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

Sy123 · Jul 6, 2015

Re: MX2 2015 Integration Marathon

$\\ \int_{1/t}^t \frac{\tan^{-1}x}{x} \ dx \ $for$ \ t > 0$

Sy123 · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
integrand got the lower bound ages ago if you want to look for it by fowarding to when the question was posted

I think that's with seanieg's question

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ I=\int_{1/t}^t \frac{\tan^{-1}x}{x} \ dx \ $for$ \ t > 0$

$let x= \frac{1}{u}$

$I=\int_{\frac{1}{t}}^{t}{\frac{\tan^{-1}(\frac{1}{u})}{u}}$

By dummy variables:

$2I = \int{\frac{1}{x}(tan^{-1}x + tan^{-1}\frac{1}{x}})$

$I= {\frac{\pi}{4}}}\int_{\frac{1}{t}}^t{\frac{1}{x}}$

$I=\frac{\pi ln(t)}{2}$

[]

braintic · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Alright lets see if this works:

$$ Show that: $ \int_0^{\frac{\pi}{2}}{\frac{dx}{1+sinx \cdot cos\alpha}} =\frac{2}{sin\alpha}tan^{-1}{\frac{sin\alpha}{1+cos\alpha}$

I hope people considered two cases ...... because of course √(1-cos²α) does not equal sinα.

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

braintic said:
I hope people considered two cases ...... because of course √(1-cos²α) does not equal sinα.

why not
EDIT: shit...

braintic · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
why not
EDIT: shit...

Don't worry ... it still works for both cases. But not considering that is one mark off in an exam! ... unless there is another method which bypasses the issue.

Drsoccerball · Jul 6, 2015

Re: MX2 2015 Integration Marathon

braintic said:
Don't worry ... it still works for both cases. But not considering that is one mark off in an exam!

I made the answer based on the 1-c^2=s^2 fact so wouldnt that also make the answer wrong

braintic · Jul 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I made the answer based on the 1-c^2=s^2 fact so wouldnt that also make the answer wrong

Inverse tan and sine are both odd functions.

So (ignoring all the extraneous rubbish):

2/sin(-α) times arctan sin(-α) = ....... ?

glittergal96 · Jul 7, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
I wasn't able to think of a way to approximate the lower bound as $\sqrt{2\pi}$ (though technically if you allow for a strict lower bound like 2pi as an answer to the question, then a strict lower bound of 1 would be sufficient too and easy to prove, though such a bound would not be useful).

How did you get it?

You don't need to find a lower bound of sqrt(2*pi) though, I just asked for any positive constant in my original question.

So to summarise, if Q(n) is the quotient in the middle.

i) Show there are positive constants A,B such that A < Q(n) < B for all n. (My original question. You found a B, using the trapezoidal approximation, but can you find an A?)

ii) Show that Q(n) is monotone. (I just did this roughly, but I think it's not too hard.)

iii) Since the first two parts show that Q(n) -> c, you may assume this. Can you then find c? (This is distinct from i), you don't have to do it in an immediate sandwiching way, that sounds impossibly hard. Fortunately calculus gives us pretty powerful tools for computing things, especially when we don't have to worry about convergence!)

glittergal96 · Jul 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
integrand got the lower bound ages ago if you want to look for it by fowarding to when the question was posted

That was to seanieg89's question (I think an upper bound). And in the end, that exponent turned out to be the wrong one, so Integrand's bound wasn't very sharp. (Not that what he did was wrong.)

Drsoccerball · Jul 7, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
That was to seanieg89's question (I think an upper bound). And in the end, that exponent turned out to be the wrong one, so Integrand's bound wasn't very sharp . (Not that what he did was wrong.)

Pun intended ? sean said he was right tho ?

Drsoccerball · Jul 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Alright lets see if this works:

$$ Show that: $ \int_0^{\frac{\pi}{2}}{\frac{dx}{1+sinx \cdot cos\alpha}} =\frac{2}{sin\alpha}tan^{-1}{\frac{sin\alpha}{1+cos\alpha}$

Bump

glittergal96 · Jul 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Pun intended ? sean said he was right tho ?

What pun? lol.

