HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

U MAD BRO · Aug 12, 2012

Re: MX2 Integration Marathon

nightweaver066 said:
He added the definite integrals of $xcot(x)$ and $(\pi/2 -x)tan(x)$ together.

As they're both the same, it becomes 2I = etc.

He then put them both under the same denominator and multiplied numerator and denominator by 2.

thanks, that makes sense.

D94 · Aug 12, 2012

Re: MX2 Integration Marathon

Shadowless said:
could someone explain to me what happened between the 1st line and the 2nd line?

Sean added the two equivalent integrands together. You should notice this by the 2I. If you add the first integrand to the second, and through a bit of manipulation (i.e. multiply top and bottom by 2cos²x) then you will get to the result in the second line.

EDIT: eh, too slow.

Shadowless · Aug 12, 2012

Re: MX2 Integration Marathon

nightweaver066 said:
he added the definite integrals of $xcot(x)$ and $(\pi/2 -x)tan(x)$ together.

As they're both the same, it becomes 2i = etc.

He then put them both under the same denominator and multiplied numerator and denominator by 2.

thanks

Shadowless · Aug 12, 2012

Re: MX2 Integration Marathon

seanieg89 said:
U MAD BRO keep your integrals at the level of the MX2 integration topic. If you are interested in how to do these yourself, do outside research / ask in extracurricular. This thread is an inappropriate place to post them.

@Rolpsy nice question!

$I=\int_0^{\pi/2}x\cot(x)\,dx=\int_0^{\pi/2}(\pi/2-x)\tan(x)\,dx\\ \Rightarrow 2I=\int_0^{\pi/2}\frac{2x\cos^2(x)+2(\pi/2-x)\sin^2(x)}{2\sin(x)\cos(x)}\, dx\\=\int_0^{\pi/2}\frac{2x\cos(2x)+\frac{\pi}{2}(1-\cos(2x))}{\sin(2x)}\,dx\\\Rightarrow 4I=\int_0^{\pi}\frac{x\cos(x)+\frac{\pi}{2}(1-\cos(x))}{\sin(x)}\, dx\\=\int_0^{\pi/2}x\cot(x)\,dx+\frac{\pi}{2}\int_0^{\pi/2}\csc(x)-\cot(x)\,dx+\int_{\pi/2}^\pi (x-\pi)\cot(x)\,dx+\frac{\pi}{2}\int_{\pi/2}^\pi\csc(x)+\cot(x)\,dx\\=2I+\pi\int_0^{\pi/2}\csc(x)-\cot(x)\,dx\\=2I-\pi\log|1+\cos(x)|]^{\pi/2}_0\\\Rightarrow I=\frac{\pi}{2}\log(2).$

from the 3rd line to the 4th line... what identity is that? (limits changed?)

nightweaver066 · Aug 12, 2012

Re: MX2 Integration Marathon

Shadowless said:
from the 3rd line to the 4th line... what identity is that? (limits changed?)

I don't know how to explain it otherwise, but if you use the substitution u = 2x, you get that next line.

D94 · Aug 12, 2012

Re: MX2 Integration Marathon

It's a u-substitution concept, so he let u = 2x, but 'u' is as valid as using 'x' or 'y' or 'z', so he really let x = 2x.

Shadowless · Aug 12, 2012

Re: MX2 Integration Marathon

nightweaver066 said:
I don't know how to explain it otherwise, but if you use the substitution u = 2x, you get that next line.

okay i have the 4th line... BUT HOW DO WE GET THE 5TH ONE?... -_- why is this integral so difficult

(BTW: is this MX2 LEVEL? because I have never really encountered any past HSC questions like this...?)

rolpsy · Aug 12, 2012

Re: MX2 Integration Marathon

Shadowless said:
okay i have the 4th line... BUT HOW DO WE GET THE 5TH ONE?... -_- why is this integral so difficult

(BTW: is this MX2 LEVEL? because I have never really encountered any past HSC questions like this...?)

5th line should be (note limits of last two integrals)

$\\\int^{\frac{\pi}{2}}_0 x\cot(x) \, \mathrm{d}x + \frac{\pi}{2}\int^{\frac{\pi}{2}}_0 \csc(x) - \cot(x) \, \mathrm{d}x + \int^{\frac{\pi}{2}}_0 (x-\pi)\cot(x) \, \mathrm{d}x\\+ \frac{\pi}{2}\int^{\frac{\pi}{2}}_0 \csc(x) + \cot(x) \, \mathrm{d}x$

to get there, split up the integral and rewrite the ones between pi/2 and pi with a substitution of u = pi - x
e.g.,

$\begin{align*}\int^{\pi}_{0} x\cot(x) \, \mathrm{d}x &= \int^{\frac{\pi}{2}}_{0} x\cot(x) \, \mathrm{d}x + \int^{\pi}_{\frac{\pi}{2}} x\cot(x) \, \mathrm{d}x\\&=\int^{\frac{\pi}{2}}_{0} x\cot(x) \, \mathrm{d}x - \int^{0}_{\frac{\pi}{2}} (\pi - x)\cot(\pi - x) \, \mathrm{d}x\\&=\int^{\frac{\pi}{2}}_{0} x\cot(x) \, \mathrm{d}x + \int_{0}^{\frac{\pi}{2}} (x - \pi)\cot(x) \, \mathrm{d}x\end{align*}$

great solution seanieg89

asianese · Aug 13, 2012

Re: MX2 Integration Marathon

$\int \frac{\ln (\ln x)}{x\ln x} dx$

level: easy

john-doe · Aug 13, 2012

Re: MX2 Integration Marathon

asianese said:
$\int \frac{\ln (\ln x)}{x\ln x} dx$

level: easy

do u mean (int) of ln(x)/x dx???

asianese · Aug 13, 2012

Re: MX2 Integration Marathon

No. I meant what I typed.

