HSC 2012-14 MX2 Integration Marathon (archive) (3 Viewers)

Ikki · Mar 25, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int_{0}^{\pi/2} \frac{dx }{a^2\sin^2x + b^2\cos^2x}$

The usual dividing top and bottom by cos works well here.
$\int_{0}^{pi/2}\frac{dx}{a^2sin^2x+b^2cos^2x}$

$=\int_{0}^{pi/2}\frac{sec^2xdx}{a^2tan^2x+b^2}$

Let u=tanx

$=\int_{0}^{\infty} \frac{du}{a^2u^2+b^2}$

$=\frac{1}{ab}\arctan (\frac{au}{b})$

$=\frac{1}{ab}\arctan (\frac{atanx}{b})$

And yeh i run into a problem roundabout here as you know tanpi/2=infinity.

Attempting to solve between the limits...

$=\frac{1}{ab}\arctan (\frac{a\infty}{b})$

$=\frac{1}{ab}\arctan (\infty)$

$=\frac{pi}{2ab}$

Not sure if legit.

dunjaaa · Mar 25, 2014

Re: MX2 Integration Marathon

trying to think of another way of doing this without involving substitution

Sy123 · Mar 25, 2014

Re: MX2 Integration Marathon

Ikki said:
The usual dividing top and bottom by cos works well here.
$\int_{0}^{pi/2}\frac{dx}{a^2sin^2x+b^2cos^2x}$

$=\int_{0}^{pi/2}\frac{sec^2xdx}{a^2tan^2x+b^2}$

Let u=tanx

$=\int_{0}^{\infty} \frac{du}{a^2u^2+b^2}$

$=\frac{1}{ab}\arctan (\frac{au}{b})$

$=\frac{1}{ab}\arctan (\frac{atanx}{b})$

And yeh i run into a problem roundabout here as you know tanpi/2=infinity.

Attempting to solve between the limits...

$=\frac{1}{ab}\arctan (\frac{a\infty}{b})$

$=\frac{1}{ab}\arctan (\infty)$

$=\frac{pi}{2ab}$

Not sure if legit.

Yep that's what I did.

Shadowdude · Mar 25, 2014

Re: MX2 Integration Marathon

Ikki said:
Attempting to solve between the limits...

$=\frac{1}{ab}\arctan (\frac{a\infty}{b})$

$=\frac{1}{ab}\arctan (\infty)$

$=\frac{pi}{2ab}$ .

This part of the solution makes me cry. I'm literally cringing right now...

(substitute n in, and take the limit as n to infinity)

Carrotsticks · Mar 25, 2014

Re: MX2 Integration Marathon

Shadowdude said:
This part of the solution makes me cry. I'm literally cringing right now...

(substitute n in, and take the limit as n to infinity)

Haha me too, I felt pretty uncomfortable seeing it.

Sy123 · Mar 25, 2014

Re: MX2 Integration Marathon

$\int_0^{2\pi} \sqrt{1+\cos x} \ dx$

RealiseNothing · Mar 25, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{2\pi} \sqrt{1+\cos x} \ dx$

$1+cos(x)=2cos^2(\frac{x}{2})$

You would probably have to deal with cases of $\pm$

Kurosaki · Mar 25, 2014

Re: MX2 Integration Marathon

RealiseNothing said:
$1+cos(x)=2cos^2(\frac{x}{2})$

You would probably have to deal with cases of $\pm$

Shouldn't that, uh, just be integrating $\sqrt2|cos(x)|$ from 0 to $2\pi$ ? So then can't you (not sure what the proper term is) split up the interval of integration in a more intervals then evaluate each one? Or just, uh evaluate from 0 to $\pi$ and from there multiply by 2?

Edit: Is it $4\sqrt2$ ?
Edit 2: It should actually be $\int^{2\pi}_0\sqrt2|cos(\frac{x}{2})|dx$ . (ignore the above integral)
We can then draw a graph and notice the 'symmetry' (not sure if this is the correct term) about x= $\pi$ in the interval of integration, so we can simplify this integral to $2\sqrt2\int^{\pi}_0 cos(\frac{x}{2})dx$

Sy123 · Mar 25, 2014

Re: MX2 Integration Marathon

Kurosaki said:
Shouldn't that, uh, just be integrating $\sqrt2|cos(x)|$ from 0 to $2\pi$ ? So then can't you (not sure what the proper term is) split up the interval of integration in a more intervals then evaluate each one? Or just, uh evaluate from 0 to $2\pi$ and from there multiply by 4?

Edit: Is it $4\sqrt2$ ?

Yep that is what I got.

seanieg89 · Mar 26, 2014

Re: MX2 Integration Marathon

seanieg89 said:
In this question we compute the Dirichlet integral using high school methods.

