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

Drsoccerball · Apr 9, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
he multiplied top and bottom by root 2-x

This definitely doesnt work

VBN2470 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Then his restrictions are wrong if that substitution was used doubt he used it

Sorry, it should be $u = x+1$

Drsoccerball · Apr 9, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Sorry, it should be $u = x+1$

How did you see this

VBN2470 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
How did you see this

Secret ways

Drsoccerball · Apr 9, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Secret ways

This was in one of my school's past papers so i need to know DD:

seanieg89 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

I just did the shift by one (intuition told me this substitution would make it nicer because of the numerator/denominator symmetry) then multiplied the numerator and denominator by the numerator, to get rid of square roots up there.

Alternatively you can get rid of the numerator surd first, giving you an integrand of 1/sqrt(quadratic), which is solved by completing that quadratics square anyway.

VBN2470 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
How did you see this

Drsoccerball said:
This was in one of my school's past papers so i need to know DD:

seanieg89 said:
I just did the shift by one (intuition told me this substitution would make it nicer because of the numerator/denominator symmetry) then multiplied the numerator and denominator by the numerator, to get rid of square roots up there.

Alternatively you can get rid of the numerator surd first, giving you an integrand of 1/sqrt(quadratic), which is solved by completing that quadratics square anyway.

.

VBN2470 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

NEW Q:

$If $I_n = \int_{0}^{1} x(1-x^3)^{n}\text{ d}x$ for $n\,\geq 0$, show that $I_n = \frac{3n}{3n+2}I_{n-1}$ for $n$\,\geq 1.$

seanieg89 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

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.)

Drsoccerball · Apr 9, 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.)

Rearrange the equation too:

$C_1 < \int_0^{2\pi}{\frac{||z|-1|^a}{|z-cis\theta|^a}}d\theta < C_2$

$$ Integrate the middle expression and the values for $ P(2\pi) $ and $ p(0) $ are the values of $ C_1 $ and $ C_2$

But how do you integrate that ahahah

FrankXie · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Rearrange the equation too:

$C_1 < \int_0^{2\pi}{\frac{||z|-1|^a}{|z-cis\theta|^a}}d\theta < C_2$

good start, but you do not need to actually integrate it, you just need an estimation, why not try to apply the comparison property of definite integrals

Ekman · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Here's a question that has been plaguing me:

$\int \sqrt{x^2-2} dx$

I understand that it uses the table of standard integrals but I want to know the logic behind it.

VBN2470 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Here's a question that has been plaguing me:

$\int \sqrt{x^2-2} dx$

I understand that it uses the table of standard integrals but I want to know the logic behind it.

What exactly is troubling you?

InteGrand · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Here's a question that has been plaguing me:

$\int \sqrt{x^2-2} dx$

I understand that it uses the table of standard integrals but I want to know the logic behind it.

Basically, you want to know how to find $\int \frac{1}{\sqrt{x^2 \pm a^2}}\text{ d}x$ ?

$$For $I_1 = \int \frac{1}{\sqrt{x^2 - a^2}\text{ d}x}$ ($x^2 > a^2$), use the substitution $x=a\sec \theta$. (Suppose $a>0$ and $x>a$, so we can use $\theta \in \left[ 0,\frac{\pi}{2}\right)$, where $\tan \theta >0$). Then $\mathrm{d}x = a\sec\theta\tan \theta \text{ d}\theta$, and $x^2 - a^2 = a^2(\sec^2 \theta -1)=a^2 \tan^2 \theta\Rightarrow \sqrt{x^2 -a^2}=a\tan\theta$ and we get:$

$I_1= \int \frac{a\sec\theta \tan \theta}{a\tan \theta}\text{ d}\theta$

$=\int \sec \theta \text{ d}\theta$

$=\ln (\sec \theta +\tan \theta)+c$

$=\ln \left(\frac{x}{a}+\frac{\sqrt{x^2 - a^2}}{a}\right)+c $ (draw up a right-angled triangle to get $\sec \theta$ and $\tan \theta$ in terms of $x$, or use trig. identities)$$

$=\ln \left(\frac{1}{a}\times (x+\sqrt{x^2 - a^2})\right)+c$

$=\ln (x+\sqrt{x^2 - a^2})-\ln a+c$

$=\ln (x+\sqrt{x^2 - a^2})+C, $ where $C=c-\ln a$$$ .

