Author Topic: Interesting questions (spesh) (Read 11830 times) Tweet Share

tram · « **Reply #30 on:** September 28, 2010, 08:33:31 am »

hmmmm these questions are quite good.... still working on them tho.....

EDIT: hmmmm these questions are ~~quite good~~ FREKIN HARD

Mao · « **Reply #31 on:** September 28, 2010, 08:05:45 pm »

These are old school, but try:

$\int \sec (x) \; dx$

$\int \sec^3 (x) \; dx$

brightsky · « **Reply #32 on:** September 28, 2010, 08:35:23 pm »

Quote from: Mao on September 28, 2010, 08:05:45 pm

$\int \sec (x) \; dx$

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$\int \sec(x) dx = \int \frac{1}{\cos(x)} dx = \int \frac{\cos(x)}{\cos^2(x)} dx = \int \frac{\cos(x)}{1 - \sin^2(x)} sx$

Let $u = \sin(x)$ , then $du = \cos(x) dx$

$\implies \int \frac{1}{1 - u^2} du$

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

Partial decomposition is of the form $\frac{A}{1 + u} + \frac{B}{1 - u}$

Let $\frac{1}{(1+u)(1-u)} = \frac{A}{1 + u} + \frac{B}{1 - u}$

$1 = A(1-u) + B(1 + u)$

$1 = A - Au + B + Bu$

$1 = u(B - A) + (A+B)$

Hence $B-A = 0$ and $A + B = 1$

$A = \frac{1}{2}$ , $B = \frac{1}{2}$

So $\int \left( \frac{\frac{1}{2}}{1 + u} + \frac{\frac{1}{2}}{1 - u} \right) du$

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

$= \frac{1}{2} \int \left( \frac{1}{1 + u} + \frac{1}{1-u} \right) du$

Integrate term by term:

$= \frac{1}{2} (\log_e|1 + u| - \log_e|1 - u|) + C$

Substitute $u = \sin(x)$

$= \frac{1}{2} (\log_e|1 + \sin(x)| - \log_e|1 - \sin(x)|) + C$

Quote from: Mao on September 28, 2010, 08:05:45 pm

$\int \sec^3 (x) \; dx$

Using same technique as before:

$\int \sec^3(x) dx = \int \frac{1}{\cos^3(x)} dx = \int \frac{\cos(x)}{\cos^4(x)} dx = \int \frac{\cos(x)}{(1 - \sin^2(x))^2} dx$

Let $u = \sin(x)$ , $du = \cos(x) dx$

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

Partial decompose it and integrate term by term.

sajib_mostofa · « **Reply #33 on:** September 28, 2010, 10:25:56 pm »

You are a god Brightsky

Chavi · « **Reply #34 on:** September 28, 2010, 11:45:38 pm »

Solutions to the final problem in the first post:
$\int \frac{1}{e^{3x}+4}dx$

let $u=e^{3x}+4$ $\therefore du = 3e^{3x}dx$

$\implies \int \frac{1}{u} \frac{du}{3e^{3x}}$ but, $e^{3x}=u-4$ and $3e^{3x}=3(u-4)$

$\therefore \frac{1}{3} \int \frac{1}{u(u-4)}du$

$\implies$ cbf writing out partial fractions in latex

$\frac{1}{12} \int \frac{1}{u-4} - \frac{1}{u}du = \frac{1}{12} \ln|\frac{u-4}{u}| +c$ c is an element of R

$\therefore \int \frac{1}{e^{3x}+4}dx = \frac{1}{12} \ln|\frac{e^{3x}}{e^{3x}+4}| +c$

Martoman · « **Reply #35 on:** September 29, 2010, 12:42:37 am »

Quote from: Chavi on September 28, 2010, 11:45:38 pm

Solutions to the final problem in the first post:
$\int \frac{1}{e^{3x}+4}dx$

let $u=e^{3x}+4$ $\therefore du = 3e^{3x}dx$

$\implies \int \frac{1}{u} \frac{du}{3e^{3x}}$ but, $e^{3x}=u-4$ and $3e^{3x}=3(u-4)$

