HSC 2016 MX2 Integration Marathon (archive) (5 Viewers)

omegadot · Jan 8, 2016

Re: MX2 2016 Integration Marathon

$\noindent Turning my previous question into something a little more interesting, suppose that$

$I = \int \frac{1 + x^2}{1 - x^2} \frac{dx}{\sqrt{1 + x^4}}, \quad J = \int \frac{1 - x^2}{1 + x^2} \frac{dx}{\sqrt{1 + x^4}}.$

$\noindent By finding $I$ and $J$, use these results to find the following two integrals:$

$\noindent (a) $ \int \frac{1 + x^4}{1 - x^4} \frac{dx}{\sqrt{1 + x^4}}$, and$

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

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent Turning my previous question into something a little more interesting, suppose that$

$I = \int \frac{1 + x^2}{1 - x^2} \frac{dx}{\sqrt{1 + x^4}}, \quad J = \int \frac{1 - x^2}{1 + x^2} \frac{dx}{\sqrt{1 + x^4}}.$

$\noindent By finding $I$ and $J$, use these results to find the following two integrals:$

$\noindent (a) $ \int \frac{1 + x^4}{1 - x^4} \frac{dx}{\sqrt{1 + x^4}}$, and$

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

$\noindent After making the appropriate symmetric substitutions, $I$ is amenable to $\sqrt{2}\sinh{u}\;\mathbf{OR}\sqrt{2}\tan\theta$ and $J$ is amenable to $\sqrt{2}\cosh{u}\;\mathbf{OR}\sqrt{2}\sec{\theta}\\ \mathbf{Ir}$regardless, we proceed onwards, and $\\I = \frac{1}{\sqrt{2}}\log{\frac{x^2-1}{\sqrt{x^4+1}}},\; J = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{x^4+1}}{x^2+1}\\I = \int \frac{(1+x^2)^2}{1-x^4}\frac{\text{d}x}{\sqrt{1+x^4}}= \int \frac{1+2x^2+x^4}{1-x^4}\frac{\text{d}x}{\sqrt{1+x^4}}\\J = \int \frac{(1-x^2)^2}{1-x^4}\frac{\text{d}x}{\sqrt{1+x^4}}= \int \frac{1-2x^2+x^4}{1-x^4}\frac{\text{d}x}{\sqrt{1+x^4}}\\I+J = 2\int\frac{1+x^4}{1-x^4}\frac{dx}{\sqrt{1+x^4}},\;I-J=4\int\frac{x^2}{1-x^4}\frac{dx}{\sqrt{1+x^4}}\\$The answer then follows by substitution and simplification.$

KingOfActing · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Wait, so would we not be allowed to use hyperbolic functions in exams? Our teacher never told us that, so that's a bit worrying. Would we still be allowed to be all cheeky though and make a substitution like $x = \frac{e^u-e^{-u}}{2}$ and then just prove a few basic properties before using it, or nah?

Also slightly related, is L'Hospitals allowed? Our teacher also taught our 3U class that, and said we were allowed to use it in exams.

JusticeTackle · Jan 8, 2016

Re: MX2 2016 Integration Marathon

KingOfActing said:
Wait, so would we not be allowed to use hyperbolic functions in exams? Our teacher never told us that, so that's a bit worrying. Would we still be allowed to be all cheeky though and make a substitution like $x = \frac{e^u-e^{-u}}{2}$ and then just prove a few basic properties before using it, or nah?

Also slightly related, is L'Hospitals allowed? Our teacher also taught our 3U class that, and said we were allowed to use it in exams.

You are truly the king of acting

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

KingOfActing said:
Wait, so would we not be allowed to use hyperbolic functions in exams? Our teacher never told us that, so that's a bit worrying. Would we still be allowed to be all cheeky though and make a substitution like $x = \frac{e^u-e^{-u}}{2}$ and then just prove a few basic properties before using it, or nah?

Also slightly related, is L'Hospitals allowed? Our teacher also taught our 3U class that, and said we were allowed to use it in exams.

I think it is completely fine, but it's a fair bit to write. However, I do think that if you are sufficiently well practised with hyperbolic substitutions, then the amount of time it takes to do the working with hyperbolic sub and writing out the basic properties of the hyperbolic subs will be faster than doing it through more acceptable means.

I would suggest also practising Euler substitutions, as those are completely acceptable.

L'Hopital's Rule is not accepted. However, the proof is incredibly short, so one must wonder if one could prove it in the middle of the exam and use it...

KingOfActing · Jan 8, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
I think it is completely fine, but it's a fair bit to write. However, I do think that if you are sufficiently well practised with hyperbolic substitutions, then the amount of time it takes to do the working with hyperbolic sub and writing out the basic properties of the hyperbolic subs will be faster than doing it through more acceptable means.

