Author Topic: Interesting questions (spesh) (Read 13693 times) Tweet Share

vea · « **Reply #45 on:** September 29, 2010, 04:12:10 pm »

You guys make spesh seem so fun, can't wait til next year.

Martoman · « **Reply #46 on:** September 29, 2010, 04:37:02 pm »

Quote from: brightsky on September 29, 2010, 04:06:32 pm

Quote from: /0 on September 29, 2010, 02:47:43 am
gj Martoman xD

Quote from: brightsky on September 27, 2010, 10:20:42 pm
$\text{3.} \int^{1}_{0} \frac{\arctan(x)}{\log_e(x+1)} dx$

That doesn't have a primitive, is there a way without complex analysis? (Even that would be reaaally hard, if not impossible)

Hmm yeah my friend gave me that problem and I thought there was something wrong with it. I think it should be $\int^1_0 \frac{\arctan(x)}{1 + x}$ .

Polylogarithm functions aren't on the course (that is the integral deals with this). That one *can* be solved but its not possible at yr 12.

supernatural · « **Reply #47 on:** September 29, 2010, 04:54:31 pm »

heres one for you

$\int e^{(x)^2} dx$

letsride · « **Reply #48 on:** September 29, 2010, 04:59:32 pm »

- fml you changed it =X

Martoman · « **Reply #49 on:** September 29, 2010, 05:00:05 pm »

Quote from: supernatural on September 29, 2010, 04:54:31 pm

heres one for you

$\int e^{(x)^2}$

you can't actually solve that supernatural unless you did something.... supernatural. :buck2:

supernatural · « **Reply #50 on:** September 29, 2010, 05:05:24 pm »

http://www.wolframalpha.com/input/?i=integrate+e%5Ex%5E2

the.watchman · « **Reply #51 on:** September 29, 2010, 05:11:05 pm »

My answer would be undefined, no 'dx'

Martoman · « **Reply #52 on:** September 29, 2010, 05:12:02 pm »

Quote from: Martoman on September 29, 2010, 05:00:05 pm

Quote from: supernatural on September 29, 2010, 04:54:31 pm
heres one for you

$\int e^{(x)^2}$

you can't actually solve that supernatural unless you did something.... supernatural. :buck2:

I meant you can't integrate it in terms of elementary functions

brightsky · « **Reply #53 on:** September 29, 2010, 10:31:44 pm »

Quote from: Martoman on September 29, 2010, 04:37:02 pm

Quote from: brightsky on September 29, 2010, 04:06:32 pm
Quote from: /0 on September 29, 2010, 02:47:43 am
gj Martoman xD

Quote from: brightsky on September 27, 2010, 10:20:42 pm
$\text{3.} \int^{1}_{0} \frac{\arctan(x)}{\log_e(x+1)} dx$

That doesn't have a primitive, is there a way without complex analysis? (Even that would be reaaally hard, if not impossible)

Hmm yeah my friend gave me that problem and I thought there was something wrong with it. I think it should be $\int^1_0 \frac{\arctan(x)}{1 + x}$ .

Polylogarithm functions aren't on the course (that is the integral deals with this). That one *can* be solved but its not possible at yr 12.

There's a trick that doesn't require you to actually find the indefinite integral. (Credits to my friend who showed me this.) For those interested, the answer is $\frac{\pi}{8} \log_e(2)$ .

supernatural · « **Reply #54 on:** October 04, 2010, 02:06:13 pm »

$\int \frac{sec(x)cos(2x)}{sin(x)+sec(x)}dx$

is that $ln(sin(2x)+2)$

Martoman · « **Reply #55 on:** October 04, 2010, 04:20:47 pm »

Quote from: supernatural on October 04, 2010, 02:06:13 pm

$\int \frac{sec(x)cos(2x)}{sin(x)+sec(x)}dx$

is that $ln(sin(2x)+2)$

Bang on the money (i bet someone will post something saying +c but meh :angel: )

TrueTears · « **Reply #56 on:** October 04, 2010, 04:25:44 pm »

Quote from: supernatural on September 29, 2010, 04:54:31 pm

heres one for you

$\int e^{(x)^2} dx$

Should be $\int_{-\infty}^{\infty} e^{x^2} dx = \sqrt{\pi}$

The famous Gaussian integral, can be easily evaluated using double integrals and a change of variables.

m@tty · « **Reply #57 on:** October 04, 2010, 04:31:58 pm »

Quote from: supernatural on October 04, 2010, 02:06:13 pm

$\int \frac{sec(x)cos(2x)}{sin(x)+sec(x)}dx$

is that $ln(sin(2x)+2)$

Thought it was interesting.. So I did it...

$I=\int \frac{\sec(x)\cos(2x)}{\sin(x)+\sec(x)}dx = \int \frac{\sec(x)\cos(2x)}{\sec(x)(\cos(x)\sin(x)+1)}dx$

$=2 \int \frac{\cos(2x)}{\sin(2x)+2}dx$

$\mbox{Let }u=\sin(2x)+2 \implies du=2\cos(2x)dx$

$I=\int \frac{1}{u}du$

$=\log_e(\sin(2x)+2)+C$

Martoman · « **Reply #58 on:** October 04, 2010, 04:42:51 pm »

Quote from: TrueTears on October 04, 2010, 04:25:44 pm

Quote from: supernatural on September 29, 2010, 04:54:31 pm
heres one for you

$\int e^{(x)^2} dx$
Should be $\int_{-\infty}^{\infty} e^{x^2} dx = \sqrt{\pi}$

The famous Gaussian integral, can be easily evaluated using double integrals and a change of variables.

You mean.... e^negativex^2

TrueTears · « **Reply #59 on:** October 04, 2010, 04:43:55 pm »

guess i still dont have enough sleep

Search the forums now!

We have moved!

Author Topic: Interesting questions (spesh) (Read 13693 times) Tweet Share

vea

Re: Interesting questions (spesh)

Martoman

Re: Interesting questions (spesh)

supernatural

Re: Interesting questions (spesh)

letsride

Re: Interesting questions (spesh)

Martoman

Re: Interesting questions (spesh)

supernatural

Re: Interesting questions (spesh)

the.watchman

Re: Interesting questions (spesh)

Martoman

Re: Interesting questions (spesh)

brightsky

Re: Interesting questions (spesh)

supernatural

Re: Interesting questions (spesh)

Martoman

Re: Interesting questions (spesh)

TrueTears

Re: Interesting questions (spesh)

m@tty

Re: Interesting questions (spesh)

Martoman

Re: Interesting questions (spesh)

TrueTears

Re: Interesting questions (spesh)

Recent Posts

User:
Password: