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

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
$\int_{-\infty }^{\infty }\frac{sin(x)}{x}dx$

and this one = $\pi$

The proofs for these integrals are extremely complicated... and obviously I used a resource "an introduction to the theory of numbers by hardy" and some other calculus book... will stop posting on this forum =(

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

Lol. So much lol.

Next:

$\int \frac{\sqrt{1-x}}{1-\sqrt{x}} dx$

nightweaver066 · Aug 11, 2012

Re: MX2 Integration Marathon

asianese said:
Lol. So much lol.

Next:

$\int \frac{\sqrt{1-x}}{1-\sqrt{x}} dx$

$$Let $ x = sin^2u$

$dx = 2sinucosu du$

$\int \frac{\sqrt{1 - sin^2u}}{1 - \sqrt{sin^2u}}\times 2sinucosu du$

$= \int \frac{2sinucos^2u}{1-sinu}}du$

$= \int \frac{2sinu(1 + sinu)(1 - sinu)}{1-sinu}du$

$= \int 2sinu(1 + sinu)du$

$= \int 2sinu + 1 - cos2udu$

$= u - 2cosu - \frac{1}{2}sin2u$

$= sin^{-1}\sqrt{x} - 2cossin^{-1}\sqrt{x} - \frac{1}{2}sin2(sin^{-1}\sqrt{x})$

$= sin^{-1}\sqrt{x} - 2\sqrt{1 - x} - \sqrt{x(1 - x)}$

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

nightweaver066 said:
$\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{\sqrt{2}}} \frac{1 - 2x^2}{x^5 - x^3}dx, \; $using x = sinu$$

AAEldar · Aug 11, 2012

Re: MX2 Integration Marathon

Find this one WITHOUT using a trig substitution (and this includes hyperbolic functions, I don't want to see them). Spiral did this one a long time ago for me.

$\int \frac{dx}{\sqrt{1+x^2}}$

nightweaver066 · Aug 11, 2012

Re: MX2 Integration Marathon

AAEldar said:
Find this one WITHOUT using a trig substitution (and this includes hyperbolic functions, I don't want to see them). Spiral did this one a long time ago for me.

$\int \frac{dx}{\sqrt{1+x^2}}$

so we can use other substitutions right

$$Let $ u = 1 + x^2$

$du = 2xdx$

$\int \frac{1}{2\sqrt{u(u - 1)}}du$

$= \int \frac{1}{2\sqrt{(u - \frac{1}{2})^2 - \frac{1}{4}}}du$

$= \frac{1}{2}ln|u - \frac{1}{2} + \sqrt{u^2 - u}| + c$

$= \frac{1}{2}ln|x^2 + \frac{1}{2} + x\sqrt{x^2 + 1}| + c$

AAEldar · Aug 11, 2012

Re: MX2 Integration Marathon

nightweaver066 said:
so we can use other substitutions right

$$Let $ u = 1 + x^2$

$du = 2xdx$

$\int \frac{1}{2\sqrt{u(u - 1)}}du$

$= \int \frac{1}{2\sqrt{(u - \frac{1}{2})^2 - \frac{1}{4}}}du$

$= \frac{1}{2}ln|u - \frac{1}{2} + \sqrt{u^2 - u}| + c$

$= \frac{1}{2}ln|x^2 + \frac{1}{2} + x\sqrt{x^2 + 1}| + c$

Hahah not what I was expecting. There are a few ways to do this one! I completely forgot about that standard integral though.

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

$\int_{0}^{\infty }\frac{1}{x^{4}+1}dx$

MX2 level integral...

Final answer should be $\frac{\pi }{2\sqrt{2}}$

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

Well...1+x^4 is just testing an arduous and unpleasant partial fraction decomposition...not really getting to the core of integration.

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

asianese said:
Well...1+x^4 is just testing an arduous and unpleasant partial fraction decomposition...not really getting to the core of integration.

You cannot do that with partial fractions.

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
You cannot do that with partial fractions.

You can, actually =)

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

LOL

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
You can, actually =)

Whaaaaat? =( no way, show us

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

I'll let one of the students do the honours. Quite a popular 'Harder Partial Fractions' problem, actually, along with 1/(x^3+1).

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

COMPLETE THE SQUARE BITCHASS!

$x^4+1=x^4+2x^2+1-2x^2\\=(x^2+1)^2-2x^2\\=(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)$

Now let things be...

$\frac{ax+b}{x^2-\sqrt2 x+1}+\frac{cx+d}{x^2+\sqrt2 x+1} = \frac{1}{(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)}$

Now it turns out that $a=-\frac{1}{2\sqrt2}, c=\frac{1}{2\sqrt2}, b=d=\frac{1}{2}$

Now you can factor out a 1/2 in the integrand.

After that you need to split the ax and b term (as well as cx and d) and they all become some sort of logs or inverse tans, depending on the sign of the fraction on the bottom. Then you can put the limits in, with inverse tan going to pi/2 as x goes to infty, and the logs will eventually cancel or simplify to give the required result.

What did you have in mind, UMAD? (To solve this integral)

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

Nice work =)

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

Don't mind if I do.

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

This is how it is done bitch

$\oint \frac{1}{z^{4}+1}dz=\oint_{c}^{}f(z)dz=2\pi i\left ( \frac{1}{4\sqrt{i}}\left ( \frac{1}{i}+1 \right ) \right )=\frac{\pi \sqrt{2}}{2}\therefore \int_{-\infty }^{\infty }\frac{1}{x^{4}+1}dx+\int_{\infty }^{-\infty }\frac{1}{(x+i)^{4}+1}dx=\frac{\pi \sqrt{2}}{2}\therefore \int_{0}^{\infty }\frac{1}{x^{4}+1}dx=\frac{1}{2}\int_{-\infty }^{\infty }\frac{1}{x^{4}+1}dx=\frac{\pi }{2\sqrt{2}}$

Much more concise.

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

Carrot, if I use these techniques in the MX2 paper, do I get zero?

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

Hey hey c'mon keep it friendly and elementary.

And for the record UMAD, you didn't properly show the acquiring of the Laurent Expension and the substituting of the poles to acquire the Residue.

U MAD BRO said:
Carrot, if I use these techniques in the MX2 paper, do I get zero?

Yes.

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

UMAD?!

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