As I said in brackets, the thing he wrote down wasn't untrue, it is just weaker than the result we can prove. Sean's original question was "wrong" in that the upper bound was easy to prove and the lower bound was hard to prove (impossible in fact, because it wasn't true lol). If you add 1 to both of the exponents the question makes sense though. (See my recent proof).

leehuan · Jul 7, 2015

Re: MX2 2015 Integration Marathon

Just gonna do one case. Assume sin(α)>0

$\\ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+\sin { x } \cos { \alpha } } } \\ Let\quad t=\tan { \frac { x }{ 2 } } etc.\\ \int _{ 0 }^{ 1 }{ \frac { \frac { 2dt }{ 1+{ t }^{ 2 } } }{ 1+\frac { 2t\cos { \alpha } }{ 1+{ t }^{ 2 } } } } \\ =\int _{ 0 }^{ 1 }{ \frac { 2 }{ 1+2t\cos { \alpha } +{ t }^{ 2 } } dt } \\ =2\int _{ 0 }^{ 1 }{ \frac { dt }{ 1-\cos ^{ 2 }{ \alpha } +{ \left( t+\cos { \alpha } \right) }^{ 2 } } }$

$\\ =2{ \left[ \frac { 1 }{ \sin { \alpha } } \arctan { \left( \frac { t+\cos { \alpha } }{ \sin { \alpha } } \right) } \right] }_{ 0 }^{ 1 }\\ =\frac { 2 }{ \sin { \alpha } } \left[ \arctan { \left( \frac { 1+\cos { \alpha } }{ \sin { \alpha } } \right) } -\arctan { \left( \frac { \cos { \alpha } }{ \sin { \alpha } } \right) } \right] \\ =\frac { 2 }{ \sin { \alpha } } \arctan { \left( \frac { \frac { 1+\cos { \alpha } }{ \sin { \alpha } } -\frac { \cos { \alpha } }{ \sin { \alpha } } }{ 1+\frac { 1+\cos { \alpha } }{ \sin { \alpha } } \frac { \cos { \alpha } }{ \sin { \alpha } } } \right) }$

$\\ =\frac { 2 }{ \sin { \alpha } } \arctan { \left( \frac { \sin { \alpha } }{ \sin ^{ 2 }{ \alpha } +\cos { \alpha } +\cos ^{ 2 }{ \alpha } } \right) } \\ =\frac { 2 }{ \sin { \alpha } } \arctan { \left( \frac { \sin { \alpha } }{ 1+\cos { \alpha } } \right) } \\ $The tangent inverse identity can be derived from the tangent compound-angle formula.$$

NEXT QUESTION:
I've seen this done already using indefinite integrals. But definite integrals seem to make things more fun and relaxing.

$\int _{ \frac { 1 }{ \sqrt { 3 } } }^{ \sqrt { 3 } }{ \frac { \ln { y } }{ 1+{ y }^{ 2 } } } dy \\ $And hence show that:$ \\ \int _{ 1 }^{ 3 }{ \frac { \ln { x } }{ 3+{ x }^{ 2 } } dx } =\frac { \pi \sqrt { 3 } \ln { 3 } }{ 36 }$

InteGrand · Jul 7, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
NEXT QUESTION:
I've seen this done already using indefinite integrals. But definite integrals seem to make things more fun and relaxing.

$\int _{ \frac { 1 }{ \sqrt { 3 } } }^{ \sqrt { 3 } }{ \frac { \ln { y } }{ 1+{ y }^{ 2 } } } dy \\ $And hence show that:$ \\ \int _{ 1 }^{ 3 }{ \frac { \ln { x } }{ 3+{ x }^{ 2 } } dx } =\frac { \pi \sqrt { 3 } \ln { 3 } }{ 36 }$

The latter integral was solved on the first page of this marathon (three different approaches are there).

Drsoccerball · Jul 7, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Just gonna do one case. Assume sin(α)>0

$\\ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+\sin { x } \cos { \alpha } } } \\ Let\quad t=\tan { \frac { x }{ 2 } } etc.\\ \int _{ 0 }^{ 1 }{ \frac { \frac { 2dt }{ 1+{ t }^{ 2 } } }{ 1+\frac { 2t\cos { \alpha } }{ 1+{ t }^{ 2 } } } } \\ =\int _{ 0 }^{ 1 }{ \frac { 2 }{ 1+2t\cos { \alpha } +{ t }^{ 2 } } dt } \\ =2\int _{ 0 }^{ 1 }{ \frac { dt }{ 1-\cos ^{ 2 }{ \alpha } +{ \left( t+\cos { \alpha } \right) }^{ 2 } } }$

$\\ =2{ \left[ \frac { 1 }{ \sin { \alpha } } \arctan { \left( \frac { t+\cos { \alpha } }{ \sin { \alpha } } \right) } \right] }_{ 0 }^{ 1 }\\ =\frac { 2 }{ \sin { \alpha } } \left[ \arctan { \left( \frac { 1+\cos { \alpha } }{ \sin { \alpha } } \right) } -\arctan { \left( \frac { \cos { \alpha } }{ \sin { \alpha } } \right) } \right] \\ =\frac { 2 }{ \sin { \alpha } } \arctan { \left( \frac { \frac { 1+\cos { \alpha } }{ \sin { \alpha } } -\frac { \cos { \alpha } }{ \sin { \alpha } } }{ 1+\frac { 1+\cos { \alpha } }{ \sin { \alpha } } \frac { \cos { \alpha } }{ \sin { \alpha } } } \right) }$