Drsoccerball · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Here's a question that has been plaguing me:

$\int \sqrt{x^2-2} dx$

I understand that it uses the table of standard integrals but I want to know the logic behind it.

area of a hyperbola ?

InteGrand · Apr 9, 2015

Re: MX2 2015 Integration Marathon

$$For $I_2 = \int \frac{1}{\sqrt{x^2 + a^2}}\text{ d}x$, do something similar; use the substitution $x=a\tan \theta$, $\theta \in \left(-\frac{\pi}{2},\frac{\pi}{2} \right)$ (so $\sec \theta >0\Rightarrow \sqrt{\sec^2 \theta}=\sec \theta$)$.$

Ekman · Apr 9, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Basically, you want to know how to find $\int \frac{1}{\sqrt{x^2 \pm a^2}}\text{ d}x$ ?

$$For $I_1 = \int \frac{1}{\sqrt{x^2 - a^2}\text{ d}x}$ ($x^2 > a^2$), use the substitution $x=a\sec \theta$. Suppose $a>0$ and $x>a$, so we can use $\theta \in \left[ 0,\frac{\pi}{2})$, where $\tan \theta >0$). Then $\mathrm{d}x = a\sec\theta\tan \theta \text{ d}\theta$, and $x^2 - a^2 = a^2(\sec^2 \theta -1)=a^2 \tan^2 \theta\Rightarrow \sqrt{x^2 -a^2}=a\tan\theta$ and we get:$

$I_1= \int \frac{a\sec\theta \tan \theta}{a\tan \theta}\text{ d}\theta$

$=\int \sec \theta \text{ d}\theta$

$=\ln (\sec \theta +\tan \theta)+c$

$=\ln \left(\frac{x}{a}+\frac{\sqrt{x^2 - a^2}}{a}\right)+c $ (draw up a right-angled triangle to get $\sec \theta$ and $\tan \theta$ in terms of $x$, or use trig. identities)$$

$=\ln \left(\frac{1}{a}\times (x+\sqrt{x^2 - a^2})\right)+c$

$=\ln (x+\sqrt{x^2 - a^2})-\ln a+c$

$=\ln (x+\sqrt{x^2 - a^2})+C, $ where $C=c-\ln a$$$ .

Just a little alteration:

$\int \frac{x^2}{\sqrt{x^2 \pm a^2}}\text{ d}x$

Is there a quicker way without doing IBP?

InteGrand · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Just a little alteration:

$\int \frac{x^2}{\sqrt{x^2 \pm a^2}}\text{ d}x$

Yeah, once you've gotten to that stage, you add and subtract $\pm a^2$ on the numerator, and split the fraction (and use $\frac{u}{\sqrt{u}}=\sqrt{u}$ ) to get $\underbrace{\sqrt{x^2 \pm a^2}}_\text{This is what we were integrating before, so you get I on the RHS}\mp \underbrace{\frac{a^2}{\sqrt{x^2 \pm a^2}}}_{\text{So I just showed how to find this}}$ , and you get an equation with I on both the LHS and RHS, so you can just solve for I.

InteGrand · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Just a little alteration:

$\int \frac{x^2}{\sqrt{x^2 \pm a^2}}\text{ d}x$

Is there a quicker way without doing IBP?

Yeah, just use the appropriate substitution $x=a\sec\theta$ or $x=a\tan \theta$ at the start.

InteGrand · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Actually wait, since you're given the standard integrals sheet, integration by parts is probably faster.

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