$\therefore \frac{1}{3} \int \frac{1}{u(u-4)}du$

$\implies$ cbf writing out partial fractions in latex

$\frac{1}{12} \int \frac{1}{u-4} - \frac{1}{u}du = \frac{1}{12} \ln|\frac{u-4}{u}| +c$ c is an element of R

$\therefore \int \frac{1}{e^{3x}+4}dx = \frac{1}{12} \ln|\frac{e^{3x}}{e^{3x}+4}| +c$

A far more elegant solution exists I'm hoping someone other than brightsky will see what to do.

Quote from: brightsky on September 28, 2010, 08:35:23 pm

Quote from: Mao on September 28, 2010, 08:05:45 pm
$\int \sec (x) \; dx$
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$\int \sec(x) dx = \int \frac{1}{\cos(x)} dx = \int \frac{\cos(x)}{\cos^2(x)} dx = \int \frac{\cos(x)}{1 - \sin^2(x)} sx$

Let $u = \sin(x)$ , then $du = \cos(x) dx$

$\implies \int \frac{1}{1 - u^2} du$

lol this is the exact way I went about it i'm now doing it in terms of $e^{ix}$ ~~for fun~~

scrath that i seem to be doing something wrong :S

We know that $cos(x) = \frac{e^{ix}+e^{-ix}}{2}$

$\int \frac{1}{cos(x)} = \int \frac{1}{\frac{e^{ix}+e^{-ix}}{2}} =2 \int \frac{1}{e^{ix}+e^{-ix}}$

Using uber leet hacks we multiply the bottom and top by $e^{-ix}$ (incidentially Chavi this is a HUGE hint for your problem)

$2 \int \frac{e^{-ix}}{1+e^{-2ix}}$

clearly this is in the form of the arctan integration so making $u = e^{-ix}$

$du = -ie^{-ix}dx$

$\frac{-2}{i}tan^{-1}(e^{-ix})$ = $-2i \times tan^{-1}(e^{-ix})$

how can i proceed?

/0 · « **Reply #36 on:** September 29, 2010, 01:24:46 am »

Have a go at these:

$\int \frac{1}{1+\sin{x}-\cos{x}}\,dx$

$\int \frac{1+2x}{\sqrt{1-3x-2x^2}}\,dx$

$\int_{-\infty}^\infty \frac{\sin^3{x}}{x^2}\,dx$ (take for granted that $\lim_{x \to 0}\frac{sin{x}}{x}=0$ .)

Also, the following aren't that interesting, but you should know them

$\int \frac{1}{\sqrt{ax^2+bx+c}}\,dx$ , $\int \frac{1}{ax^2+bx+c}\,dx$

Chavi · « **Reply #37 on:** September 29, 2010, 01:49:19 am »

I'm really liking these "uber leet hacks"

$\frac{1}{e^{3x}+4}$ multimply numerator and demonator by $e^{-3x}$ so that $\frac{1}{e^{3x}+4}$ = $\frac{e^{-3x}}{1+4e^{-3x}}$

$\therefore \frac{-1}{12} \int \frac{-12e^{-3x}}{1+4e^{-3x}}dx$ = $\frac {-1}{12} \ln (1+4e^{-3x})$ = $\frac{1}{12} \ln (\frac{e^{3x}}{e^{3x}+4})$

Still attempting Brightsky's 3rd problem integrating arctan and log

Martoman · « **Reply #38 on:** September 29, 2010, 01:55:30 am »

Quote from: Chavi on September 29, 2010, 01:49:19 am

I'm really liking these "uber leet hacks"

$\frac{1}{e^{3x}+4}$ multimply numerator and demonator by $e^{-3x}$ so that $\frac{1}{e^{3x}+4}$ = $\frac{e^{-3x}}{1+4e^{-3x}}$

