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[Edited on September 24, 2007 at 6:43 PM. Reason : .]

9/24/2007 6:19:07 PM

your last name is iran?

9/24/2007 6:29:22 PM

it's Liran... Iran would be kinda freaky.... don't you think?

9/24/2007 6:35:00 PM

first you must prove your worth. We'll start out easy,

Int ( x / (x-1) )dx

9/24/2007 9:23:24 PM

you have the number and you have the email.
I'll show you a few ways to approach the problem.
feel free to contact

9/25/2007 10:42:32 AM

Its a scam if he can't answer that.

9/25/2007 11:18:17 AM

Quote :

"first you must prove your worth. We'll start out easy, Int ( x / (x-1) )dx"

um... (x-1) + ln(abs(x-1)) + "C"

[Edited on September 25, 2007 at 4:36 PM. Reason : +C]

9/25/2007 4:36:04 PM

^WRONG!

x + ln(x-1)

9/25/2007 6:52:24 PM

it's entirely correct. the substitution i used produces (x-1), however, that -1 can be assigned to part of the "C" constant yielding:

x + ln(abs(x-1)) + "C"

and yes, the absolute value DOES belong there

9/25/2007 9:19:06 PM

Yea i dont htink you can do it without the abs.

9/25/2007 9:32:39 PM

ha, if you were a good tutor you would have already answered the question. See even TWW is better for free!

9/25/2007 10:07:57 PM

To be fair I'll give you another chance, this time we adjust the level of skill needed, level up as it were,

Int ( sec(x) dx )

good luck to you sir.

9/25/2007 11:42:03 PM

how is this one any harder?


By the way, I'll tutor calculus 1, 2, 3, differential equations, statisticsor just about any other undergraduate level applied math course (no number theory) in exchange for lunch.

[Edited on September 26, 2007 at 12:34 AM. Reason : -]

9/26/2007 12:21:42 AM

ln (abs (sec x + tan x)) + C

TI-89 FTW




[Edited on September 26, 2007 at 1:13 AM. Reason : ]

9/26/2007 1:11:04 AM

Quote :

"it's entirely correct. the substitution i used produces (x-1), however, that -1 can be assigned to part of the "C" constant yielding:"

I shoulda rephrased, yes it's right, but we all know 99.9% of calc teachers would of marked something off.

[Edited on September 26, 2007 at 7:46 AM. Reason : ]

9/26/2007 7:45:48 AM

Would have marked wrong on your answer.

9/26/2007 8:28:08 AM

^^ your answer on the other hand was ACTUALLY wrong

9/26/2007 4:19:15 PM

yeah I know I didn't put +C or abs value, blah blah blah.

9/26/2007 5:44:57 PM

those abs value bars are basically junk anyhow.

they just |clutterin| up |shit|

who needs em?

9/26/2007 7:23:02 PM

joe_schmoe, hey, calculators are cheating.

Of course the TI-89 can do much more than most of us.

LimpyNuts you think the integral of sec(x) is not much harder than
the integral of x/(x-1) ? How would you do the integral of sec(x) ?
As far as I know sec(x) is not that easy w/o a trick.
x/(x-1) on the other hand is like 2 lines.

anyway apparently jobarc does not have the goods.

9/26/2007 7:28:09 PM

^^ integrate 1/x from -5 to -3 w/o the absolute value bars, see how that goes for you.

9/26/2007 7:29:13 PM

it's a joke son, a joke Ah say!





(* i think it's too quick for him *)

9/26/2007 7:53:32 PM

Quote :

"LimpyNuts you think the integral of sec(x) is not much harder than the integral of x/(x-1) ? How would you do the integral of sec(x) ? As far as I know sec(x) is not that easy w/o a trick. x/(x-1) on the other hand is like 2 lines."

sec(x)*(sec(x)+tan(x))/(sec(x)+tan(x)) dx = dY/Y where Y = sec(x)+tan(x)
... ---> ln(abs(sec(x)+tan(x)))

you can do it in your head... all it requires is that you know what the derivative of sec(x) is.

9/26/2007 9:08:51 PM

^yeah that's the trick, where did you learn it ?
That integral is like a half-page w/o that very nonintuitive substitution.

and to joe_schmoe I knew you were joking, just speaking to the abs-value haters out there.

