HSC 2016 MX2 Integration Marathon (archive) (6 Viewers)

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

KingOfActing said:
I did it a kinda similar way, but with just a lot of substitutions (I like substitutions okay).

$\begin{align*}\text{Let } x + \frac{1}{2} = \frac{\sqrt3}{2}\sinh{u}\\\int \frac{1}{x+\sqrt{x^2+x+1}}\,dx &= \int \frac{\sqrt{3}\cosh{u}}{\sqrt3 e^u-1}\,du\\\text{Now, let }u = \ln{z}\\&=\frac{1}{2} \int \frac{z^2+1}{z^2(z-\frac{1}{\sqrt3})}\,dz\\\text{Now, by partial fraction decomposition: }\\&=\frac{1}{2}\left(\int\frac{-3}{z}\,dz - \int\frac{\sqrt3}{z^2}\,dz + \int\frac{4}{z-\frac{1}{3}}\,dz\right)\\&= \frac{-3}{2}\ln|z| + \frac{\sqrt3}{2z} + 2\ln|z - \frac{1}{\sqrt3}|+ C\\\text{Now, just back substituting twice: }\\&=\sqrt{x^2+x+1} - x + 2\ln|x+\sqrt{x^2+x+1}| - \frac{3}{2}\sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C\end{align*}$

That is a very nice way of doing it, I only refer to inverse hyperbolic functions because I can be bothered saying log(insertshithere).
I don't use hyperbolic subs since it's not in the syllabus, but if it was, I would not hold back for even one second.

$\noindent \mathbf{NEXT QUESTION}$

$$Using the substitution $y=\frac{4}{x}$, evaluate $\int_0^\infty \frac{\log{x}}{x^2+2x+4}$d$x$

omegadot · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Of course developing and using the hyperbolic functions indirectly in any given question could be expected, but employing a hyperbolic substitution straight off the bat in order to find an integral is what I was really asking about.

InteGrand · Jan 8, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
Of course developing and using the hyperbolic functions indirectly in any given question could be expected, but employing a hyperbolic substitution straight off the bat in order to find an integral is what I was really asking about.

Yeah, these aren't in the syllabus.

omegadot · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent \mathbf{NEXT QUESTION}$

$$Using the substitution $y=\frac{4}{x}$, evaluate $\int_0^\infty \frac{\log{x}}{x^2+2x+4}$d$x$

$$\noindent Mmm...improper integrals. Not sure if they are still in the syllabus either. I know they were asked way back when, and I know a section on them can still be found in Coroneos' 4 Unit book. But it is such a nice question, so here we go.$

$$\noindent Let $I = \int^\infty_0 \frac{\ln x}{x^2 + 2x + 4} \, dx.$ Using the provided substitution of $y = \frac{4}{x}$ one has $dx = - \frac{4}{y^2} \, dy.$

$$\noindent For the limits: As $x \rightarrow 0^+, y \rightarrow +\infty$ and as $x \rightarrow +\infty, y \rightarrow 0^+$

$$\noindent So$\\\begin{align*}I &= - \int^0_{\infty} \frac{\ln \left (\frac{4}{y} \right )}{\left (\frac{4}{y} \right )^2 + 2 \left (\frac{4}{y} \right ) + 4} \cdot \frac{4}{y^2} \, dy\\&= \ln (4) \int^\infty_0 \frac{dy}{y^2 + 2y^2 + 4} - \int^\infty_0 \frac{\ln y}{y^2 + 2y + 4} \, dy\\\Rightarrow 2I &= \ln (4) \int^\infty_0 \frac{dy}{(y + 1)^2 + 3}\\\Rightarrow I &= \ln 2 \left [\frac{1}{\sqrt{3}} \tan^{-1} \left (\frac{y + 1}{\sqrt{3}} \right ) \right ]^\infty_0\\&= \frac{\pi \ln 2}{3 \sqrt{3}}\end{align*}$

$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{dx}{(x^2 - 2) \sqrt{x^2 - 1}}.$

InteGrand · Jan 8, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Mmm...improper integrals. Not sure if they are still in the syllabus either. I know they were asked way back when, and I know a section on them can still be found in Coroneos' 4 Unit book. But it is such a nice question, so here we go.$

$$\noindent Let $I = \int^\infty_0 \frac{\ln x}{x^2 + 2x + 4} \, dx.$ Using the provided substitution of $y = \frac{4}{x}$ one has $dx = - \frac{4}{y^2} \, dy.$

$$\noindent For the limits: As $x \rightarrow 0^+, y \rightarrow +\infty$ and as $x \rightarrow +\infty, y \rightarrow 0^+$

$$\noindent So$\\\begin{align*}I &= - \int^0_{\infty} \frac{\ln \left (\frac{4}{y} \right )}{\left (\frac{4}{y} \right )^2 + 2 \left (\frac{4}{y} \right ) + 4} \cdot \frac{4}{y^2} \, dy\\&= \ln (4) \int^\infty_0 \frac{dy}{y^2 + 2y^2 + 4} - \int^\infty_0 \frac{\ln y}{y^2 + 2y + 4} \, dy\\\Rightarrow 2I &= \ln (4) \int^\infty_0 \frac{dy}{(y + 1)^2 + 3}\\\Rightarrow I &= \ln 2 \left [\frac{1}{\sqrt{3}} \tan^{-1} \left (\frac{y + 1}{\sqrt{3}} \right ) \right ]^\infty_0\\&= \frac{\pi \ln 2}{3 \sqrt{3}}\end{align*}$

