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

porcupinetree · Jun 29, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
NEXT QUESTION:
$\int { { x }^{ 8 }\sin { \left( { x }^{ 3 } \right) } \cos { \left( { x }^{ 3 } \right) } \tan { \left( { x }^{ 3 } \right) } dx }$

https://imgur.com/BnJZmcb

porcupinetree · Jun 29, 2015

Re: MX2 2015 Integration Marathon

Lets get some recurrence formulae up in here.

Next question:

If In = integral between 0 and 1 of (x^n)sqrt(1-x)dx show that In = 4^(n+1) . n!(n+1)!/(2n+3)!

Sorry for my lack of LaTeX

leehuan · Jun 29, 2015

Re: MX2 2015 Integration Marathon

Well that was exhaustive! Anyway, I usually use this to do my LaTeX: Daum Equation Editor

$\\ { I }_{ n }=\int _{ 0 }^{ 1 }{ { x }^{ n }\sqrt { 1-x } dx } \\ ={ \left[ -\frac { 2{ x }^{ n }{ \left( 1-x \right) }^{ \frac { 3 }{ 2 } } }{ 3 } \right] }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ -\frac { 2n{ x }^{ n-1 }{ \left( 1-x \right) }^{ \frac { 3 }{ 2 } } }{ 3 } dx } \\ =\left( -\frac { 2{ \left( 1 \right) }^{ n }{ \left( 1-1 \right) }^{ \frac { 3 }{ 2 } } }{ 3 } +\frac { 2{ \left( 0 \right) }^{ n }{ \left( 1-0 \right) }^{ \frac { 3 }{ 2 } } }{ 3 } \right) +\frac { 2n }{ 3 } \int _{ 0 }^{ 1 }{ { x }^{ n-1 }\left( 1-x \right) \sqrt { 1-x } dx } \\ =\frac { 2n }{ 3 } \int _{ 0 }^{ 1 }{ { x }^{ n-1 }\sqrt { 1-x } dx } -\frac { 2n }{ 3 } \int _{ 0 }^{ 1 }{ { x }^{ n }\sqrt { 1-x } dx }$

$\\ { I }_{ n }=\frac { 2n }{ 3 } { I }_{ n-1 }-\frac { 2n }{ 3 } { I }_{ n }\\ \frac { 2n+3 }{ 3 } { I }_{ n }=\frac { 2n }{ 3 } { I }_{ n-1 }\\ { I }_{ n }=\frac { 2n }{ 2n+3 } { I }_{ n-1 }\\ \therefore { I }_{ n }=\frac { 2n }{ 2n+3 } \left( \frac { 2n-2 }{ 2n+1 } { I }_{ n-2 } \right) \\ { I }_{ n }=\frac { 2n\left( 2n-2 \right) }{ \left( 2n+3 \right) \left( 2n+1 \right) } { I }_{ n-2 }\\ { I }_{ n }=\frac { 2n\left( 2n-2 \right) \left( 2n-4 \right) }{ \left( 2n+3 \right) \left( 2n+1 \right) \left( 2n-1 \right) } { I }_{ n-3 }\\ \vdots$

$\\ { I }_{ n }=\frac { 2n\left( 2n-2 \right) \left( 2n-4 \right) \dots 6\cdot 4\cdot 2 }{ \left( 2n+3 \right) \left( 2n+1 \right) \left( 2n-1 \right) \dots 9\cdot 7\cdot 5 } { I }_{ 0 }\\ ={ 2 }^{ n }\frac { n! }{ \left( 2n+3 \right) \left( 2n+1 \right) \left( 2n-1 \right) \dots 9\cdot 7\cdot 5 } \cdot \frac { \left( 2n+2 \right) \left( 2n \right) \left( 2n-2 \right) \dots 6\cdot 4\cdot 2 }{ \left( 2n+2 \right) \left( 2n \right) \left( 2n-2 \right) \dots 6\cdot 4\cdot 2 } \int _{ 0 }^{ 1 }{ \sqrt { 1-x } dx } \\ ={ 2 }^{ n }\cdot { 2 }^{ n+1 }\frac { 3\cdot n!\left( n+1 \right) ! }{ \left( 2n+3 \right) \left( 2n+2 \right) \left( 2n+1 \right) \dots 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 } { \left[ -\frac { 2{ \left( 1-x \right) }^{ \frac { 3 }{ 2 } } }{ 3 } \right] }_{ 0 }^{ 1 }$

$\\ ={ 2 }^{ 2n+1 }\cdot \frac { n!\left( n+1 \right) ! }{ \left( 2n+3 \right) ! } \left( -2{ \left( 1-1 \right) }^{ \frac { 3 }{ 2 } }+2{ \left( 1-0 \right) }^{ \frac { 3 }{ 2 } } \right) \\ ={ 2 }^{ 2n+1 }\cdot \frac { n!\left( n+1 \right) ! }{ \left( 2n+3 \right) ! } \left( 2 \right) \\ ={ 2 }^{ 2\left( n+1 \right) }\cdot \frac { n!\left( n+1 \right) ! }{ \left( 2n+3 \right) ! } \\ ={ 4 }^{ n+1 }\cdot \frac { n!\left( n+1 \right) ! }{ \left( 2n+3 \right) ! }$