9/26/2007 9:47:34 PM

9/26/2007 10:09:50 PM

^^ i didn't learn the "trick" anywhere. it was pretty apparent to me. of course after you're given the answer taking the derivative gives you the expression.

9/26/2007 10:14:41 PM

seriously, you just look at Int( sec(x)dx ) and think well maybe I should try Y = sec(x) + tan(x) ?

That's obvious ?

That's what I teach my students, but I doubt anybody would dream that up w/o some help.

If you did then I'm impressed, seriously.

To compare, something like Int( x*exp(x^2) dx ) just screams u = x^2 in your face.

The question is why try Y= sec(x) + tan(x) ?

The answer "it works" is insufficient.

[Edited on September 27, 2007 at 12:00 AM. Reason : .]

9/26/2007 11:58:10 PM

My line of thought:

hmm.. trig substitution?
*thinks about triangles for a second*
no...
well d(tan(x))=sec^2(x)
what's the derivative of sec(x) again? oh yeah sec(x)*tan(x)
*click*
sec(x)*(sec(x)+tan(x)) = d(sec(x)+tan(x))/dx


maybe i'm just special? when i seen an integral that isn't obvious i just start trying to work backwards. it usually works for integrals with simple expressions. but once you know the trick you can get:

int(csc(x)) = -ln(abs(csc(x)+cot(x))) ----- d(csc(x)) = csc(x)cot(x)
int(tan(x)) = -ln(abs(cos(x)) ... etc.
int(cot(x)) = ln(sin(x))

9/27/2007 1:31:47 AM

^ nice way of thinking. But, are you using the answer to guess the u-sub. ?
Interesting idea. I learned the trick from a friend. We didn't do that sort of thing in my calc II so I only learned of it when I started teaching.

9/27/2007 11:15:50 PM

for the sec(x) problem i didn't know the answer before i figured out how to get it. what i meant was that i applied the same thought process to obtain the other 3 integrals, which are also not immediately apparent.

9/28/2007 12:15:18 AM

Nice work indeed.^

Let me see if I can come up with something a little harder,

Int ( 1 / [1 + sin(x) + 2*cos(x) ] )dx

Or to make it more fun,

Int ( [ A*cos(x) + B*sin(x) + C ] / [ D*cos(x) + E*sin(x) + F ] )dx

for arbitrary, but sensible, constants A,B,C,D,E,F.

9/28/2007 12:58:31 AM

^not difficult, per se, but extremely time consuming.

inverse tangents, FTL

9/30/2007 7:03:20 PM

something like that, but I don't think I'd classify it not difficult.

What do you consider difficult integration?

btw the silence of jobarc is deafening.

9/30/2007 7:29:52 PM

well this thread failed

9/30/2007 7:41:08 PM

^^ difficult in this context would be something that the average person (of the intended audience... i.e. calculus tutors) would not recognize how to solve immediately.

That said, I would say the previous problem and the sec(x) problem are of moderate difficulty. ln(cosh(x)), for example could be considered very difficult. While I know it has an answer, I've never been able to attain it.

9/30/2007 9:59:44 PM

^^ no the thread fails you.
^ hmm... interesting. I'll ask my math muse later.

9/30/2007 10:49:13 PM

The only simple method for Int(ln(cosh(x)) would be to cheat and go here:

http://integrals.wolfram.com/index.jsp

The result is certianly non-obvious. My gut feeling is to expand everything in terms of taylor series and integrate term by term. You could at least get an approximate solution. If you guys can think of a more elegant way, please share. I never did take functional analysis.

10/1/2007 10:24:06 AM

you wouldn't learn that in functional analysis anyway.

10/1/2007 11:47:59 AM

^^ hmm... polylogarithms... this is a special functions problem then. So your gut feeling is probably spot on, I've still got a lot to learn about special functions myself.

Seeing ln(cosh(x)) makes me think some sort of ma 513 trickery may help. But the poylog's in the answer tell me it's above my current skill level.

^ yeah really. You don't really learn how to calculate anything in functional analysis. Unless you count
calculating the norm of an infinite dimensional vector, ok I guess that's something. Anyway, functional analysis is really about learning how to avoid the paradoxes the can arise from sloppy math-think.

10/1/2007 1:15:53 PM