HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

porcupinetree · Feb 22, 2016

Re: MX2 2016 Integration Marathon

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

Reschedule it, integration is more important

omegadot · Feb 22, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Let $n = -k$ where $k \in \mathbb{N}.$

$\begin{align*}\int \frac{dx}{1 - \sqrt[n]{x}} &= \int \frac{dx}{1 - x^{-1/k}} = \int \frac{x^{1/k}}{x^{1/k} - 1} \, dx.\end{align*}$

$\noindent Now let $u = x^{1/k}, dx = k u^{k - 1} \, du.$ Thus$

$\begin{align*}\int \frac{dx}{1 - \sqrt[n]{x}} &= k \int \frac{- u^k}{1 - u} \, du\\ &= k \int \frac{1 - u^k}{1 - u} \, du - k \int \frac{du}{1 - u}\end{align*}$

$$\noindent Noting that the term $\frac{1 - u^k}{1 - u}$ is the sum of the following geometric series $\sum^k_{m = 1} u^{m - 1}$, we have$

$\begin{align*}\int \frac{dx}{1 - \sqrt[n]{x}} &= k \sum^k_{m = 1} \int u^{m - 1} \, du + k \ln |1 - u| + \cal{C}\\&= k \sum^k_{m = 1} \frac{u^m}{m} + k \ln |1 - u| + \cal{C}\\&= k \sum^k_{m = 1} \frac{x^{\frac{m}{k}}}{m} + k \ln |1 - x^{1/k}| + {\cal{C}}, \qquad k \in \mathbb{N}\end{align*}$

$\noindent Replacing $k$ with $-n$ where $n$ is a negative integer gives the desired result.$

omegadot · Feb 22, 2016

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, dx$.$

omegadot · Feb 22, 2016

Re: MX2 2016 Integration Marathon

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.

$\noindent If possible, could someone please pick up and post a copy of the questions to this forum after it has been held. I would be interested in seeing the types of questions they ask. Thanks.$

KingOfActing · Feb 22, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent If possible, could someone please pick up and post a copy of the questions to this forum after it has been held. I would be interested in seeing the types of questions they ask. Thanks.$

I'm pretty sure the 2014 questions are up online somewhere, they didn't look too bad.

Paradoxica · Feb 22, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, dx$.$

$\noindent By inspection...$

Ok I'll show my actual thought process before Carrotsticks decides to roast me again.

$$\noindent To simplify the integrand immensely, we seek to remove the denominator by cancellation. The only terms we can manipulate in order to do that are the numerator terms. Since we have singular terms, we only have sums to products to help us reach the desired cancellation. Observe that $\frac{5+1}{2} = 3,\, \frac{4+2}{2} = 3$, and the corresponding differences are integers. Thus, we have $ \cos{5x} + \cos{x} = 2\cos{3x}\cos{2x}$ and $\cos{4x}+\cos{2x}=2\cos{3x}\cos{x} \\\\ \int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, $d$x \equiv \int \frac{-\cos{2x}+2\cos{3x}\cos{2x}-\cos{x}+2\cos{3x}\cos{x}}{1-2\cos{3x}}\\\int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, $d$x \equiv -\int \frac{(\cos{2x}+\cos{x})(1-2\cos{3x})}{1-2\cos{3x}}\,$d$x\\\int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, $d$x \equiv -\int \cos{2x}+\cos{x}\,$d$x = -\left(\frac{\sin{2x}}{2}+\sin{x}\right) + \cal{C}$

InteGrand · Feb 22, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent By inspection...$

Ok I'll show my actual thought process before Carrotsticks decides to roast me again.

$$\noindent To simplify the integrand immensely, we seek to remove the denominator by cancellation. The only terms we can manipulate in order to do that are the numerator terms. Since we have singular terms, we only have sums to products to help us reach the desired cancellation. Observe that $\frac{5+1}{2} = 3,\, \frac{4+2}{2} = 3$, and the corresponding differences are integers. Thus, we have $ \cos{5x} + \cos{x} = 2\cos{3x}\cos{2x}$ and $\cos{4x}+\cos{2x}=2\cos{3x}\cos{x} \\\\ \int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, $d$x \equiv \int \frac{-\cos{2x}+2\cos{3x}\cos{2x}-\cos{x}+2\cos{3x}\cos{x}}{1-2\cos{3x}}\\\int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, $d$x \equiv -\int \frac{(\cos{2x}+\cos{x})(1-2\cos{3x})}{1-2\cos{3x}}\,$d$x\\\int \frac{\cos 5x + \cos 4x}{1 - 2 \cos 3x} \, $d$x \equiv -\int \cos{2x}+\cos{x}\,$d$x = -\left(\frac{\sin{2x}}{2}+\sin{x}\right) + \cal{C}$

When you used do integrals here by inspection, were they usually actually by inspection, or did you mainly say that to save having to type out lots of tedious working out / LaTeX code?

Drsoccerball · Feb 22, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
When you used do integrals here by inspection, were they usually actually by inspection, or did you mainly say that to save having to type out lots of tedious working out / LaTeX code?

Obviously he is joking when he says that...

omegadot · Feb 22, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, dx.$

$\noindent I hope this one requires a little bit more work than ``inspection'' to find.$

InteGrand · Feb 22, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Obviously he is joking when he says that...

