HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

glittergal96 · Sep 19, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Another way to look at it is this: Invert the function, restricting the range so that $0<y\leq \frac{\pi}{2}$ where $y=\sin^{-1}{e^x}$
For small theta, $\sin{\theta} \approx \theta$ , and as a corollary, $\theta \approx \sin^{-1}{\theta}$
This means that any improper integral "all the way out over there" to negative infinity will be very close to the improper integral from "all the way out there" to negative infinity of $e^x$ . Since the improper integral of e^x in this given bound is finite, the improper integral of our case is also finite, otherwise approximation would fail to exist. Since the rest of the integral from "all the way out there" to here at x=0 is also finite, then that means the improper integral over the entire domain is finite. Ergo, our original integral must also be finite since it is identical.

Sure, this is just performing an additional substitution on mine. You could do other substitutions as well if you preferred for whatever reason.

All that matters is that the integrand is asymptotic to log(x) near the origin, and that this is integrable.

leehuan · Sep 20, 2015

Re: MX2 2015 Integration Marathon

Haven't tried it yet so I'm not sure if it's easier or harder than the one I did identical to this.

$\int { \sqrt [ 3 ]{ \tan { x } } dx }$

Paradoxica · Sep 20, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Haven't tried it yet so I'm not sure if it's easier or harder than the one I did identical to this.

$\int { \sqrt [ 3 ]{ \tan { x } } dx }$

It's doable. I've done it before, but it is incredibly messy and non simplifiable. In fact, any specific case of $\sqrt[n]{\tan{x}}dx$ is expressable in terms of elementary functions, but the general case requires the hypergeometric function (non-elementary)
Edit: hmmm... we should have a specific case as a multiple choice question... lol

snail489 · Sep 20, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Haven't tried it yet so I'm not sure if it's easier or harder than the one I did identical to this.

$\int { \sqrt [ 3 ]{ \tan { x } } dx }$

We get this horrid thing:

By rewriting it as 3u^3/(u^6+1) where u^3=tanx, and then decomposing it into a giant partial fraction of quadratic factors.

leehuan · Sep 21, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
It's doable. I've done it before, but it is incredibly messy and non simplifiable. In fact, any specific case of $\sqrt[n]{\tan{x}}dx$ is expressable in terms of elementary functions, but the general case requires the hypergeometric function (non-elementary)
Edit: hmmm... we should have a specific case as a multiple choice question... lol

Makes sense if they can all be expressed in elementary functions I suppose.
_______________________________
I have a feeling it's been asked already but

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ \sqrt { 1+\cos { x } } \sqrt { \sin { x } +\cos { x } } } }$

porcupinetree · Sep 21, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
I have a feeling it's been asked already but

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ \sqrt { 1+\cos { x } } \sqrt { \sin { x } +\cos { x } } } }$

Could you pls link me to an image (or something) of that? LaTeX images are stuffing up for me on my computer atm, all I see is a little white/blue/green square. Thx

Paradoxica · Sep 21, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Could you pls link me to an image (or something) of that? LaTeX images are stuffing up for me on my computer atm, all I see is a little white/blue/green square. Thx

Second that. Not sure why, and nobody can explain it.

Ekman · Sep 21, 2015

Re: MX2 2015 Integration Marathon

snail489 said:
We get this horrid thing:
View attachment 32418

By rewriting it as 3u^3/(u^6+1) where u^3=tanx, and then decomposing it into a giant partial fraction of quadratic factors.

Just a heads up, you can reduce 3u^3/(u^6+1) into 3/2 * x/(x^3+1) by letting u^2 = x, so it would be a little more cleaner.

leehuan · Sep 21, 2015

Re: MX2 2015 Integration Marathon

It was working when I first posted it.

If it doesn't work then... whatever

1/sqrt((sinx+cosx)(1+cosx)); x=0; x=pi/2

Ekman · Sep 21, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
View attachment 32426I thought that was just me, didn't think it'd be everyone.