I would suggest also practising Euler substitutions, as those are completely acceptable.

L'Hopital's Rule is not accepted. However, the proof is incredibly short, so one must wonder if one could prove it in the middle of the exam and use it...

Alright, thanks. By the way, which proof are you referring to? The one I know only applies to limits to a real number in the case $\frac{0}{0}$ .

Paradoxica · Jan 8, 2016

Re: MX2 2016 Integration Marathon

KingOfActing said:
Alright, thanks. By the way, which proof are you referring to? The one I know only applies to limits to a real number in the case $\frac{0}{0}$ .

Yes, the one used for that exact indeterminate form.

Paradoxica · Jan 9, 2016

Re: MX2 2016 Integration Marathon

$$\noindent$\mathbf{NEXT \; QUESTION: show}\\ \int_0^{\frac{\pi}{4}} \left (\frac{(1-x^2)\log(1+x^2) +(1+x^2) -(1-x^2)\log(1-x^2)}{(1-x^4)(1+x^2)} \right )xe^{\left(\frac{x^2-1}{x^2+1}\right)}$d$x = -\frac{1}{4}e^{-\left(\frac{16-\pi^2}{16+\pi^2}\right)}\log{\left(\frac{16-\pi^2}{16+\pi^2} \right )}$

KingOfActing · Jan 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent$\mathbf{NEXT \; QUESTION: show}\\ \int_0^{\frac{\pi}{4}} \left (\frac{(1-x^2)\log(1+x^2) +(1+x^2) -(1-x^2)\log(1-x^2)}{(1-x^4)(1+x^2)} \right )xe^{\left(\frac{x^2-1}{x^2+1}\right)}$d$x = -\frac{1}{4}e^{-\left(\frac{16-\pi^2}{16+\pi^2}\right)}\log{\left(\frac{16-\pi^2}{16+\pi^2} \right )}$

$\begin{align*} \text{Let } u = \frac{1-x^2}{1+x^2},\,du = \frac{-4x}{(x^2+1)^2}\,dx\\&\int_0^{\frac{\pi}{4}} \left (\frac{(1-x^2)\log(1+x^2) +(1+x^2) -(1-x^2)\log(1-x^2)}{(1-x^4)(1+x^2)} \right )xe^{\left(\frac{x^2-1}{x^2+1}\right)}\,dx \\&= \int_0^{\frac{\pi}{4}} \left (\frac{(1-x^2)\log(\frac{1+x^2}{1-x^2}) +(1+x^2)}{(1-x^2)(1+x^2)^2} \right )xe^{\left(\frac{x^2-1}{x^2+1}\right)}\,dx\\&=\int_0^{\frac{\pi}{4}} \left (\frac{\log(\frac{1+x^2}{1-x^2}) +\frac{(1+x^2)}{(1-x^2)}}{(1+x^2)^2} \right )xe^{\left(\frac{x^2-1}{x^2+1}\right)}\,dx\\&=\frac{-1}{4}\int_{1}^{\frac{16-\pi^2}{16+\pi^2}} \left(-\log{u}+\frac{1}{u}\right )e^{-u}\,du\\&=\frac{-1}{4}\left(e^{-u}\log{u} \Big |_{1}^{\frac{16-\pi^2}{16+\pi^2}}\right )\\&=-\frac{1}{4}e^{-\left( \frac{16-\pi^2}{16+\pi^2}\right )}\ln{\left(\frac{16-\pi^2}{16+\pi^2} \right )}\end{align*}$

InteGrand · Jan 9, 2016

Re: MX2 2016 Integration Marathon

$\textbf{NEW QUESTION}$

$\noindent Find $\int \frac{\sin^2 x-4\sin x \cos x+3\cos^2 x}{\sin x + \cos x}\text{ d}x$.$

Paradoxica · Jan 9, 2016

Re: MX2 2016 Integration Marathon

That integral is so messy... just like my life.

Paradoxica · Jan 9, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$\noindent Find $\int \frac{\sin^2 x-4\sin x \cos x+3\cos^2 x}{\sin x + \cos x}\text{ d}x$.$

$$\noindent$ I_1 = \int \frac{-3\sin^2{x}-4\sin{x}\cos{x}-\cos^2{x}+4}{\sin{x}+\cos{x}}$d$x=\int \frac{(\sin{x}+\cos{x})(-3\sin{x}-\cos{x})+4}{\sin{x}+\cos{x}}$d$x\\I_1=\int -3\sin{x}-\cos{x}+\frac{4}{\sin{x}+\cos{x}}$d$x = 3\cos{x}-\sin{x}+I_2\\I_2 = \int \frac{4(1+t^2)}{1+2t-t^2}\frac{\text{2d}t}{1+t^2}=\int \frac{8\text{d}(t-1)}{2-(t-1)^2}=2\sqrt{2}\log{\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}}\\ \therefore I_1 = 3\cos{x}-\sin{x}+2\sqrt{2}\log{\frac{(\sqrt{2}-1)\cos{x}+\sin{x}}{(\sqrt{2}+1)\cos{x}-\sin{x}}}+C$

