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

HeroicPandas · Sep 21, 2012

Re: MX2 Integration Marathon

Sy123 said:
Oops, lol. Hmm..

So the domain is for all x
$\sin x + \cos x \geq 0$
Using auxilliary angles

$\sqrt{2} \sin (x + \frac{\pi}{4}) \geq 0$

So our domain is

$k\pi \leq x+ \pi/4 \leq (k+1) \pi \\ \\ k\pi - \frac{\pi}{4} \leq x \leq (k+1)\pi -\frac{\pi}{4} \ \ \ \ k \ \epsilon \ \ \mathbb{Z}$

?

I agree!

BUT I found my domain by drawing the graph

How did you know it was from kπ to (k+1)π

I don't quite understand this part:

$k\pi \leq x+\pi /4\leq (k+1)\pi$

Please ignore the attachment

Sy123 · Sep 21, 2012

Re: MX2 Integration Marathon

From the 'ASTC' rule, or rather knowing that the sine graph is positive for:

$0 \leq x \leq \pi \ \ 2\pi \leq x \leq 3\pi \ \ \ \ etc$

So I generalised the statement with k integer. However when I saw from 0 to pi and 0 to 2pi above, I dont mean x, I mean for any function within x, that is the domain. Then we go and find the true domain of x by itself by rearranging. We dont want this to get off topic heh

bleakarcher · Sep 21, 2012

Re: MX2 Integration Marathon

Sy123 said:
Oops, lol. Hmm..

So the domain is for all x
$\sin x + \cos x \geq 0$
Using auxilliary angles

$\sqrt{2} \sin (x + \frac{\pi}{4}) \geq 0$

So our domain is

$k\pi \leq x+ \pi/4 \leq (k+1) \pi \\ \\ k\pi - \frac{\pi}{4} \leq x \leq (k+1)\pi -\frac{\pi}{4} \ \ \ \ k \ \[B]epsilon [/B]\ \ \mathbb{Z}$

?

Sy, why did you use epsilon lol?

HeroicPandas · Sep 21, 2012

Re: MX2 Integration Marathon

Sy123 said:
From the 'ASTC' rule, or rather knowing that the sine graph is positive for:

$0 \leq x \leq \pi \ \ 2\pi \leq x \leq 3\pi \ \ \ \ etc$

So I generalised the statement with k integer. However when I saw from 0 to pi and 0 to 2pi above, I dont mean x, I mean for any function within x, that is the domain. Then we go and find the true domain of x by itself by rearranging. We dont want this to get off topic heh

Ohh alright thanks BUT 1 more thing and we will end it

why can it be from 0 to 2pi

Sy123 · Sep 21, 2012

Re: MX2 Integration Marathon

bleakarcher said:
Sy, why did you use epsilon lol?

Lol isnt that what used to say element of? I dont know much about the math notation, I just didnt want to write k is an integer in latex, and wanted it to do it fancily heh.

HeroicPandas said:
Ohh alright thanks BUT 1 more thing and we will end it

why can it be from 0 to 2pi

Because from 0 to 2pi we have some parts of the sine function which is less than zero, since we squaring than square rooting, we get an always positive value (absolute value). So we must solve the inequality
sin x + cos x =>0

From 0 to 2pi there is part of the domain which is less than zero, and we want to exclude it. Sure 0 to pi is always positive, but so is 2pi to 3pi, and 4pi to 5pi and so on, so its just better to simply generalise it, but looking back at the generalisation it is wrong

$2k\pi - \frac{\pi}{4} \leq x \leq (2k+1)\pi - \frac{\pi}{4}$

What I did before just covered everything and didnt make sense, this domain is the true one

SpiralFlex · Sep 21, 2012

Re: MX2 Integration Marathon

Sy123 said:
Lol isnt that what used to say element of? I dont know much about the math notation, I just didnt want to write k is an integer in latex, and wanted it to do it fancily heh.

Because from 0 to 2pi we have some parts of the sine function which is less than zero, since we squaring than square rooting, we get an always positive value (absolute value). So we must solve the inequality
sin x + cos x =>0

From 0 to 2pi there is part of the domain which is less than zero, and we want to exclude it. Sure 0 to pi is always positive, but so is 2pi to 3pi, and 4pi to 5pi and so on, so its just better to simply generalise it, but looking back at the generalisation it is wrong

$2k\pi - \frac{\pi}{4} \leq x \leq (2k+1)\pi - \frac{\pi}{4}$

What I did before just covered everything and didnt make sense, this domain is the true one

Usually Epsilon is used usually in Calculus to denote a small positive quantity of a measure. The is an element of symbol looks different.