Sy123 · Jun 29, 2015

Re: MX2 2015 Integration Marathon

$\\ \int x^2(\sin \ln x + \cos \ln x) \ dx$

kawaiipotato · Jun 29, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ \int x^2(\sin \ln x + \cos \ln x) \ dx$

$let u = sin(lnx) + cos(lnx) dv = x^{2} dx$

$du = \frac{dx}{x}(cos(lnx) - sin(lnx)) v = \frac{x^3}{3}$

$$ I = $ \frac{x^3}{3}(sin(lnx) + cos(lnx)) + \frac{1}{3} \int x^2(sin(lnx) - cos(lnx)) dx$

$Let new integral = J$

$J: let u = sin(lnx) - cos(lnx) dv = x^2 dx$

$du = \frac{dx}{x}(cos(lnx) + sin(lnx)) v = \frac{x^3}{3}$

$J = \frac{x^3}{3}(sin(lnx) - cos(lnx)) - \frac{I}{3}$

$$ I = $ \frac{x^3}{3}(sin(lnx) + cos(lnx)) + \frac{1}{3} [\frac{x^3}{3}(sin(lnx) - cos(lnx)) - \frac{I}{3}$

$I = \frac{3}{4} [\frac{x^3}{3}(sin(lnx) + cos(lnx)) + \frac{1}{3} (\frac{x^3}{3}(sin(lnx) - cos(lnx))]$

porcupinetree · Jun 29, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Well that was exhaustive! Anyway, I usually use this to do my LaTeX: Daum Equation Editor

Thx for the link. Just had a go at redoing my solution earlier on it:
$\\ \int { { x }^{ 8 }\sin { { x }^{ 3 } } \cos { { x }^{ 3 }\tan { { x }^{ 3 } } dx } } \\ let\quad u\quad =\quad { x }^{ 3 }\\ \Rightarrow \quad du\quad =\quad 3{ x }^{ 2 }dx\\ \therefore \quad we\quad have:\\ \int { { u }^{ 2 } } { x }^{ 2 }\sin { u } \cos { u } \tan { u } \frac { du }{ 3{ x }^{ 2 } }$

$\\ =\quad \frac { 1 }{ 3 } \int { { u }^{ 2 }\sin { u } \cos { u } \frac { \sin { u } }{ \cos { u } } du } \\ =\quad \frac { 1 }{ 3 } \int { { u }^{ 2 } } \sin ^{ 2 }{ u } du\\ =\quad \frac { 1 }{ 6 } \int { { u }^{ 2 } } (1-\cos { 2u } )du\\ =\frac { 1 }{ 6 } \int { { u }^{ 2 } } du\quad -\quad \frac { 1 }{ 6 } \int { { u }^{ 2 }\cos { 2u } du } \\ Perform\quad IBP$

$\\ =\quad \frac { { u }^{ 3 } }{ 18 } -\frac { 1 }{ 6 } \left\{ { u }^{ 2 }\frac { 1 }{ 2 } \sin { 2u } \quad -\quad \int { 2u\frac { 1 }{ 2 } \sin { 2u } du } \right\} \\ And\quad again...\\ =\quad \frac { { u }^{ 3 } }{ 18 } -\frac { { u }^{ 2 } }{ 12 } \sin { 2u } +\quad \frac { 1 }{ 6 } \left\{ u\frac { -1 }{ 2 } \cos { 2u } \quad -\quad \int { \frac { -1 }{ 2 } \cos { 2u } du } \right\} \\ =\quad \frac { { x }^{ 9 } }{ 18 } \quad -\quad \frac { { x }^{ 6 } }{ 12 } \sin { 2{ x }^{ 3 } } \quad -\quad \frac { { x }^{ 3 } }{ 12 } \cos { 2{ x }^{ 3 } } \quad +\quad \frac { 1 }{ 24 } \sin { 2{ x }^{ 3 } } \quad +\quad C$

I think it's safe to say I won't be using it in the future, too fiddly for someone as lazy as me

clbaker · Jul 1, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
= sqrt(sin^2 x + cos^2 x + 2sinxcosx) dx

= sqrt((sinx + cosx)^2) dx
= |sinx+cosx|

And so you'll now continue by actually doing an integration???

glittergal96 · Jul 1, 2015

Re: MX2 2015 Integration Marathon

This question is based on your ability to approximate sums by integrals, hence its location in this marathon thread.

