viewers)

Paradoxica · Feb 19, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$\noindent Find $\int \frac{1}{\left(x^2 -1 \right)^2}\text{ d}x$.$

$$ \noindent The integral is expected to contain an inverse hyperbolic tangent and the quadratic denominator reduced by a power of one. Assume the integral is of the form $ A \log{\frac{1+x}{1-x}} + \frac{Bx+C}{x^2-1}+\cal{C}\\$Differentiating all of the above gives $\frac{2A-B-(2A+B)x^2 -2Cx}{(x^2-1)^2}\\$Clearly, $C=0$, $2A-B =1$, $2A+B=0 \Leftrightarrow A=\frac{1}{4}$, $B=-\frac{1}{2} \\ \therefore \int \frac{\text{d}x}{(x^2-1)^2} = \frac{1}{4}\log{\frac{1+x}{1-x}}-\frac{x}{2(x^2-1)} +\cal{C}$

Paradoxica · Feb 19, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent$ \int \frac{\sqrt{1-x}}{1-\sqrt{x}}\,$d$x \\ \int \frac{1-\sqrt{x}}{\sqrt{1-x}} \, $d$x\\$You need only evaluate one of them, then the other one can be obtained rather easily.$

Second hint: Once you find either, add the two integrals and simplify. Make your conclusion from there.

porcupinetree · Feb 20, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Second hint: Once you find either, add the two integrals and simplify. Make your conclusion from there.

Sent from my iPhone using Tapatalk

Paradoxica · Feb 20, 2016

Re: MX2 2016 Integration Marathon

$\noindent a) Evaluate $\int \frac{\text{d}x}{\sqrt[n]{x}-1}$ for $n= 3,4,5$. $\\$b) (Challenge) Hence evaluate the general form $\forall n \in \mathbb{N}$, leaving your final answer in sigma notation.$

Drsoccerball · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent a) Evaluate $\int \frac{\text{d}x}{\sqrt[n]{x}-1}$ for $n= 3,4,5$. $\\$b) (Challenge) Hence evaluate the general form $\forall n \in \mathbb{N}$, leaving your final answer in sigma notation.$

$I= 2n(\sum_{k=3}^n{\frac{x}{(2k-2)x^{\frac{1}{k}}}}) -n\ln{|1-x^{\frac{1}{n}}|} + C$

Paradoxica · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$I= 2n(\sum_{k=3}^n{\frac{x}{(2n-2)x^{\frac{1}{n}}}}) -n\ln{|1-x^{\frac{1}{n}}|} + C$

$I= 2n(\sum_{k=3}^n{\frac{x}{(2k-2)x^{\frac{1}{k}}}}) -n\ln{|1-x^{\frac{1}{n}}|} + C$

$Next time, check you answer before you post it.$

$$\noindent Further Challenge: Evaluate the above integral for negative $n.\\$For convenience, multiply the integral by $-1.$

Carrotsticks · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$I= 2n(\sum_{k=3}^n{\frac{x}{(2k-2)x^{\frac{1}{k}}}}) -n\ln{|1-x^{\frac{1}{n}}|} + C$

$Next time, check you answer before you post it.$

$$\noindent Further Challenge: Evaluate the above integral for negative $n.\\$For convenience, multiply the integral by $-1.$

I'm just gonna say this because nobody's called you out on it yet.

You have a very condescending and arrogant tone when you say things like "Next time, check your answer before you post it" or pointing out trivialities. It is very clear what Drsoccerball meant.

This is most certainly not the first time, as I remember you were trying to call out seanieg89 over something as silly as vocabulary.

Tone it down before you make yourself unpopular.

Paradoxica · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Carrotsticks said:
I'm just gonna say this because nobody's called you out on it yet.

You have a very condescending and arrogant tone when you say things like "Next time, check your answer before you post it" or pointing out trivialities. It is very clear what Drsoccerball meant.

This is most certainly not the first time, as I remember you were trying to call out seanieg89 over something as silly as vocabulary.

Tone it down before you make yourself unpopular.

The first time was a joke, and then it wasn't.

As a side note, if you saw my desktop background, you'd very quickly understand why I don't control my ego at times.

Drsoccerball · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Carrotsticks said:
I'm just gonna say this because nobody's called you out on it yet.

You have a very condescending and arrogant tone when you say things like "Next time, check your answer before you post it" or pointing out trivialities. It is very clear what Drsoccerball meant.

This is most certainly not the first time, as I remember you were trying to call out seanieg89 over something as silly as vocabulary.

Tone it down before you make yourself unpopular.