Drongoski · Sep 21, 2012

Re: MX2 Integration Marathon

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

Just one I thought of on top of my head

$= \int \frac {1}{tan^{-1}x} dtan^{-1}x = ln \left | tan^{-1}x \right | + C$

seanieg89 · Sep 21, 2012

Re: MX2 Integration Marathon

Re: the restricted region of integration, keep in mind that you CAN integrate that function over any interval in R, it is just that your formula will only be valid in some places. Similarly, a slightly different formula would allow you to evaluate the integral over intervals elsewhere. I think it wouldn't be a great hsc question as there is no "correct" choice of primitive and corresponding region of integration.

bleakarcher · Sep 21, 2012

Re: MX2 Integration Marathon

seanieg89 said:
Re: the restricted region of integration, keep in mind that you CAN integrate that function over any interval in R, it is just that your formula will only be valid in some places. Similarly, a slightly different formula would allow you to evaluate the integral over intervals elsewhere. I think it wouldn't be a great hsc question as there is no "correct" choice of primitive and corresponding region of integration.

What's this in response to?

EDIT, disregard this.

Shadowdude · Sep 21, 2012

Re: MX2 Integration Marathon

SpiralFlex said:
Usually Epsilon is used usually in Calculus to denote a small positive quantity of a measure. The is an element of symbol looks different.

use "in"

$\in$

as "in"

$5 \in \mathbb{Z}$

abecina · Sep 25, 2012

Re: MX2 Integration Marathon

$\int_2^4\ \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}$

juantheron · Sep 25, 2012

Re: MX2 Integration Marathon

answer..........................................

juantheron · Sep 25, 2012

Re: MX2 Integration Marathon

abecina said:
$\int_2^4\ \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}$

Let $I = \int_2^4\ \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} = \int_2^4\ \frac{\sqrt{\ln(9-(2+4-x))} dx}{\sqrt{\ln(9-(2+4-x))}+\sqrt{\ln((2+4-x)+3)}}$

Using $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$

$I = \int_2^4\ \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} = \int_2^4\ \frac{\sqrt{\ln(x+3)} dx}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}}$

$2I = \int_{2}^{4}1.dx = 2\Leftrightarrow I = 1$

So $I = \int_2^4\ \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} = 1$

juantheron · Sep 25, 2012

Re: MX2 Integration Marathon

$\int_{\sqrt{\ln(2)}}^{\sqrt{\ln(3)}}\frac{x.\sin(x^2)}{\sin(x^2)+\sin(\ln(6)-x^2)}dx$

juantheron · Sep 25, 2012

Re: MX2 Integration Marathon

$\int_{0}^{\pi}\frac{\sin (2013x)}{\sin (x)}dx$

rolpsy · Sep 25, 2012

Re: MX2 Integration Marathon

juantheron said:
$\int_{\sqrt{\ln(2)}}^{\sqrt{\ln(3)}}\frac{x.\sin(x^2)}{\sin(x^2)+\sin(\ln(6)-x^2)}dx$

$\begin{align*}I &= \int^{\sqrt{\ln(3)}}_{\sqrt{\ln(2)}} \frac{x\sin(x^2)}{\sin(x^2) + \sin(\ln(6) - x^2)} \, \mathrm{d}x\\&= \frac{1}{2}\int^{\ln(3)}_{\ln(2)} \frac{\sin(x)}{\sin(x) + \sin(\ln(6) - x)} \, \mathrm{d}x\\&= \frac{1}{2}\int^{\ln(3)}_{\ln(2)} \frac{\sin(\ln(6) - x)}{\sin(\ln(6) - x) + \sin(x)} \, \mathrm{d}x\\\implies 2I &= \frac{1}{2}\int^{\ln(3)}_{\ln(2)} \mathrm{d}x\\I &= \frac{1}{4}\ln\left ( \frac{3}{2} \right ).\end{align*}$

