# HSC 2013 Maths Marathon (archive) (1 Viewer)

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#### RealiseNothing

##### what is that?It is Cowpea
Re: HSC 2013 2U Marathon

I think this is right?

$\bg_white \int \frac {1}{xln(x)} dx \\ \\ \indent = x \int \frac {\frac {1}{x}}{xln(x)} dx \\ \\ \indent = xln(xln(x)) + c$
1) $\bg_white x$ is not a constant, you can't just take it outside the integral.

2) $\bg_white \frac{d}{dx} x \ln(x) \neq \frac{1}{x}$

#### nakruf

##### New Member
Re: HSC 2013 2U Marathon

I think this is right?

$\bg_white \int \frac {1}{xln(x)} dx \\ \\ \indent = x \int \frac {\frac {1}{x}}{xln(x)} dx \\ \\ \indent = xln(xln(x)) + c$
$\bg_white \int \frac {1}{x} \times \frac{1}{lnx} dx$

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#### Shazer2

##### Member
Re: HSC 2013 2U Marathon

I'm not sure, all I know is 1/x integrated is ln(x), not sure about 1/ln(x).

#### nakruf

##### New Member
Re: HSC 2013 2U Marathon

I'm not sure, all I know is 1/x integrated is ln(x), not sure about 1/ln(x).
hint: $\bg_white \int \frac{f'(x)}{f(x)} dx = ln(f(x))+c$

#### ocatal

##### Active Member
Re: HSC 2013 2U Marathon

$\bg_white \int \frac{1}{xlnx} \ dx$

$\bg_white \int \frac {1}{x} \times \frac{1}{lnx} \ dx$

$\bg_white \int \frac{\frac{1}{x}}{lnx} \ dx$

$\bg_white Since \ y = lnx, \ then \ y' = \frac{1}{x}$

$\bg_white \therefore \int \frac{1}{xlnx} \ dx = ln(lnx) + c$

#### Shazer2

##### Member
Re: HSC 2013 2U Marathon

Woah that's heaps neat!

#### braintic

##### Well-Known Member
Re: HSC 2013 2U Marathon

I think this is right?

$\bg_white \int \frac {1}{xln(x)} dx \\ \\ \indent = x \int \frac {\frac {1}{x}}{xln(x)} dx \\ \\ \indent = xln(xln(x)) + c$
Oh dear. Four days to get this out of your system.

#### JJ345

##### Member
Re: HSC 2013 2U Marathon

$\bg_white \int_0^\frac{\pi}{2} \frac{1}{sin(x)} dx$
The area under a cosecx graph between 0 and pi/2?
I'm getting infinity :/ Which is wrong.

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

The area under a cosecx graph between 0 and pi/2?
I'm getting infinity :/ Which is wrong.
$\bg_white i posted this because i didnt know how to integrate it, all i know is \ \ cosecx = \sqrt{1+cot^2 (x)} which isnt even close$

#### HeroicPandas

##### Heroic!
Re: HSC 2013 2U Marathon

$\bg_white \int_0^\frac{\pi}{2} \frac{1}{sin(x)} dx$
Too hard lol

But i do remember something...when integrating a cosecx, you multiply top and bottom by (csc^2 - cot^2) or (csc^2 + cot^2) or (csc + cot) or (csc - cot) then you use $\bg_white \int \frac{P'(x)}{P(x)}dx = ln[P(x)] +C$

#### JJ345

##### Member
Re: HSC 2013 2U Marathon

$\bg_white i posted this because i didnt know how to integrate it, all i know is \ \ cosecx = \sqrt{1+cot^2 (x)} which isnt even close$
You would integrate it by writing 1/sinx as sinx/sinx^2 ie. sinx/1-cosx^2
Then perform substitution u=cosx , partial fractions etc. (too hard for 2U). Its just that the limits in your question stuffs things up.

#### HeroicPandas

##### Heroic!
Re: HSC 2013 2U Marathon

$\bg_white \frac{1}{sinx} = cosecx$

$\bg_white \int cosecx * \frac{cotx - cosec x}{cotx - cosecx}dx = \int \frac{cosecxcotx - (cosecx)^2}{cotx - cosecx}dx \\\\ \\ \rightarrow = ln(cotx - cosec x) +C$

$\bg_white as \frac{d}{dx}\left ( cotx - cosecx \right ) = -(cosecx)^2 + cosecxcotx$ <------change csc and cot into sine and cosine, apply quotient rule to differentiate

In the end, this is just a memory exercise (unless somehow you knew to multiply top and bottom by that)

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

Too hard lol

But i do remember something...when integrating a cosecx, you multiply top and bottom by (csc^2 - cot^2) or (csc^2 + cot^2) or (csc + cot) or (csc - cot) then you use $\bg_white \int \frac{P'(x)}{P(x)}dx = ln[P(x)] +C$
so many options

You would integrate it by writing 1/sinx as sinx/sinx^2 ie. sinx/1-cosx^2
Then perform substitution u=cosx , partial fractions etc. (too hard for 2U). Its just that the limits in your question stuffs things up.
so say for instance we change the limits to pi/3 to p/2 the area wouldnt be infinite but is there anyway the function can be differentiated using 2U or 3U even?

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

$\bg_white \frac{1}{sinx} = cosecx$

$\bg_white \int cosecx * \frac{cotx - cosec x}{cotx - cosecx}dx = \int \frac{cosecxcotx - (cosecx)^2}{cotx - cosecx}dx \\\\ \\ \rightarrow = ln(cotx - cosec x) +C$

$\bg_white as \frac{d}{dx}\left ( cotx - cosecx \right ) = -(cosecx)^2 + cosecxcotx$ <------change csc and cot into sine and cosine, apply quotient rule to differentiate

In the end, this is just a memory exercise (unless somehow you knew to multiply top and bottom by that)
ohh theres the answer how gay is that maths just turned into a guessing game

#### Sy123

##### This too shall pass
Re: HSC 2013 2U Marathon

ohh theres the answer how gay is that maths just turned into a guessing game
nah mate just stick to your

$\bg_white \int_1^2 2x \ dx$

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

nah mate just stick to your

$\bg_white \int_1^2 2x \ dx$
wow your questions are so amazing, i cant solve that teach me please, teach me your ways sy

#### Menomaths

##### Exaı̸̸̸̸̸̸̸̸lted Member
Re: HSC 2013 2U Marathon

nah mate just stick to your

$\bg_white \int_1^2 2x \ dx$
what's wrong with that

#### nakruf

##### New Member
Re: HSC 2013 2U Marathon

wow your questions are so amazing, i cant solve that teach me please, teach me your ways sy
do it yourself

#### -o-

##### New Member
Re: HSC 2013 2U Marathon

this thread is stressing my out so much dont know how to do half these questions...

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

do it yourself
but i cant because sy's questions are hard

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