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

seanieg89 · Aug 26, 2012

Re: MX2 Integration Marathon

Shadowdude said:
Well, my method uses integrating over a contour in the complex plane and taking real and imaginary parts.

Which is pretty cool because... well, it is.

if i survive until honours - i might like to dabble in this sort of thing.

In what sort of thing? Using contour integration to evaluate real integrals isn't exactly an honours project...

Shadowdude · Aug 26, 2012

Re: MX2 Integration Marathon

seanieg89 said:
In what sort of thing? Using contour integration to evaluate real integrals isn't exactly an honours project...

In what? Complex analysis.

It's fascinating. Obviously not going to write on... second year topics. But if the deeper stuff is just as fun and great as what I'm learning now - then... why not? It should be awesome fun

seanieg89 · Aug 26, 2012

Re: MX2 Integration Marathon

Shadowdude said:
In what? Complex analysis.

It's fascinating. Obviously not going to write on... second year topics. But if the deeper stuff is just as fun and great as what I'm learning now - then... why not? It should be awesome fun

Cool, yep complex is very nice!

asianese · Aug 29, 2012

Re: MX2 Integration Marathon

$\begin{align} &\int \sqrt{\csc x-\sin x}\; dx\\& \int \frac{dx}{\sqrt{e^x-1}} \end{align}$

nightweaver066 · Aug 29, 2012

Re: MX2 Integration Marathon

asianese said:
$\begin{align} &\int \sqrt{\csc x-\sin x}\; dx\\& \int \frac{dx}{\sqrt{e^x-1}}\;dx\end{align}$

$\mathbf{1.} \int \sqrt{\csc x - \sin x} dx$

$= \int \sqrt{\frac{1 - \sin ^2x}{\sin x}}dx$

$= \int \frac{\cos x}{\sqrt{\sin x}}dx$

$= 2\sqrt{\sin x} + c$

$\mathbf{2.} \int \frac{dx}{\sqrt{e^x - 1}}dx$

$$Let $ u^2 = e^x - 1$

$2udu = e^xdx$

$\frac{2u}{u^2 + 1}du = dx$

$\therefore \int \frac{1}{u} \times \frac{2u}{u^2 + 1}du$

$= \int \frac{2}{u^2 + 1}du$

$= 2tan^{-1}u$

$= 2tan^{-1}(\sqrt{e^x - 1}) + c$

asianese · Aug 29, 2012

Re: MX2 Integration Marathon

Good job

Carrotsticks · Aug 29, 2012

Re: MX2 Integration Marathon

Here's a decent one.

$\int \frac{dx}{\sqrt{\sin x}}$

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Here's a decent one.

$\int \frac{dx}{\sqrt{\sin x}}$

$\frac{2\sqrt{sinx}}{cosx}} + C?$

barbernator · Aug 29, 2012

Re: MX2 Integration Marathon

RealiseNothing said:
$\frac{2\sqrt{sinx}}{cosx}} + C?$

yep

Carrotsticks · Aug 29, 2012

Re: MX2 Integration Marathon

Explanations explanations.

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Explanations explanations.

Fine I'll LaTeX it up.

Carrotsticks · Aug 29, 2012

Re: MX2 Integration Marathon

Nah no need to latex, a brief explanation will do.

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

$\int \frac{dx}{\sqrt{\sin x}}$

$Let~u = \sqrt{sinx}$

$dx = \frac{2\sqrt{sinx}du}{cosx}$

$dx = \frac{2udu}{cosx}$

$\int \frac{1}{u} \frac{2udu}{cosx}$

$\int \frac{2}{cosx}du$

$\frac{2u}{cosx} + C$

$\frac{2\sqrt{sinx}}{cosx} + C$

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Nah no need to latex, a brief explanation will do.

lol it took me less than 3 minutes to LaTeX that up (look at post times).

LaTeX speed winning.

barbernator · Aug 29, 2012

Re: MX2 Integration Marathon

RealiseNothing said:
lol it took me less than 3 minutes to LaTeX that up (look at post times).

LaTeX speed winning.

look at the integrand lol

Carrotsticks · Aug 29, 2012

Re: MX2 Integration Marathon

RealiseNothing said:
lol it took me less than 3 minutes to LaTeX that up (look at post times).

LaTeX speed winning.

Your cos and sine functions didn't have \ in front of them, 2/10.

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

barbernator said:
look at the integrand lol

What? I don't see anything wrong with it

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Your cos and sine functions didn't have \ in front of them, 2/10.

$sinx$

$\sinx$

What do you mean? The slash doesn't work?

Carrotsticks · Aug 29, 2012

Re: MX2 Integration Marathon

\sin x m8

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

$\sin x$

Testing.

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