juantheron said:
$\int_{0}^{\pi}\frac{\sin (2013x)}{\sin (x)}dx$

$\\\textup{Let } I_{n} = \int^{\pi}_{0} \frac{\sin(nx)}{\sin(x)} \, \mathrm{d}x, n = 1,2\dots\\\begin{align*}I_{n} - I_{n-2} &= \int^{\pi}_{0} \frac{\sin(nx) - \sin((n-2)x)}{\sin(x)} \, \mathrm{d}x\\&= \int^{\pi}_{0} \frac{2\sin(x)\cos((n-1)x)}{\sin(x)} \, \mathrm{d}x\\&=2\int^{\pi}_{0} \cos((n-1)x) \, \mathrm{d}x\\&=2\left [ \frac{1}{n-1} \sin((n-1)x) \right ]^\pi_0\\&=0\\\end{align*}\\\therefore I_{2013} = I_{2011} = \cdots = I_1 = \pi.$

next:

$\int^{\infty}_{0} \frac{\ln(x)}{x^2 + a^2} \, \mathrm{d}x$

abecina · Sep 27, 2012

Re: MX2 Integration Marathon

next:

$\int^{\infty}_{0} \frac{\ln(x)}{x^2 + a^2} \, \mathrm{d}x$

$\begin{align*}\textrm{Let} \;x &= a\tan{\theta}\\I:= \int_{0}^{\infty}\frac{\ln{x}}{x^2 + a^2} \textrm{d}x &=\int_{0}^{\frac{\pi}{2}}\frac{\ln(a\tan{\theta})}{a^2\tan^2{\theta}+a^2} a\sec^2{\theta} d\theta\\& = \int_{0}^{\frac{\pi}{2}}\frac{\ln(a\tan{\theta})}{a} d\theta\;\;\;\;\;(1) \\\textrm{Since} \int_{0}^{\frac{\pi}{2}} f(\theta) d\theta &= \int_{0}^{\frac{\pi}{2}} f\left(\frac{\pi}{2}-\theta \right) \textrm{d}\theta\\I&= \int_{0}^{\frac{\pi}{2}}\frac{\ln(a\cot{\theta})}{a} d\theta\;\;\;\;\;(2)\\(1)+(2)\Rightarrow 2I &= \int_{0}^{\frac{\pi}{2}}\frac{\ln(a\tan{\theta})}{a} d\theta + \int_{0}^{\frac{\pi}{2}}\frac{\ln(a\cot{\theta})}{a} d\theta\\&= \int_{0}^{\frac{\pi}{2}}\frac{\ln(a^2\tan{\theta}\cot{\theta})}{a} d\theta\\&= 2\int_{0}^{\frac{\pi}{2}}\frac{\ln{a}}{a} d\theta\\I &= \frac{\pi}{2a}\ln{a}\\\end{align*}$

juantheron · Sep 28, 2012

Re: MX2 Integration Marathon

$\displaystyle \int_{0}^{\infty}\frac{\ln(x)}{(a^2+x^2)^3}dx$

where $a>0$

seanieg89 · Sep 28, 2012

Re: MX2 Integration Marathon

juantheron said:
$\displaystyle \int_{0}^{\infty}\frac{\ln(x)}{(a^2+x^2)^3}dx$

where $a>0$

$I=\frac{\pi(3\log a-4)}{16a^5}.$

Consider the integrand over the complex plane with a branch cut on the non-negative imaginary axis. Then use the residue theorem with the contour consisting of an epsilon upper semicircle about 0, an R upper semicircle about 0 and the two segments of the real axis required to close this contour off.

As R->inf and epsilon->0, the contributions from the semicircular segments tend to zero by pretty easy estimates.

One could also obtain this result by using repeated differentiation under the integral sign on the integral computed by abecina above.

Shadowdude · Sep 28, 2012

Re: MX2 Integration Marathon

seanieg89 said:
$I=\frac{\pi(3\log a-4)}{16a^5}.$

Consider the integrand over the complex plane with a branch cut on the non-negative imaginary axis. Then use the residue theorem with the contour consisting of an epsilon upper semicircle about 0, an R upper semicircle about 0 and the two segments of the real axis required to close this contour off.

As R->inf and epsilon->0, the contributions from the semicircular segments tend to zero by pretty easy estimates.

One could also obtain this result by using repeated differentiation under the integral sign on the integral computed by abecina above.

ooh we're doing this now

so so interesting

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