$1. Show that: \\ \\$\int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin(x)}\, dx = \frac{\pi}{2}.$\\ \\ 2. Let: \\ \\$I_n:=\int_0^{\frac{(2n+1)\pi}{2}}\frac{\sin(x)}{x}\, dx$\\ and show that $D_n=\frac{\pi}{2}-I_n \rightarrow 0$ by using 1.\\ \\ 3. Hence, given that the limit: $I:=\lim_{T\rightarrow \infty}\left(\int_0^T \frac{\sin(x)}{x}\, dx\right) $ exists, find it.$

Note: n ranges over the non-negative integers.

bump.

Ikki · Mar 26, 2014

Re: MX2 Integration Marathon

Carrotsticks said:
Haha me too, I felt pretty uncomfortable seeing it.

lol soz guys never done a q like that before with infinities, would u care to show me the correct working?

dunjaaa · Mar 31, 2014

Re: MX2 Integration Marathon

Screen shot 2014-03-25 at 5.45.37 PM.png

Ikki · Apr 1, 2014

Re: MX2 Integration Marathon

On the third last line it's meant to be a u not x but other than that thanks

@SYAMAN: Brah that frog is so distracting LOL.

trapizi · Apr 1, 2014

Re: MX2 Integration Marathon

Ikki · Apr 2, 2014

Re: MX2 Integration Marathon

$$Well isn't that nice... $ \\ \\ \int_{0}^{2014\pi} \sqrt{1-(1-2sin^2x)} \\ =\int_{0}^{2014\pi} \sqrt{2}sinx $ dx $ \\ =\sqrt{2} \left [ -cosx \right ]_{0}^{2014\pi}$ dx $ \\ =\sqrt{2}(-1+1) \\ =0$

But i feel i remember doing something like this before, and its not that straight forward... Nope it seems about legit.

Hey guys how do u guys do the integration brackets?

trapizi · Apr 2, 2014

Re: MX2 Integration Marathon

^
Sorry mate wrong answer

Kurosaki · Apr 2, 2014

Re: MX2 Integration Marathon

Ikki said:
$$Well isn't that nice... $ \\ \\ \int_{0}^{2014\pi} \sqrt{1-(1-2sin^2x)} \\ =\int_{0}^{2014\pi} \sqrt{2}sinx $ dx $ \\ =\sqrt{2} \left [ -cosx \right ]_{0}^{2014\pi}$ dx $ \\ =\sqrt{2}(-1+1) \\ =0$

But i feel i remember doing something like this before, and its not that straight forward... Nope it seems about legit.

Hey guys how do u guys do the integration brackets?

It's not 0, the reason is because the graph is |sin(x)|. So what you do is draw a graph, calculate a single interval and multiply.
Leading on from your working, the integral is $\int^{2014\pi}_{0}\sqrt2 |sin(x)|$ . Notice that the same pattern repeats for every $n\pi$ , n being integral values, when graphing.
Then we can simplify this to $2014\sqrt2\int^{\pi}_{0}sin(x)$ . Integrating, we find $4028\sqrt2$ ?

Ikki · Apr 2, 2014

Re: MX2 Integration Marathon

when do you know the simplification is to be |absolute| ?

Ah I get it now, but... Why wouldn't it be 2014 not 4028 in front of the integral? If you say for example take the boundaries between 0 and 4pi. The integral of 0 to pi multiplied by 4 is the solution for sinx.
Therefore final solution 4028sqrt(2)?

EDIT: Oh k cool you changed it

Kurosaki · Apr 2, 2014

Re: MX2 Integration Marathon

Ikki said:
when do you know the simplification is to be |absolute| ?

Ah I get it now, but... Why wouldn't it be 2014 not 4028 in front of the integral? If you say for example take the boundaries between 0 and 4pi. The integral of 0 to pi multiplied by 4 is the solution for sinx.
Therefore final solution 4028sqrt(2)?

EDIT: Oh k cool you changed it

When you do something like, say, $\sqrt{x^2}$ , it's not equal to x, it's |x|. You can examine this by using values for x<0
Edit: I realised I didn't actually answer your question. Hmmm, generally when there is a square root, you should start thinking absolute value, unless it's obvious that in the interval of integration, it won't be required, e.g. the one we just did, but with bounds from 0 to $\frac{\pi}{2}$

trapizi · Apr 2, 2014

Re: MX2 Integration Marathon

Nice. There is another way using this integration method:

HSC 2012-14 MX2 Integration Marathon (archive) (3 Viewers)

Active Member

Active Member

This too shall pass

Cult of Personality

Retired

This too shall pass

what is that?It is Cowpea

True Fail Kid

This too shall pass

Well-Known Member

Active Member

Active Member

Active Member

╰(゜Д゜)╯

Attachments

Active Member

╰(゜Д゜)╯

True Fail Kid

Active Member

True Fail Kid

╰(゜Д゜)╯

Users Who Are Viewing This Thread (Users: 0, Guests: 3)