Can never be too sure…

leehuan · Feb 22, 2016

Re: MX2 2016 Integration Marathon

porcupinetree said:
Reschedule it, integration is more important

I already had to reschedule it once because I had a meeting with my old school's alumni on Friday. I'm not rescheduling again rip

Paradoxica · Feb 22, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, dx.$

$$\noindent The following identities are left as exercises for the reader:\\$1-\cos{3x} \equiv (1-\cos{x})(1+2\cos{x})^2,\, \cos{2x}-\cos{x} \equiv (2\cos{x}+1)(\cos{x}-1)\\$Following in a manner similar to the previous integral, $\\\cos{5x}-\cos{4x} \equiv 2\cos{3x}\cos{2x}-2\cos{3x}\cos{x}+\cos{2x}-\cos{x}\\\cos{5x}-\cos{4x} \equiv (1+2\cos{3x})(\cos{2x}-\cos{x})\\ \cos{5x}-\cos{4x} \equiv (1+2\cos{3x})(2\cos{x}+1)(\cos{x}-1)\\\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, $d$x \equiv \int \frac{(1+2\cos{3x})(2\cos{x}+1)(\cos{x}-1)}{(1-\cos{x})(1+2\cos{x})^2}\,$d$x\\\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, $d$x \equiv -\int \frac{1+2\cos{3x}}{1+2\cos{x}}\,$d$x \equiv -\int \frac{1+(2\cos{x})^3 -6\cos{x}}{1+2\cos{x}}$

$\noindent$\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, $d$x \equiv -\int 1-2\cos{x}+4\cos^2{x}-3+\frac{3}{1+2\cos{x}}\,$d$x\\\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, $d$x \equiv \int 2\cos{x}-2\cos{2x}-\frac{3}{1+2\cos{x}}\,$d$x\\\int \frac{\cos 5x - \cos 4x}{1 - \cos 3x} \, $d$x = 2\sin{x}-\sin{2x}+\sqrt{3}\log{\left(\frac{\sqrt{3}\sin{x}-\cos{x}+1}{\sqrt{3}\sin{x}+\cos{x}-1}\right)}+\cal{C}\\$The last integral is left as an exercise for the reader.$

Paradoxica · Feb 22, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
When you used do integrals here by inspection, were they usually actually by inspection, or did you mainly say that to save having to type out lots of tedious working out / LaTeX code?

Drsoccerball said:
Obviously he is joking when he says that...

InteGrand said:
Can never be too sure…

I do it to mock the teachers and professors who always use "by inspection" as their justification for anything. It's patronising, demoralising, and a form of proof by intimidation.

Paradoxica · Feb 22, 2016

Re: MX2 2016 Integration Marathon

$$\noindent$ \int \frac{\text{d}x}{x^{n+1} \sqrt{x^n-a^2}} \\ \int \frac{\text{d}x}{x^{n+1} \sqrt{x^n+a^2}}\\ \int \frac{\text{d}x}{x \sqrt{x^n-a^2}}\\ \int \frac{\text{d}x}{x \sqrt{x^n+a^2}}\\$Universal Hint: Substitute$\\$It is assumed $n \in \mathbb{N}$

leehuan · Feb 22, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
I do it to mock the teachers and professors who always use "by inspection" as their justification for anything. It's patronising, demoralising, and a form of proof by intimidation.

Usually when my teacher said by inspection it was by inspection to my MX2 class as well.

Paradoxica · Feb 22, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Usually when my teacher said by inspection it was by inspection to my MX2 class as well.

Not everyone is that quick.

KingOfActing · Feb 22, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent$ \int \frac{\text{d}x}{x^{n+1} \sqrt{x^n-a^2}}$

$\begin{align*}&\text{First substitution: } x^n = u^2 + a^2\\&\int \frac{dx}{x^{n+1} \sqrt{x^n-a^2}} = \frac{2}{n}\int\frac{du}{(u^2+a^2)^2}\\&\text{Second substitution: }u = a\tan\theta\\&\frac{2}{n}\int\frac{du}{(u^2+a^2)^2} = \frac{2}{a^3n}\int \cos^2{\theta}\,d\theta = \frac{1}{a^3n}(\frac{1}{2}\sin{2\theta} + \theta) + C\\&\text{Assuming I didn't mistakes in back substitution:}\\&\int \frac{dx}{x^{n+1} \sqrt{x^n-a^2}} = \frac{1}{a^3n}\left( \frac{\sqrt{x^n-a^2}}{x^n} + \arctan{\frac{\sqrt{x^n-a^2}}{a}} \right) + C\end{align*}$

leehuan · Feb 23, 2016

Re: MX2 2016 Integration Marathon

$\int{\left(\sqrt{\tan{x}}+\sqrt{\cot{x}}\right) dx}$

Drsoccerball · Feb 25, 2016

Re: MX2 2016 Integration Marathon

Can someone give me a hint in this : $\int_0^1{\frac{1}{(x^2-x+1)(e^{2x-1}+1)}}dx$

Nevermind got it.

Drsoccerball · Feb 25, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int_0^1 \frac{\text{d}x}{\lfloor 1-\log_2{(1-x)} \rfloor}$

I think I posted this and got told off because app it wasn't elementary...

HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

not actually a porcupine

Active Member

Active Member

Active Member

lukewarm mess

-insert title here-

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

-insert title here-

-insert title here-

-insert title here-

Well-Known Member

-insert title here-

lukewarm mess

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)