Prove that there exist positive constants $C_1,C_2$ such that

$C_1\leq\frac{n!}{n^{n+1/2}e^{-n}}\leq C_2$

for all positive integers $n.$

(In fact, the ratio in this question tends to an exact constant, but proving this convergence without guidance is perhaps a bit much to ask. It is a reasonable enough followup exercise though to calculate this exact constant, given that the ratio does in fact converge.)

Drsoccerball · Jul 1, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
This question is based on your ability to approximate integrals, hence its location in this marathon thread.

Prove that there exist positive constants $C_1,C_2$ such that

$C_1\leq\frac{n!}{n^{n+1/2}e^{-n}}\leq C_2$

for all positive integers $n.$

(In fact, the ratio in this question tends to an exact constant, but proving this convergence without guidance is perhaps a bit much to ask. It is a reasonable enough followup exercise though to calculate this exact constant, given that the ratio does in fact converge.)

A similar question was asked by sean still unsolved

glittergal96 · Jul 1, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
A similar question was asked by sean still unsolved

Bump it so I can have a crack

.

glittergal96 · Jul 1, 2015

Re: MX2 2015 Integration Marathon

seanieg89 said:
Here is a question that concerns inequalities arising from integration. (It is harder conceptually than most questions in this thread, but easier than a lot of them in terms of how technically demanding the required manipulations are.)

$Let $\alpha>0$ be a positive parameter.\\ Show that there exist positive constants $C_1,C_2$ such that: \\ \\ $C_1||z|-1|^{-\alpha}<\int_0^{2\pi} |z-\textrm{cis}(\theta)|^{-\alpha} \, d\theta < C_2||z|-1|^{-\alpha}$ for $|z|\neq 1$.$

(This integral clearly blows up as $z$ approaches the unit circle. The point of the question is to quantify how quickly this happens, which is generally a useful thing to know.)

Probably this one. Will try later today.

glittergal96 · Jul 1, 2015

Re: MX2 2015 Integration Marathon

That lower bound seems weird, I think the exponents for both bound should be larger...perhaps $1-\alpha$ ? The part of the integration that is blowing up as $d^{-\alpha}$ is shrinking in size...

Anyway, will properly look at it when I get home later.

InteGrand · Jul 1, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
That lower bound seems weird, I think the exponents for both bound should be larger...perhaps $1-\alpha$ ? The part of the integration that is blowing up as $d^{-\alpha}$ is shrinking in size...

Anyway, will properly look at it when I get home later.

Well the upper bound has $C_2=2\pi$ sufficient (upper bound has already been established in an earlier post).

Also, can the constants depend on $\alpha$ ?

glittergal96 · Jul 1, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Well the upper bound has $C_2=2\pi$ sufficient (upper bound has already been established in an earlier post).

Also, can the constants depend on $\alpha$ ?

Sure, but these inequalities are only "good" if the growth rates of the upper and lower bound match...my suspicion is that $d^{-\alpha}$ blows up faster than this integral does. (So we can still find an upper bound of sean's form as you say, but I don't think we can find such a lower bound...so maybe the correct exponent is greater, which sharpens the upper bound we have to prove.)

Edit: Yeah, it looks like the alpha is fixed throughout the question, so everything can depend on it.

Sy123 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
This question is based on your ability to approximate sums by integrals, hence its location in this marathon thread.

Prove that there exist positive constants $C_1,C_2$ such that

$C_1\leq\frac{n!}{n^{n+1/2}e^{-n}}\leq C_2$

for all positive integers $n.$

(In fact, the ratio in this question tends to an exact constant, but proving this convergence without guidance is perhaps a bit much to ask. It is a reasonable enough followup exercise though to calculate this exact constant, given that the ratio does in fact converge.)

$C_2 = e$ by approximating $\int_1^n \ln x$ with n trapeziums each of width 1, I'll post a proper proof later and for the lower bound

Sy123 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

$\int_0^1 \frac{\ln(x+1)}{\ln(2+x-x^2)} \ dx$

Drsoccerball · Jul 4, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\int_0^1 \frac{\ln(x+1)}{\ln(2+x-x^2)} \ dx$

$I= \int_0^1{\frac{ln(x+1)}{ln(2+x-x^2)}}= \int_0^1{\frac{ln(2-x)}{ln(2+x-x^2)}}$

$2I= \int_0^1 1$

$I= \frac{1}{2}$

Sy123 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

$\\ \ $Find in terms of$ \ n \ \int_0^{\pi} \frac{x\sin x}{(n-1)(n+1) + \sin^2x} \ dx$

VBN2470 · Jul 4, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ \ $Find in terms of$ \ n \ \int_0^{\pi} \frac{x\sin x}{(n-1)(n+1) + \sin^2x} \ dx$

$$\frac{\pi}{2n}\ln{\frac{n-1}{n+1}}$$ ?

Drsoccerball · Jul 4, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
$$\frac{\pi}{2n}\ln{\frac{n-1}{n+1}}$$ ?

i got $\frac{\pi}{2n}ln|\frac{n+1}{n-1}|$

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