$\therefore \frac{-1}{12} \int \frac{-12e^{-3x}}{1+4e^{-3x}}dx$ = $\frac {-1}{12} \ln (1+4e^{-3x})$ = $\frac{1}{12} \ln (\frac{e^{3x}}{e^{3x}+4})$

Still attempting Brightsky's 3rd problem integrating arctan and log

I thought you'd like it

fuck that first one is good /0.
SPOILERS
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Ok so you have to recognise dops

$\int \frac{1}{1+sin(x)-cos(x)}dx = \frac{1-sin(x)-cos(x)}{(1+sin(x)-cos(x))(1-sin(x)-cos(x))}$

Sexy dops leads to:

$\frac{1-sin(x)-cos(x)}{1^2-(sin(x)-cos(x))^2}$

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

Magic identity leads to

$\int \frac{1-sin(x)-cos(x)}{1-1 +2sin(x)cos(x)}dx$

$\int \frac{1-sin(x)-cos(x)}{2sin(x)cos(x)}dx$ splitting up the terms we get

$\int \frac{1}{2sin(x)cos(x)}dx -\int \frac{sin(x)}{2sin(x)cos(x)}dx -\int \frac{cos(x)}{2sin(x)cos(x)}dx$

Indentity skillz lead to

$\int \frac{1}{sin(2x)}dx - \frac{1}{2} \int \frac{1}{cos(x)}dx - \frac{1}{2} \int \frac{1}{sin(x)}dx$

(now it follows the same logic as Mao's post... this is why I like this it contains Mao's post and some nice trig!)

$\int \frac{sin(2x)}{1-cos^2(2x)}dx - \frac{1}{2} \int \frac{cos(x)}{1-sin^2(x)}}dx - \int \frac{sin(x)}{1-cos^2(x)}}dx$

From left to right

$u = cos(2x), v = sin(x), w = cos(x)$ and as the King says: etc etc etc :smitten:

/0 · « **Reply #39 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)

Martoman · « **Reply #40 on:** September 29, 2010, 03:06:34 am »

This is what i was thinking it didn't feel like it too me and ty /0 luv ya

Mao · « **Reply #41 on:** September 29, 2010, 03:49:38 am »

Tut tut martoman, got a sign wrong in that epic thing.

Quote from: Martoman on September 29, 2010, 01:55:30 am

$\int \frac{1}{1+sin(x)-cos(x)}dx = \frac{1-sin(x)-cos(x)}{(1+sin(x)-cos(x))(1-sin(x)-cos(x))}$

Sexy dops leads to:

$\frac{1-sin(x)-cos(x)}{1^2-(sin(x)-cos(x))^2}$

You actually meant to multiply top and bottom by $1-\sin(x)+\cos(x)$ yeah? [It just means the final integral on your last line is a +.] Otherwise, everything else is correct (checked with software). =]

Mao · « **Reply #42 on:** September 29, 2010, 03:51:27 am »

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

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)

Mathematica seem to think that there isn't a 'exact' answer. I'm going to side with 'impossible'.

Martoman · « **Reply #43 on:** September 29, 2010, 04:18:29 am »

That second one is particularly nasty.

It involves a trig substitution. Die.

Quote from: Mao on September 29, 2010, 03:49:38 am

Tut tut martoman, got a sign wrong in that epic thing.

Quote from: Martoman on September 29, 2010, 01:55:30 am

$\int \frac{1}{1+sin(x)-cos(x)}dx = \frac{1-sin(x)-cos(x)}{(1+sin(x)-cos(x))(1-sin(x)-cos(x))}$

Sexy dops leads to:

$\frac{1-sin(x)-cos(x)}{1^2-(sin(x)-cos(x))^2}$

You actually meant to multiply top and bottom by $1-\sin(x)+\cos(x)$ yeah? [It just means the final integral on your last line is a +.] Otherwise, everything else is correct (checked with software). =]

And SIGH.... $-(x-y) = -xPLUSy$

brightsky · « **Reply #44 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 $\int^1_0 \frac{\arctan(x)}{1 + x}$ .

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