SpiralFlex · Aug 13, 2012

Re: MX2 Integration Marathon

asianese said:
$\int \frac{\ln (\ln x)}{x\ln x} dx$

level: easy

$\int \ln (\ln x)*\frac{1}{x\ln x} \;dx$

$=\int \ln (\ln x) \;d(\ln (\ln x))$

$=\frac{\ln^2 (\ln x)}{2}+C$

Bozza555 · Aug 13, 2012

Re: MX2 Integration Marathon

asianese said:
$\int \frac{\ln (\ln x)}{x\ln x} dx$

level: easy

<a href="http://www.codecogs.com/eqnedit.php?latex=\noindent let \space\ u=lnx \\\therefore du=\frac{1}{x}dx\\ \therefore \int \frac{ln(lnx)}{xlnx}=\int \frac{lnu}{u}du\\ let\space\ I= \int \frac{lnu}{u}du\\ integrating \space\ by \space\ parts:\\ I=(lnu)^2-\int \frac{lnu}{u}du @plus;c\\ \therefore 2I=(lnu)^2@plus;c\\ I=\frac{1}{2}(ln(lnx))^2@plus;c" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\noindent let \space\ u=lnx \\\therefore du=\frac{1}{x}dx\\ \therefore \int \frac{ln(lnx)}{xlnx}=\int \frac{lnu}{u}du\\ let\space\ I= \int \frac{lnu}{u}du\\ integrating \space\ by \space\ parts:\\ I=(lnu)^2-\int \frac{lnu}{u}du +c\\ \therefore 2I=(lnu)^2+c\\ I=\frac{1}{2}(ln(lnx))^2+c" title="\noindent let \space\ u=lnx \\\therefore du=\frac{1}{x}dx\\ \therefore \int \frac{ln(lnx)}{xlnx}=\int \frac{lnu}{u}du\\ let\space\ I= \int \frac{lnu}{u}du\\ integrating \space\ by \space\ parts:\\ I=(lnu)^2-\int \frac{lnu}{u}du +c\\ \therefore 2I=(lnu)^2+c\\ I=\frac{1}{2}(ln(lnx))^2+c" /></a>

Edit: to slow haha

SpiralFlex · Aug 13, 2012

Re: MX2 Integration Marathon

$Define$\; g_{0}(x)=1 \;$and$\; g_{n}(x)=(x^2-1)^n \;$for every positive integer n.$

$$For every non-negative integer n, let$ \; I_{n}=\int_{-1}^{1} g_{n}(x)\;dx$

$$a) Express$ \;I_{n+1} \;$in terms of$\; I_{n}.$

$$Hence prove that$\; I_{n+1}=\frac{(-1)^{n+1}*2^{2n+3}*((n+1)!)^2}{(2n+3)!}$

[HR][/HR]

$$b) Prove that$\; (g_{n+1}) ^{(k)}(-1)=(g_{n+1})^{(k)}(1)=0,\; $for all$\; k=0,1,2...n.$

$($Where n is a non-negative integer$)$

[HR][/HR]

$$c)For every non-negative integer n, let$\; h_{n}(x)=\frac{(g_{n})^{(n)}(x)}{2^n *n!}$

$$i) Prove that$\; \int_{-1}^{1} p(x)h_{n+1}(x)\;dx=\frac{(-1)^{n+1}}{2^{n+1}*(n+1)!}\int_{-1}^{1} p^{n+1}(x)g_{n+1}(x) \;dx$

$($for any polynomial$\; p(x))$

$$ii) Hence or otherwise evaluate$\;\int_{-1}^{1} xh_{n}(x)h_{n+1}(x)\;dx$

Actually, last part [c ii.] don't worry about it...

lolcakes52 · Aug 13, 2012

Re: MX2 Integration Marathon

Any one who saw what was hear just forget about that time i forget the difference between odd and even functions.

sky_angel · Aug 13, 2012

Re: MX2 Integration Marathon

I don't get anything in this threadddd

barbernator · Aug 26, 2012

Re: MX2 Integration Marathon

.

Shadowdude · Aug 26, 2012

Re: MX2 Integration Marathon

Basic

How many ways can you place 5 people into 3 distinct rooms? e.g. one 'way' is 3 people in room one, 2 people in room two and none in room three

Challenging

How many ways can you place 10 people into 5 distinct rooms?

If you're game

Show that the number of ways you can place n-k people into k+1 distinct rooms is C(n,k).

kaz1 · Aug 26, 2012

Re: MX2 Integration Marathon

Shadowdude said:
Basic

How many ways can you place 5 people into 3 distinct rooms? e.g. one 'way' is 3 people in room one, 2 people in room two and none in room three

Challenging

How many ways can you place 10 people into 5 distinct rooms?

If you're game

Show that the number of ways you can place n-k people into k+1 distinct rooms is C(n,k).

tis' not integration, no wonder you failed MX2

kaz1 · Aug 26, 2012

Re: MX2 Integration Marathon

Shadowdude said:
Basic

How many ways can you place 5 people into 3 distinct rooms? e.g. one 'way' is 3 people in room one, 2 people in room two and none in room three

Challenging

How many ways can you place 10 people into 5 distinct rooms?

If you're game

Show that the number of ways you can place n-k people into k+1 distinct rooms is C(n,k).

tis' not integration, no wonder you failed MX2

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