$\\ =\frac { 2 }{ \sin { \alpha } } \arctan { \left( \frac { \sin { \alpha } }{ \sin ^{ 2 }{ \alpha } +\cos { \alpha } +\cos ^{ 2 }{ \alpha } } \right) } \\ =\frac { 2 }{ \sin { \alpha } } \arctan { \left( \frac { \sin { \alpha } }{ 1+\cos { \alpha } } \right) } \\ $The tangent inverse identity can be derived from the tangent compound-angle formula.$$

NEXT QUESTION:
I've seen this done already using indefinite integrals. But definite integrals seem to make things more fun and relaxing.

$\int _{ \frac { 1 }{ \sqrt { 3 } } }^{ \sqrt { 3 } }{ \frac { \ln { y } }{ 1+{ y }^{ 2 } } } dy \\ $And hence show that:$ \\ \int _{ 1 }^{ 3 }{ \frac { \ln { x } }{ 3+{ x }^{ 2 } } dx } =\frac { \pi \sqrt { 3 } \ln { 3 } }{ 36 }$

i) $let y= \frac{1}{u}$

$\therefore I=\int_{{\frac{1}{\sqrt{3}}}^{\sqrt{3}}){{\frac{ln(\frac{1}{u})}{u^2+1}}$

$2I= \int{0}$

$=0$

ii) $$ Basically the same thing except let $ x=\sqrt{3}u$
EDIT: notsure how to fix this latex

braintic · Jul 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
notsure how to fix this latex

Famous last words before a Pom gets his girlfriend pregnant.

leehuan · Jul 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
i) $let y= \frac{1}{u}$

$\therefore I=\int_{{\frac{1}{\sqrt{3}}}^{\sqrt{3}}){{\frac{ln(\frac{1}{u})}{u^2+1}}$

$2I= \int{0}$

$=0$

ii) $$ Basically the same thing except let $ x=\sqrt{3}u$
EDIT: notsure how to fix this latex

Oh dear. But I'll take the 2I = \Int 0 part and say yeah you got it.

Whilst I'm here. Mega oh dear at the other comment.
EDIT: Yeah I figured you saw that.

Drsoccerball · Jul 7, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Oh dear. But I'll take the 2I = \Int 0 part and say yeah you got it.

Whilst I'm here. Mega oh dear at the other comment.

Im using dummy variables i need to add the first I and the last I to let it =0 otherwise you lose a mark for fudging...

glittergal96 · Jul 8, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
You don't need to find a lower bound of sqrt(2*pi) though, I just asked for any positive constant in my original question.

So to summarise, if Q(n) is the quotient in the middle.

i) Show there are positive constants A,B such that A < Q(n) < B for all n. (My original question. You found a B, using the trapezoidal approximation, but can you find an A?)

ii) Show that Q(n) is monotone. (I just did this roughly, but I think it's not too hard.)

iii) Since the first two parts show that Q(n) -> c, you may assume this. Can you then find c? (This is distinct from i), you don't have to do it in an immediate sandwiching way, that sounds impossibly hard. Fortunately calculus gives us pretty powerful tools for computing things, especially when we don't have to worry about convergence!)

So did you end up having a proof that the Q(n) is bounded below by 1 Sy?

My upper bound (for the log sum) came from the trapezoidal approximation (and the geometric observation that this approximation underestimates the integral of a concave function.)

My lower bound I did in a couple of ways.

i) We can in fact overestimate the integral of a concave function by integrating its tangent curve. A convenient (and in fact optimal, which you can test with differential calculus) way of bounding $\int_a^{a+1}\log(t)\, dt$ is by using the tangent at $a+1/2$ . If we then perform the integration of this tangent line (ie calculate the area of a trapezium lol), we get log(a+1/2). This is a tight enough upper bound to be able to sum such approximations and get an appropriate lower bound for the logsum.

ii) We can study the trapezoidal approximation in a little more detail (wish we learned this in high school). Try to find a bound for the error of the trapezoidal approximation on $[a,a+1]$ in terms of $M=\max_{x\in[a,a+1]}|f''(x)|.$ Then this means the trapezoidal bound is good enough, as we can overestimate the logintegral by (its trapezoidal sum) + (a partial sum of the basel series). Knowing that the latter is finite is good enough to get a lower bound for the logsum.

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