As much as I appreciate that you stood up for me I'm sure that wasn't Paradoxica's intention.

lita1000 · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Carrotsticks said:
I'm just gonna say this because nobody's called you out on it yet.

You have a very condescending and arrogant tone when you say things like "Next time, check your answer before you post it" or pointing out trivialities. It is very clear what Drsoccerball meant.

This is most certainly not the first time, as I remember you were trying to call out seanieg89 over something as silly as vocabulary.

Tone it down before you make yourself unpopular.

Carrotsticks, do you know who Paradoxica reminds me of?

A guy called hayabusaboston lol, do you remember him?

They're like exactly the same lol.

Drsoccerball · Feb 21, 2016

Re: MX2 2016 Integration Marathon

lita1000 said:
Carrotsticks, do you know who Paradoxica reminds me of?

A guy called hayabusaboston lol, do you remember him?

They're like exactly the same lol.

lita ! You left something undone. Please solve that inequalities question on the 4u advanced marathon.

Paradoxica · Feb 21, 2016

Re: MX2 2016 Integration Marathon

lita1000 said:
Carrotsticks, do you know who Paradoxica reminds me of?

A guy called hayabusaboston lol, do you remember him?

They're like exactly the same lol.

You'd have to highlight similarities for me to be able to affirm that.

Nailgun · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
The first time was a joke, and then it wasn't.

As a side note, if you saw my desktop background, you'd very quickly understand why I don't control my ego at times.

pls pm me lol im curious

RealiseNothing · Feb 21, 2016

Re: MX2 2016 Integration Marathon

lita1000 said:
Carrotsticks, do you know who Paradoxica reminds me of?

A guy called hayabusaboston lol, do you remember him?

They're like exactly the same lol.

100% agree lol

SpiralFlex · Feb 21, 2016

Re: MX2 2016 Integration Marathon

Hey guys not an integration Q but there will be an integration bee @ O'week UNSW. Feel free to rock up.

Macauley Theatre (QUAD 1027) Thurs 25th Feb 2016 at 11am.

InteGrand · Feb 21, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{NEW QUESTION}$

$\noindent Find $\int \frac{\sin x}{\sin x+ \cos x}\text{ d}x$.$

Carrotsticks · Feb 22, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent \textbf{NEW QUESTION}$

$\noindent Find $\int \frac{\sin x}{\sin x+ \cos x}\text{ d}x$.$

Let the original integral be I_1, then define I_2 to be the same except with cos in the numerator.

I_1 + I_2 = ....

I_1 - I_2 = ....

Solve simultaneously.

Rationale: The expression is semi-symmetric, so we can somehow force a more 'complete' symmetry (constructing the new integral) and then obtain something from that.

I think it might be an idea for people to provide their rationale behind their methods here, so that others may learn from it.

EDIT: Oops didn't see this was the normal marathon. My point above still stands though.

Carrotsticks · Feb 22, 2016

Re: MX2 2016 Integration Marathon

$\\ $To add to the marathon, find $ \int \frac{x^2-1}{x^3+x}dx$

Paradoxica · Feb 22, 2016

Re: MX2 2016 Integration Marathon

Carrotsticks said:
$\\ $To add to the marathon, find $ \int \frac{x^2-1}{x^3+x}dx$

$=\int \frac{1-\frac{1}{x^2}}{x+\frac{1}{x}} $d$x = \log{\left(x+\frac{1}{x}\right)}+\cal{C}$

What can I even rationalise about this? You just have to learn to spot possible symmetric rational substitutions.

$\noindent Evaluate $\int \frac{\text{d}x}{1-\sqrt[n]{x}}$ for negative $n$, leaving your answer with sigma sum notation.$

leehuan · Feb 22, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$=\int \frac{1-\frac{1}{x^2}}{x+\frac{1}{x}} $d$x = \log{\left(x+\frac{1}{x}\right)}+\cal{C}$

What can I even rationalise about this? You just have to learn to spot possible symmetric rational substitutions.

It was probably intended for the average 2016er to do.

SpiralFlex said:
Hey guys not an integration Q but there will be an integration bee @ O'week UNSW. Feel free to rock up.

Macauley Theatre (QUAD 1027) Thurs 25th Feb 2016 at 11am.

I am sad about this. I have an orthodontist appointment at this date and time.

HSC 2016 MX2 Integration Marathon (archive) (3 Viewers)

-insert title here-

-insert title here-

not actually a porcupine

-insert title here-

Well-Known Member

-insert title here-

Retired

-insert title here-

Well-Known Member

Member

Well-Known Member

-insert title here-

Cole World

what is that?It is Cowpea

Well-Known Member

Well-Known Member

Retired

Retired

-insert title here-

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)