InteGrand · Jan 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent$ I_1 = \int \frac{-3\sin^2{x}-4\sin{x}\cos{x}-\cos^2{x}+4}{\sin{x}+\cos{x}}$d$x=\int \frac{(\sin{x}+\cos{x})(-3\sin{x}-\cos{x})+4}{\sin{x}+\cos{x}}$d$x\\I_1=\int -3\sin{x}-\cos{x}+\frac{4}{\sin{x}+\cos{x}}$d$x = 3\cos{x}-\sin{x}+I_2\\I_2 = \int \frac{4(1+t^2)}{1+2t-t^2}\frac{\text{2d}t}{1+t^2}=\int \frac{8\text{d}(t-1)}{2-(t-1)^2}=2\sqrt{2}\log{\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}}\\ \therefore I_1 = 3\cos{x}-\sin{x}+2\sqrt{2}\log{\frac{(\sqrt{2}-1)\cos{x}+\sin{x}}{(\sqrt{2}+1)\cos{x}-\sin{x}}}+C$

$Well done!$

$\noindent \textbf{NEW QUESTION}$

$\noindent Find $\int \frac{1}{x^8 + 1}\text{ d}x$.$

Paradoxica · Jan 9, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$Well done!$

$\noindent \textbf{NEW QUESTION}$

$\noindent Find $\int \frac{1}{x^8 + 1}\text{ d}x$.$

$$\noindent My formula on the extracirricular marathon has been reviewed and corrected after I realised this integral involves tedious decompositions.$

InteGrand · Jan 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent My formula on the extracirricular marathon has been reviewed and corrected after I realised this integral involves tedious decompositions.$

$\noindent Nice.$

$\noindent Here's a more HSC-friendly one: Find $\int \frac{1}{1 + \sin x }\text{ d}x$.$

Paradoxica · Jan 9, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Nice.$

$\noindent Here's a more HSC-friendly one: Find $\int \frac{1}{1 + \sin x }\text{ d}x$.$

$$\noindent This is one of those few questions where the half angle substitution actually makes things easier, not harder.$ \\ I = \int \frac{\frac{2\text{d}t}{1+t^2}}{1+\frac{2t}{1+t^2}}=\int\frac{dt}{1+2t+t^2}=\int -$d$\left(\frac{1}{1+t}\right) = -\frac{1}{1+\tan{\frac{x}{2}}}+C = \frac{-2}{1+\frac{1-\cos{x}}{\sin{x}}}+C=\frac{-2\sin{x}}{1+\sin{x}-\cos{x}}+C$

$\mathbf{NEXT QUESTION}$

$\int_0^{\frac{\pi}{2}}\frac{\text{d}x}{1+\tan^{a}{x}}\;\; \forall a \in \mathbb{R}$

Drsoccerball · Jan 9, 2016

Re: MX2 2016 Integration Marathon

Really? ^

$I= \int{\frac{1-sinx}{cos^2x}}dx$

$I= \int{ sec^2x -tanxsecx} dx$

$I=tanx -secx +C$

Paradoxica · Jan 9, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Really? ^

$I= \int{\frac{1-sinx}{cos^2x}}dx$

$I= \int{ sec^2x -tanxsecx} dx$

$I=tanx -secx +C$

M8 Inspection is fine too, but there are others around here.

Now do my question.

InteGrand · Jan 9, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Really? ^

$I= \int{\frac{1-sinx}{cos^2x}}dx$

$I= \int{ sec^2x -tanxsecx} dx$

$I=tanx -secx +C$

$\noindent I think Paradoxica meant that the $t$-substitution ends up making things nicer than what they started with (which isn't always the case).$

Drsoccerball · Jan 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\mathbf{NEXT QUESTION}$

$\int_0^{\frac{\pi}{2}}\frac{\text{d}x}{1+\tan^{a}{x}}\;\; \forall a \in \mathbb{R}$

$$ First take note that: $ \int_0^{Paradoxica}{f(x)}dx=\int_0^{paradoxica}{f((paradoxica)-x)}dx$

$I= \int_0^{\frac{\pi}{2}}{\frac{cos^ax}{sin^ax+cos^ax}}dx$

$I= \int_0^{\frac{\pi}{2}}{\frac{sin^ax}{sin^ax+cos^ax}}dx$

$2I= \int_0^{\frac{\pi}{2}}{\frac{cos^ax + sin^ax}{sin^ax+cos^ax}}dx$

$I= \frac{\pi}{4}$

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