HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

InteGrand · Jun 8, 2015

Re: MX2 2015 Integration Marathon

There are interesting extensions to this: http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_π#Proof_that_355.2F113_exceeds_.CF.80

leehuan · Jun 8, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
There are interesting extensions to this: http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_π#Proof_that_355.2F113_exceeds_.CF.80

Basically what was used. Lol.

leehuan · Jun 8, 2015

Re: MX2 2015 Integration Marathon

$\int _{ 0 }^{ 1 }{ \frac { \arctan { x } }{ x+1 } } dx$

Start with an IBP

Ekman · Jun 8, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int _{ 0 }^{ 1 }{ \frac { \arctan { x } }{ x+1 } } dx$

Start with an IBP

I believe this question has been asked before:

$\int_{0}^{1} \frac{\tan^{-1}{x}}{1+x} dx = \left[\tan^{-1}{x} \ln{|1+x|}\right]_{0}^{1} - \int_{0}^{1} \frac{\ln{|1+x|}}{1+x^2} dx$

$$ Let $ x=\tan{\theta} \therefore \int_{0}^{1} \frac{\ln{|1+x|}}{1+x^2} dx = \int_{0}^{\frac{\pi}{4}} \ln{|1+\tan{\theta}|}d\theta$

$I=\int_{0}^{\frac{\pi}{4}} \ln{|1+\tan{\theta}|}d\theta =\int_{0}^{\frac{\pi}{4}} \ln{2} - \ln{|1+\tan{\theta}|}d\theta$

$2I = \int_{0}^{\frac{\pi}{4}} \ln{2} d\theta = \frac{\pi\ln{2}}{4}$

$\therefore \int_{0}^{1} \frac{\tan^{-1}{x}}{1+x} dx = \frac{\pi\ln{2}}{4} - \frac{\pi\ln{2}}{8} = \frac{\pi\ln{2}}{8}$

leehuan · Jun 8, 2015

Re: MX2 2015 Integration Marathon

Probably was before I had really started being active on this thread.

$\int { \frac { { x }^{ 3 }+1 }{ { x }^{ 8 }+1 } } dx$

Bored now. Someone make the questions for me please...

InteGrand · Jun 8, 2015

Re: MX2 2015 Integration Marathon

Show that $\int _0 ^{\infty} t^{n}\mathrm{e}^{-t}\text{ d}t=n!$ for $n=0,1,2,...$ .

leehuan · Jun 8, 2015

Re: MX2 2015 Integration Marathon

$Let\quad { I }_{ n }=\int _{ 0 }^{ \infty }{ { t }^{ n }{ e }^{ -t }dt } \\ =\lim _{ k\rightarrow \infty }{ { \left[ -{ t }^{ n }{ e }^{ -t } \right] }_{ 0 }^{ k } } -\int _{ 0 }^{ \infty }{ -{ nt }^{ n-1 }{ e }^{ -t }dt } \\ =\left( -0--0 \right) +n{ I }_{ n-1 }\\ So\quad { I }_{ n }=n{ I }_{ n-1 }\\ So\quad { I }_{ n }=n\left( n-1 \right) { I }_{ n-2 }\\ So\quad { I }_{ n }=n\left( n-1 \right) \left( n-2 \right) { I }_{ n-3 }$
$\vdots \\ { I }_{ n }=n\left( n-1 \right) \left( n-2 \right) \left( n-3 \right) \cdots \left( 3 \right) \left( 2 \right) \left( 1 \right) { I }_{ 0 }\\ =n!\int _{ 0 }^{ \infty }{ { e }^{ -t }dt } \\ =n!\lim _{ k\rightarrow \infty }{ { \left[ -{ e }^{ -t } \right] }_{ 0 }^{ k } } \\ =n!\left( -0--1 \right) \\ =n!$

for n>=0

dunjaaa · Jun 9, 2015

Re: MX2 2015 Integration Marathon

dunjaaa · Jun 9, 2015

Re: MX2 2015 Integration Marathon

FrankXie · Jun 9, 2015

Re: MX2 2015 Integration Marathon

dunjaaa said:
View attachment 32112

nice questions, I'll post my solution for question one.

