HSC 2013 Maths Marathon (archive) (2 Viewers)

RealiseNothing · Oct 17, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
I think this is right?

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

1) $x$ is not a constant, you can't just take it outside the integral.

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

nakruf · Oct 17, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
I think this is right?

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

this may help you see the answer more easily
$\int \frac {1}{x} \times \frac{1}{lnx} dx$

Shazer2 · Oct 17, 2013

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 · Oct 17, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
I'm not sure, all I know is 1/x integrated is ln(x), not sure about 1/ln(x).

hint: $\int \frac{f'(x)}{f(x)} dx = ln(f(x))+c$

ocatal · Oct 17, 2013

Re: HSC 2013 2U Marathon

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

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

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

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

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

Shazer2 · Oct 17, 2013

Re: HSC 2013 2U Marathon

Woah that's heaps neat!

braintic · Oct 17, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
I think this is right?

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

Oh dear. Four days to get this out of your system.

JJ345 · Oct 17, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$\int_0^\frac{\pi}{2} \frac{1}{sin(x)} dx$

The area under a cosecx graph between 0 and pi/2?
I'm getting infinity :/ Which is wrong.

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

JJ345 said:
The area under a cosecx graph between 0 and pi/2?
I'm getting infinity :/ Which is wrong.

$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 · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$\int_0^\frac{\pi}{2} \frac{1}{sin(x)} dx$

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 $\int \frac{P'(x)}{P(x)}dx = ln[P(x)] +C$

JJ345 · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$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 · Oct 18, 2013

Re: HSC 2013 2U Marathon

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

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

$$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 · Oct 18, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
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 $\int \frac{P'(x)}{P(x)}dx = ln[P(x)] +C$

so many options

JJ345 said:
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 · Oct 18, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
$\frac{1}{sinx} = cosecx$

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

$$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 · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
ohh theres the answer how gay is that maths just turned into a guessing game

nah mate just stick to your

$\int_1^2 2x \ dx$

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
nah mate just stick to your

$\int_1^2 2x \ dx$

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

Menomaths · Oct 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
nah mate just stick to your

$\int_1^2 2x \ dx$

what's wrong with that

nakruf · Oct 18, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
wow your questions are so amazing, i cant solve that teach me please, teach me your ways sy

do it yourself

-o- · Oct 18, 2013

Re: HSC 2013 2U Marathon

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

mahmoudali · Oct 18, 2013

Re: HSC 2013 2U Marathon

nakruf said:
do it yourself

but i cant because sy's questions are hard

HSC 2013 Maths Marathon (archive) (2 Viewers)

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