Integrating Logarithmic Functions (1 Viewer)

LoveHateSchool · Jul 17, 2012

^As a 2U browsing this, I'm really interested in how to answer this in a two unit context.

twinklegal19 · Jul 17, 2012

LoveHateSchool said:
^As a 2U browsing this, I'm really interested in how to answer this in a two unit context.

this

Although I understand carrot's alternative solution, I wonder if there is a more systematic way that involves dividing by cos^2(x) like the question asked? I don't believe there has been an appropriate solution for that yet.

Not to say that this is a typical 2U question anyway, doubt it would be in the hsc paper lol

Carrotsticks · Jul 17, 2012

RealiseNothing · Jul 17, 2012

I'm liking the part where that solution started being 2U carrot.

SpiralFlex · Jul 17, 2012

$Method 1: Can be done with 2U knowledge, well almost.$

$Division by cos squared yields$

$\int \frac{sec^2 x}{9-\tan^2 x}$

$$Let$\; u = \tan x$

$du=sec^2 x \;dx$

$\int_{0}^{\frac{\pi}{4}}\frac {du}{9-u^2}$

$$New limits -> [1,0]$

$For 2U students an integral sheet has been provided flip to the last integrals. Quickly fill in this integral that should be on the integral list:$

$\int \frac{dx}{a^2-x^2}$

$= \frac{1}{2a} \ln |\frac{a+x}{a-x}|+C, x^2 \neq a^2$

$Then use it!$

$=\frac{1}{2*3}\ln \frac{3+1}{3-0}$

$=\frac{1}{6}\ln 2$

$Method 2: The Spiral Method$

$Be back after assignment to type$

$Method 3:$

Carrotsticks · Jul 17, 2012

RealiseNothing said:
I'm liking the part where that solution started being 2U carrot.

Haha woops! I totally forgot that. I can see at least 3 different methods atm.

LoveHateSchool · Jul 17, 2012

You're amazing robot, I still struggle to understand this question though, it's so hard! I hope it's really not a 2U question.

SpiralFlex · Jul 17, 2012

LoveHateSchool said:
You're amazing robot, I still struggle to understand this question though, it's so hard! I hope it's really not a 2U question.

Definitely not an 2U question. But I have devised a plan that will accommodate the markers. By adding a standard integral there will be less explanations so you can go straight to the answer in the exam without proof. Kekeke.

Carrotsticks · Jul 17, 2012

SpiralFlex said:
Definitely not an 2U question. But I have devised a plan that will accommodate the markers. By adding a standard integral there will be less explanations so you can go straight to the answer in the exam without proof. Kekeke.

What was the point of the hyperbolic tan... it's there but you didn't use it.

SpiralFlex · Jul 17, 2012

Carrotsticks said:
What was the point of the hyperbolic tan... it's there but you didn't use it.

Ma bad lol, I always right it because my teacher loves it.

twinklegal19 · Jul 17, 2012

SpiralFlex said:
Definitely not an 2U question. But I have devised a plan that will accommodate the markers. By adding a standard integral there will be less explanations so you can go straight to the answer in the exam without proof. Kekeke.

ahahaha

you're the best Spiral, thanks

twinklegal19 · Jul 17, 2012

Spiral I just checked the list of standard integrals

They only have one for $\int \frac{1}{a^2+x^2}dx$ but not for $\int \frac{1}{a^2-x^2}dx$

Carrotsticks · Jul 17, 2012

twinklegal19 said:
Spiral I just checked the list of standard integrals

They only have one for $\int \frac{1}{a^2+x^2}dx$ but not for $\int \frac{1}{a^2-x^2}dx$

(That's the joke).

twinklegal19 · Jul 17, 2012

Carrotsticks said:
(That's the joke).

oooh...

:S

lol, k then

RealiseNothing · Jul 17, 2012

hahah spiral, love it.

Sanjeet · Jul 17, 2012

None of those solutions are 2U, and even 3U never requires you to come up with your own substitution (that's 4U), they always give it to you

QuantumIllusion · Jul 17, 2012

Try this: $\int_{0}^{\frac{\pi }{4}} \frac{dx}{a^2sin^2x-b^2cos^2x}$

Integrating Logarithmic Functions (1 Viewer)

LoveHateSchool

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what is that?It is Cowpea

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what is that?It is Cowpea

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