$$ Denote $ I=\int_0^\infty\frac{\ln(1+x)}{(1+x)\sqrt{x}}dx. $ By the substitution $ x=\tan^2\theta, I=-4\int_0^{\frac{\pi}{2}}\ln\cos\theta d\theta$

$$ Now let $ J=\int_0^{\frac{\pi}{2}}\ln\cos\theta d\theta=\int_0^{\frac{\pi}{2}}\ln\sin\theta d\theta, $ and let $ K=\int_0^{\frac{\pi}{2}}\ln\sin2\theta d\theta. $ By the substitution $ u=2\theta, K=\frac{1}{2}\int_0^{\pi}\ln\sin u du=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\sin u du+\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}\ln\sin u du=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\sin u du+\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\sin t dt ( $ applying the substitution $ u=\pi-t $ for the second integral)$

$$ Therefor $ K=J. $ and $ 2J=\int_0^{\frac{\pi}{2}}\ln\cos\theta d\theta+\int_0^{\frac{\pi}{2}}\ln\sin\theta d\theta=\int_0^{\frac{\pi}{2}}\ln\frac{\sin2\theta}{2} d\theta=K-\frac{\pi}{2}\ln2. $ So $ J=-\frac{\pi}{2}\ln2, $ and hence $ I=2\pi\ln2$

Ekman · Jun 9, 2015

Re: MX2 2015 Integration Marathon

dunjaaa said:
View attachment 32113

Q1.
$$ Let $ x=\sin{\theta}$

$\therefore \int_{0}^{1} \ln{(1+\sqrt{1-x^2})}dx = \int_{0}^{\frac{\pi}{2}} \cos{\theta}\ln{(1+\cos{\theta})} d\theta$

$$ Via IBP: $ \int_{0}^{\frac{\pi}{2}} \cos{\theta}\ln{(1+\cos{\theta})} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2}{\theta}}{1+\cos{\theta}} d\theta$

$\int_{0}^{\frac{\pi}{2}} 1-\cos{\theta} d\theta$

$= \frac{\pi}{2} -1$

kawaiipotato · Jun 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Question 1 answers 0 let u =ln(1+x)
Question 2 answers 0 since its an odd function

q2 is an even function

Ekman · Jun 9, 2015

Re: MX2 2015 Integration Marathon

dunjaaa said:
View attachment 32112

Q2.

$$ Let: $ u^2=1+x$

$\therefore \int_{-1}^{1} \frac{1}{\sqrt{x+1} +\sqrt{1-x} +2} dx = \int_{0}^{\sqrt{2}} \frac{2udu}{u+2 + \sqrt{2-u^2}}$

$\int_{0}^{\sqrt{2}} \frac{2u(u+2 - \sqrt{2+u^2})}{(u+2)^2-2+u^2} du = \int_{0}^{\sqrt{2}} \frac{u(u+2 - \sqrt{2+u^2})}{(u+1)^2}$

$I= \int_{0}^{\sqrt{2}} 1 - \frac{1}{(1+u)^2} - \frac{u\sqrt{2-u^2}}{(u+1)^2} du$

$\int_{0}^{\sqrt{2}} 1 - \frac{1}{(1+u)^2} du = 2\sqrt{2} - 2$

$\int_{0}^{\sqrt{2}} \frac{u\sqrt{2-u^2}}{(u+1)^2} du = \int_{0}^{\sqrt{2}} \frac{\sqrt{2-u^2}}{u+1} - \frac{u^2}{(u+1)\sqrt{2-u^2}} du $ (Via IBP)$$

$= 2\int_{0}^{\sqrt{2}} \frac{1-u}{\sqrt{2-u^2}}du = \pi-2\sqrt{2}$

$\therefore I = 4\sqrt{2} - 2 -\pi$

InteGrand · Jun 9, 2015

Re: MX2 2015 Integration Marathon

dunjaaa said:
View attachment 32113

For Q2, $\mathcal{I}=\int \frac{\text{d}x}{(1-x)\sqrt{1+x^2}}$ .

Let $x=\tan \theta, \text{d}x=\sec^2 \theta \text{ d}\theta, \sqrt{1+x^2}=\sec \theta$ , then

$\mathcal{I}=\int \frac{\sec^2 \theta \text{ d}\theta}{(1-\tan \theta)\sec \theta} = \int \frac{\sec \theta}{1-\tan \theta}\text{ d}\theta = \int \frac{\text{d}\theta}{\cos \theta - \sin \theta}$ .

Now either use t-formulae or auxiliary angle method, and replace $\theta$ 's appropriately with functions of x.

FrankXie · Jun 9, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Q2.