$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{dx}{(x^2 - 2) \sqrt{x^2 - 1}}.$

$\noindent If I am not mistaken, improper integrals aren't explicitly in the syllabus now. But they have asked Q's in the current syllabus asking for an area under a curve (or maybe it was a volume) over an infinite interval, by asking to find the area from $0$ to $N$, and then asking for the limit as $N\to \infty$. Were Euler substitutions for dealing with integrands that are rational functions of $x$ and $\sqrt{ax^2 + bx + c}$ ever in the syllabus?$

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent If I am not mistaken, improper integrals aren't explicitly in the syllabus now. But they have asked Q's in the current syllabus asking for an area under a curve (or maybe it was a volume) over an infinite interval, by asking to find the area from $0$ to $N$, and then asking for the limit as $N\to \infty$. Were Euler substitutions for dealing with integrands that are rational functions of $x$ and $\sqrt{ax^2 + bx + c}$ ever in the syllabus?$

No, I've yet to see an ancient paper that used an euler substitution.

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

$$\noindent Find $\int \frac{dx}{(x^2 - 2) \sqrt{x^2 - 1}}.$

$\noindent By inspection, the answer is $\frac{1}{2\sqrt{2}}\log{\frac{\sqrt{2}\sqrt{x^2-1}-x}{\sqrt{2}\sqrt{x^2-1}+x}}+C\\$I kid. After substituting $x=\sec\theta$, you can then substitute $s=\sin\theta$ and obtain a primitive that formerly had a standard form in the integral tables.$

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Go to uni, start using hyperbolic functions everywhere... probably me...

omegadot · Jan 8, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Were Euler substitutions for dealing with integrands that are rational functions of $x$ and $\sqrt{ax^2 + bx + c}$ ever in the syllabus?$

$$\noindent Euler substitutions, no. But `coming up' with a clever substitution (which of course does not involve the hyperbolic functions), yes ;-)$

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Btw, is there like a next question?

omegadot · Jan 8, 2016

Re: MX2 2016 Integration Marathon

$$\noindent Yes, here are two in the spirit of using a symmetric substitution as invoked in an earlier question of Paradoxica's.$

$$\noindent By using symmetric substitutions, find$

$$\noindent 1. $\int \frac{x^5 - x}{x^8 + 1} \, dx$

$$\noindent 2. $\int \frac{1 + x^2}{1 - x^2} \frac{dx}{\sqrt{1 + x^4}}$

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Yes, here are two in the spirit of using a symmetric substitution as invoked in an earlier question of Paradoxica's.$

$$\noindent By using symmetric substitutions, find$

$$\noindent 1. $\int \frac{x^5 - x}{x^8 + 1} \, dx$

$$\noindent 2. $\int \frac{1 + x^2}{1 - x^2} \frac{dx}{\sqrt{1 + x^4}}$

$\noindent For the first, $u=x^2$, then $v=u+\frac{1}{u}$. For the second, $u=\frac{1}{x}-x$, then do an Euler, Hyperbolic or Trigonometric substitution. Whatever floats your boat.$

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Btw, is there like a next question?

Okay, but you asked for it...

$\int_3^6 \left(\sqrt{x+\sqrt{12x-36}} +\sqrt{x-\sqrt{12x-36}}\right)$d$x$

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Surely I've seen this before

Sent from my iPhone using Tapatalk

InteGrand · Jan 8, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Surely I've seen this before

Sent from my iPhone using Tapatalk

Was this in a Carrotsticks Trial recently?

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Was this in a Carrotsticks Trial recently?

I thought I saw it on the 2015 thread lol

InteGrand · Jan 8, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
I thought I saw it on the 2015 thread lol

Haha wasn't it in a Carrotsticks trial as well? Or maybe one of the 'scrapped' questions.

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

$$Leehuan, I think you may be going mad. The answer is definitely $6\sqrt{3}.$

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$Leehuan, I think you may be going mad. The answer is definitely $6\sqrt{3}.$

Don't worry lol I found my mistake. My wrong working of 4sqrt(3) ended up as 6sqrt(3) but I'll post the right working shortly.

leehuan · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Mistake: Factorised the perfect square wrongly so when I took square roots one of the positive absolute value cases should've been a negative

Sent from my iPhone using Tapatalk

omegadot · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent For the first, $u=x^2$, then $v=u+\frac{1}{u}$. For the second, $u=\frac{1}{x}-x$, then do an Euler, Hyperbolic or Trigonometric substitution. Whatever floats your boat.$

$$\noindent Or for the first you could just use $u = x^2 + \frac{1}{x^2}$ straight off the bat.$

$$\noindent The second question is a slight variation of a question asked in Spivak's \textit{Calculus} (3rd edition). In Q7 (v) on page 381 you are asked to find $\int \frac{x^2 - 1}{x^2 + 1} \frac{dx}{\sqrt{1 + x^4}}.$

HSC 2016 MX2 Integration Marathon (archive) (6 Viewers)

-insert title here-

Active Member

Well-Known Member

Active Member

Well-Known Member

-insert title here-

Well-Known Member

Active Member

Well-Known Member

Active Member

-insert title here-

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 6)