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

seanieg89 · Sep 28, 2012

Re: MX2 Integration Marathon

There is almost certainly a way to do this using mx2 level tricks but sledgehammers are fun to wield.

hemayunk · Apr 30, 2013

Re: MX2 Integration Marathon

I have no idea to solve it. Could anybody do it for me.
∫tan x)/ln cos x dx

Carrotsticks · Apr 30, 2013

Re: MX2 Integration Marathon

hemayunk said:
I have no idea to solve it. Could anybody do it for me.
∫tan x)/ln cos x dx

$\\ $Let $ u= \ln (\cos x) \Rightarrow du = - \frac{\sin x}{\cos x} dx = - \tan x dx \Rightarrow - du = \tan x dx \\\\\ $So substituting terms, we have $ \\\\ I = \int \frac{- du}{u} = - \ln u + C = - \ln (\ln (\cos x)+C$

bleakarcher · Apr 30, 2013

Re: MX2 Integration Marathon

lol this thread

SpiralFlex · Apr 30, 2013

Re: MX2 Integration Marathon

Bump!

Sy123 · Apr 30, 2013

Re: MX2 Integration Marathon

Do people want to use this thread too specifically for integration?

SpiralFlex · Apr 30, 2013

Re: MX2 Integration Marathon

From what I've seen in other maths forums, there's always a thread separately for integrals. It's just convention I guess?

Capt Rifle · Apr 30, 2013

Re: MX2 Integration Marathon

Calculus is basically the crux of mathematics

Capt Rifle · Apr 30, 2013

Re: MX2 Integration Marathon

Just imagine solving complex math problems, without the utilisation of calculus, would be pretty tough...

asianese · Apr 30, 2013

Re: MX2 Integration Marathon

$\int x^{2x}(1+\ln x)dx$

asianese · Apr 30, 2013

Re: MX2 Integration Marathon

Capt Rifle said:
Calculus is basically the crux of mathematics

Perhaps it is more a cornerstone, from which other things are built upon.

SpiralFlex · Apr 30, 2013

Re: MX2 Integration Marathon

asianese said:
$\int x^{2x}(1+\ln x)dx$

By inspecting your underwear,

$\frac{1}{2}x^{2x}+C$

SpiralFlex · Apr 30, 2013

Re: MX2 Integration Marathon

$\int \frac{\sqrt[4]{1+\sqrt[3]{x}}}{x}\;dx$

asianese · Apr 30, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
By inspecting your underwear,

$\frac{1}{2}x^{2x}+C$

How rude!

seanieg89 · Apr 30, 2013

Re: MX2 Integration Marathon

Capt Rifle said:
Calculus is basically the crux of mathematics

Not really, its just one branch.

Carrotsticks · Apr 30, 2013

Capt Rifle said:
Just imagine solving complex math problems, without the utilisation of calculus, would be pretty tough...

This is not really true, as there are many other branches too. However, I can understand why you think this way. The HSC course is heavily calculus based so it would naturally lead students to believe that it is essentially everything.

Sy123 · Apr 30, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
$I=\int \frac{\sqrt[4]{1+\sqrt[3]{x}}}{x}\;dx$

$u^3=x$

$I=3\int \frac{(1+u)^{1/4}}{u} \ du$

$m^4=1+u$

$I = 3\int 4 + \frac{4}{m^4-1} \ dm$

$I = 3 \int 4 + \frac{1}{m-1} - \frac{1}{m+1} - \frac{2}{m^2+1} \ dm$

$I = 12m + 3\ln(\frac{m-1}{m+1}) - 6\tan^{-1}m + c$

$I = 12(1+x^{1/3})^{1/4} + 3\ln \frac{(1+x^{1/3})^{1/4}-1}{(1+x^{1/3})^{1/4}+1} - 6\tan^{-1}((1+x^{1/3})^{1/4}) + \epsilon$

err.... I can't see any mistakes?

SpiralFlex · Apr 30, 2013

Re: MX2 Integration Marathon

That is correct.

$u^4=1+\sqrt[3]{x}$

Save space

Sy123 · Apr 30, 2013

Re: MX2 Integration Marathon

$\int \sqrt{\frac{e^x-1}{e^x+1}} \ dx$

asianese · Apr 30, 2013

Re: MX2 Integration Marathon

Let's all get exotic with our constants! (what)

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

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