$\therefore \int_{-1}^{1} \frac{1}{\sqrt{x+1} +\sqrt{1-x} +2} dx = \int_{0}^{\sqrt{2}} \frac{2udu}{u+2 + \sqrt{\textbf{(2-u)(2+u)}}}$

the algebra is wrong here

I used the substitution $x=\sin2\theta$ and got an answer of $4\sqrt{2}-\pi-2$

InteGrand · Jun 10, 2015

Re: MX2 2015 Integration Marathon

dunjaaa said:
View attachment 32113

For Q4, one way may be to substitute $x=\tan ^3 \theta$ , and note that then $1+x^2=1^3 + (\tan^2 \theta)^3 = (1+\tan^2 \theta)(1 - \tan \theta + \tan^2 \theta)=\sec^2 \theta (1-\tan \theta + \tan^2 \theta)$ . Implementing these substitutions and making some cancellations will give $I=\int _0 ^{\frac{\pi}{2}}\frac{3\tan ^3 \theta}{\tan ^2 \theta - \tan \theta + 1}\text{ d}\theta$ . To evaluate this, we may write $\frac{3\tan ^3 \theta}{\tan ^2 \theta - \tan \theta + 1} = \frac{3\tan \theta (\sec^2 \theta - 1)}{\tan ^2 \theta - \tan \theta + 1}= \frac{3\tan \theta}{\tan ^2 \theta - \tan \theta + 1}\sec^2 \theta - \frac{3\tan \theta }{\tan ^2 \theta - \tan \theta + 1}=\frac{3\tan \theta}{\tan ^2 \theta - \tan \theta + 1}\sec^2 \theta - \frac{3\tan \theta }{\tan ^2 \theta - \tan \theta + 1}\frac{\sec^2 \theta}{\tan^2 \theta + 1}=\frac{3\tan \theta}{\tan ^2 \theta - \tan \theta + 1}\sec^2 \theta - \frac{3\tan \theta }{(\tan ^2 \theta - \tan \theta + 1)(\tan^2 \theta + 1)}\cdot \sec^2 \theta$ . (The purpose of this is to get $\sec^2 \theta$ involved so we can integrate things with tan.) For the $\frac{3\tan \theta }{(\tan ^2 \theta - \tan \theta + 1)(\tan^2 \theta + 1)}$ , we can use partial fractions. However, this method may be quite tedious and they may be faster methods.

FrankXie · Jun 10, 2015

Re: MX2 2015 Integration Marathon

there is also algebra mistake: $1+x^6=(1+x^2)(1+x^4-x^2)$ not $1+x^6=(1+x^2)(1+x^2-x)$

so I would go
$I=3\int_0^\infty\frac{x^3}{1+x^6}dx=\frac{3}{2} \int_0^\infty\frac{x^2dx^2}{(1+x^2)(1+x^4-x^2)}=\frac{3}{2}\int_0^\infty\frac{udu}{(1+u)(1+u^2-u)}$
and partial fractions from there
$=\frac{1}{2}\int_0^\infty\left(\frac{u+1}{1+u^2-u}-\frac{1}{1+u}\right)du=\frac{1}{2}\left[\ln\frac{\sqrt{1+u^2-u}}{1+u}+\sqrt{3}\tan^{-1}\frac{2u-1}{\sqrt{3}}\right]_0^\infty=\frac{\sqrt{3}\pi}{3}$

Ekman · Jun 10, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
the algebra is wrong here

I used the substitution $x=\sin2\theta$ and got an answer of $4\sqrt{2}-\pi-2$

Whoops, fixed it

porcupinetree · Jun 10, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
$\frac{1}{2}\left[\ln\frac{\sqrt{1+u^2-u}}{1+u}+\sqrt{3}\tan^{-1}\frac{2u-1}{\sqrt{3}}\right]_0^\infty$

Can someone give me an overview of how I should approach 'subbing in infinity' like you do while evaluating this integral? e.g. what to do when there's fractions involved, etc

InteGrand · Jun 10, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Can someone give me an overview of how I should approach 'subbing in infinity' like you do while evaluating this integral? e.g. what to do when there's fractions involved, etc

The rule is to take limits as $u\to\infty$ . The logarithm's argument tends to 1 as $u\to\infty$ , so by continuity of ln, the log tends to ln(1) = 0. Also, the inverse tangent will tend to $\frac{\pi}{2}$ because $\lim _{X\to \infty}\tan^{-1} X = \frac{\pi}{2}$ . And we use that the limit of a sum is the sum of the limits for continuous functions.

But improper integrals like this wouldn't come up in the HSC probably, and if they did, they'd tell you the rule (i.e. you need to take limits), and they'd probably get you to show what the limits are first.

HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Active Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

not actually a porcupine

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 5)