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01-07-2011, 11:59 PM
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#1
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Senior Member
Join Date: Mar 2008
Location: Acton, MA
Posts: 1,687
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Need help with a calculus II problem...
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I had the bright idea to take an intensive Calculus II class over winter break. 5 days a week, 9 hours a day, 2 weeks = 4 credits
Anyhow, we found something really strange at the end of the day today while checking a problem we had worked out, and my professor told us to look into it over the weekend.
Anyhow, we were doing integration of trig functions using trig identities for u/du replacements. We solved the problem below, and got the second to last answer without a problem, however we don't understand the last line of the problem.
We can call (-ln|sin(x)|) part of C since it shows up on the last two lines of the solution.
So this means that -cot^2(x) = -csc^2(x) when X has restricted values? What we really want to know is how Wolfram Alpha knew to put a limit on the integral.
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01-08-2011, 01:09 AM
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#2
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Senior Member
Join Date: Nov 2008
Location: Kansas City
Posts: 3,657
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Isn't that only true when x = 0?
This seems to be some sort of Alpha idiosyncracity.
I don't see why you would restate the integral in a way that is not simpler and is only true for certain values of x...
When I took calculus back in the first decade of the 21st century, they made us take integrals by hand in first year calculus 
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01-08-2011, 01:41 AM
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#3
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Senior Member
Join Date: Jan 2009
Location: Richardson, TX
Posts: 1,175
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As far as I can tell,
cot^2(x) = csc^2(x) - 1 for all x
The only limitation needed as far as I can see is that log(sin(x)) is only defined (for real numbers) for sin(x) >=0; i.e., 0<=x<=pi, so maybe that's what they mean by restricted values?
EDIT: Maybe they're trying to eliminate x=0 and x=pi. At those two points, cot^2 and csc^2 go to inifinity, so that would have infinity^2 = infinity^2 - 1 
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01-08-2011, 02:36 AM
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#4
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Senior Member
Join Date: Nov 2008
Location: Kansas City
Posts: 3,657
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Quote:
Originally Posted by DeafSmith
As far as I can tell,
cot^2(x) = csc^2(x) - 1 for all x
The only limitation needed as far as I can see is that log(sin(x)) is only defined (for real numbers) for sin(x) >=0; i.e., 0<=x<=pi, so maybe that's what they mean by restricted values?
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Yeah, that's right.
Still not sure why with an answer expresses at cotangent you would re-express as cosecant with no simplification (with the obvious assumption that you can evaluate either function computationally in a trivial amount of time).
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01-08-2011, 09:54 PM
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#5
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turf monkey supreme
Join Date: Jan 2009
Location: Ames, Iowa
Posts: 3,075
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3, the answer is 3.
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01-08-2011, 10:17 PM
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#6
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Senior Member
Join Date: Nov 2008
Location: Jay, Adirondack Mountains, NY
Posts: 1,737
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Quote:
Originally Posted by android
3, the answer is 3.
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Wrong. It's banana.
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"...Careful, man...there's a beverage here!..."
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01-09-2011, 01:24 AM
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#7
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Senior Member
Join Date: Mar 2008
Location: Acton, MA
Posts: 1,687
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Thats the real issue, why would you restate the answer of an indefinite integral with limitations for real numbers. I agree that it's just Wolfram being wolfram. My professor knows Dr. Wolfram and said he'd talk to him if we can come up with a decent reason for them to remove that last step.
__________________
I'd rather have a bottle in front of me, than a frontal lobotomy.
--Tom Waits
You can't be a real country unless you have a beer and an airline - it helps if you have some kind of a football team, or some nuclear weapons, but at the very least you need a beer.
--Frank Zappa
My Cheap and Easy Stirplate
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01-09-2011, 01:25 AM
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#8
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Registered User
Join Date: Jun 2010
Location: gone
Posts: 388
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I always go with "all of the above"
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01-10-2011, 01:48 PM
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#9
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Senior Member
Join Date: Feb 2008
Location: Reed City, MI
Posts: 15,578
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42
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01-10-2011, 01:48 PM
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#10
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Senior Member
Join Date: Feb 2008
Location: Reed City, MI
Posts: 15,578
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Oops, sorry, wrong question!
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