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

FrankXie · Nov 12, 2014

Re: MX2 Integration Marathon

not sure this one posted before?

$\int\frac{\tan x+\tan 2x+\tan 3x}{\tan x\tan 2x\tan 3x}dx$

Carrotsticks · Nov 12, 2014

Re: MX2 Integration Marathon

FrankXie said:
not sure this one posted before?

$\int\frac{\tan x+\tan 2x+\tan 3x}{\tan x\tan 2x\tan 3x}dx$

Probably has.

If not in here, probably in the advanced Extension two thread.

If not there, then maybe in the normal 4U thread.

If not there, then maybe the last year's marathon thread.

If not there, then maybe the year before.

Carrotsticks · Nov 12, 2014

Re: MX2 Integration Marathon

So what I'm saying is that I'm pretty sure every single possible Extension two level question ever created is lurking somewhere in our Maths forums.

seanieg89 · Nov 12, 2014

Re: MX2 Integration Marathon

Sy123 said:
Is there a reason for that specific list of functions to be considered elementary?

What if I say

Elementary = anything formed recursively (using the operations of +,-,*,/,^ and composition) from: the reals, the monomial x, the 3 trig funcs and their inverses, the 3 hyp trig funcs and their inverses, the exponential and the logarithm, AND the Gamma function

Is there a reason why my this addition to the definition is wrong? Is the concept of elementary functions purely a categorization of mathematicians?

It's just convention, and what people have pretty much stuck to.

By calling the functions I described as *elementary*, we aren't making any mathematical claims about them, we are just deciding on a name for a certain class of functions that come up a lot in low level mathematics (and combinations thereof).

There is nothing *wrong* with your definition other than the fact it is different from what seems to be the broadly accepted definition.

Sy123 · Nov 12, 2014

Re: MX2 Integration Marathon

Carrotsticks said:
Probably has.

If not in here, probably in the advanced Extension two thread.

If not there, then maybe in the normal 4U thread.

If not there, then maybe the last year's marathon thread.

If not there, then maybe the year before.

Don't forget

$\int e^{x+e^{x}} \ dx$

FrankXie · Nov 12, 2014

Re: MX2 Integration Marathon

Sy123 said:
Don't forget

$\int e^{x+e^{x}} \ dx$

lol

$e^{x+e^x}=e^x\cdot e^{e^x}=\frac{d}{dx}e^{e^x}$

HeroicPandas · Nov 12, 2014

Re: MX2 Integration Marathon

Sy123 said:
Don't forget

$\int e^{x+e^{x}} \ dx$

Best question in the world

Carrotsticks · Nov 12, 2014

Re: MX2 Integration Marathon

Sy123 said:
Don't forget

$\int e^{x+e^{x}} \ dx$

Haha, a nice integral that's absolutely been done to death on these forums.

glittergal96 · Nov 12, 2014

Re: MX2 Integration Marathon

$\int e^{e^{e^{e^x}}+e^{e^x}+e^x+x} \, dx$

(Nesting in power towers is done from the right to left.)

FrankXie · Nov 12, 2014

Re: MX2 Integration Marathon

glittergal96 said:
$\int e^{e^{e^{e^x}}+e^{e^x}+e^x+x} \, dx$

(Nesting in power towers is done from the right to left.)

lol and this

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

integral95 · Nov 12, 2014

Re: MX2 Integration Marathon

glittergal96 said:
$\int e^{e^{e^{e^x}}+e^{e^x}+e^x+x} \, dx$

(Nesting in power towers is done from the right to left.)

FrankXie said:
lol and this

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

What da shiet is dis

Anyway I'm guess it's like $e^{e^{e^{e^x}}$ and $ln(ln(ln(ln(x))))$

FrankXie · Nov 22, 2014

Re: MX2 Integration Marathon

$$ Find $ \int\frac{dx}{\sqrt[3]{(x-1)(x+1)^2}}$

turntaker · Nov 22, 2014

Re: MX2 Integration Marathon

Anyone solved the root tanx one

FrankXie · Nov 22, 2014

Re: MX2 Integration Marathon

turntaker said:
Anyone solved the root tanx one

which? $\int\sqrt{\tan x}dx$ ?

turntaker · Nov 22, 2014

Re: MX2 Integration Marathon

FrankXie said:
which? $\int\sqrt{\tan x}dx$ ?

Yes.

FrankXie · Nov 22, 2014

Re: MX2 Integration Marathon

$$ Let $ u=\sqrt{\tan x}, $ then $ x=\tan^{-1}u^2 $ and $ dx=\frac{2u}{1+u^4}du$

$$ Therefore $ I=\int\sqrt{\tan x}dx=\int\frac{2u^2}{1+u^4}du$

the integral is converted into the integral of a rational function, which can be always solved. However the solution is very long and boring.

would love to see better method

aDimitri · Nov 22, 2014

Re: MX2 Integration Marathon

FrankXie said:
$$ Let $ u=\sqrt{\tan x}, $ then $ x=\tan^{-1}u^2 $ and $ dx=\frac{2u}{1+u^4}du$

$$ Therefore $ I=\int\sqrt{\tan x}dx=\int\frac{2u^2}{1+u^4}du$

the integral is converted into the integral of a rational function, which can be always solved. However the solution is very long and boring.

would love to see better method

i got this question from a mate when i started the course and this was exactly how i did it

Drongoski · Nov 22, 2014

Re: MX2 Integration Marathon

glittergal96 said:
$\int e^{e^{e^{e^x}}+e^{e^x}+e^x+x} \, dx$

(Nesting in power towers is done from the right to left.)

= $e^{e^{e^{e^x}}} + C$ ??

Sy123 · Nov 22, 2014

Re: MX2 Integration Marathon

$\\ \int_0^1 \sin \ln x \ dx$

FrankXie · Nov 22, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ \int_0^1 \sin \ln x \ dx$

$$ At $ x=0, \ln x $ has no definition, so the lower limit $ x=0 $ is understood in the sense of $ x\rightarrow0^+$

$$ Due to the boundedness of sine and cosine function, we have $ \lim_{x\rightarrow0^+}x\sin\ln x=\lim_{x\rightarrow0^+}x\cos\ln x=0$

Now applying integration by parts twice,

$I=\int_0^1\sin\ln xdx=x\sin\ln x|_0^1-\int_0^1x(\cos\ln x)\frac1xdx$

$=0-0-\int_0^1\cos\ln x dx=-x\cos\ln x|_0^1+\int_0^1x(-\sin\ln x)\frac1xdx$

$=-1-0-\int_0^1\sin\ln xdx=-1-I$

$$ Therefore, $ I=-\frac12$

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