Still not working, the link is apparently invalid

Paradoxica · Sep 21, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
View attachment 32427

It was working when I first posted it.

If it doesn't work then... whatever

1/sqrt((sinx+cosx)(1+cosx)); x=0; x=pi/2

'Kay, got the correct answer, but I find myself unable to justify removing the $\sin^{-1}(\frac{\infty-1}{\sqrt{2}})$ (I used t-substitution)

Drsoccerball · Sep 21, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
'Kay, got the correct answer, but I find myself unable to justify removing the $\sin^{-1}(\frac{\infty-1}{\sqrt{2}})$ (I used t-substitution)

Im pretty sure this is from BOS trial 2014 the only way i think it could be solved is t method

leehuan · Sep 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Im pretty sure this is from BOS trial 2014 the only way i think it could be solved is t method

Lol he got me. (Whilst you're here where are the answers I can't find them in the search D: )

____________________
Yeah, I attempted it by t-substutition as well. I was wondering if there were any possible ways when I posted it.

rand_althor · Sep 21, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Lol he got me. (Whilst you're here where are the answers I can't find them in the search D: )

____________________
Yeah, I attempted it by t-substutition as well. I was wondering if there were any possible ways when I posted it.

They are in this thread: Official BOS Trial 2015 Thread (right before the introduction). Here is a direct link: BOS Trials.zip.

leehuan · Sep 21, 2015

Re: MX2 2015 Integration Marathon

rand_althor said:
They are in this thread: Official BOS Trial 2015 Thread (right before the introduction). Here is a direct link: BOS Trials.zip.

I'm so dumb, not thinking that the .zip file would be updated.

Paradoxica · Sep 26, 2015

Re: MX2 2015 Integration Marathon

Evaluate:
$\int_{0}^{\infty}{\frac{\ln{(1+x^2)}}{1+x^2}}dx$

Drsoccerball · Sep 27, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Evaluate:
$\int_{0}^{\infty}{\frac{\ln{(1+x^2)}}{1+x^2}}dx$

Wolfram doesnt have an answer and neither do I.

Paradoxica · Sep 27, 2015

Re: MX2 2015 Integration Marathon

This is one of those Wolfram cannot do, but we can. Try a substitution.
(Also, do you always jump straight to wolfram? Shame on you... tsk tsk tsk

)

leehuan · Sep 27, 2015

Re: MX2 2015 Integration Marathon

$\\ \theta =\arctan { x } \Rightarrow d\theta =\frac { dx }{ 1+{ x }^{ 2 } } \\ Let\quad I=\int _{ 0 }^{ \infty }{ \frac { \ln { \left( 1+{ x }^{ 2 } \right) } }{ 1+{ x }^{ 2 } } dx } \\ I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sec ^{ 2 }{ \theta } \right) } d\theta } \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sec ^{ 2 }{ \left( \frac { \pi }{ 2 } -\theta \right) } \right) } d\theta } \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \csc ^{ 2 }{ \theta } \right) } d\theta } \\$

My bank of integral tricks isn't really that large, but if I try adding those two integrals up I don't get anywhere either..

And wolfram does give an answer. pi log2.

Paradoxica · Sep 27, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\\ \theta =\arctan { x } \Rightarrow d\theta =\frac { dx }{ 1+{ x }^{ 2 } } \\ Let\quad I=\int _{ 0 }^{ \infty }{ \frac { \ln { \left( 1+{ x }^{ 2 } \right) } }{ 1+{ x }^{ 2 } } dx } \\ I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sec ^{ 2 }{ \theta } \right) } d\theta } \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sec ^{ 2 }{ \left( \frac { \pi }{ 2 } -\theta \right) } \right) } d\theta } \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \csc ^{ 2 }{ \theta } \right) } d\theta } \\$

My bank of integral tricks isn't really that large, but if I try adding those two integrals up I don't get anywhere either..

And wolfram does give an answer. pi log2.

Take out the negative. Look at your domain of integration. Take out the power. I have completed a similar integral before on this thread.

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