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Old 01-07-2011, 11:59 PM   #1
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Default Need help with a calculus II problem...

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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Old 01-08-2011, 01:09 AM   #2
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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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Old 01-08-2011, 01:41 AM   #3
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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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Old 01-08-2011, 02:36 AM   #4
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Quote:
Originally Posted by DeafSmith View Post
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?
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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Old 01-08-2011, 09:54 PM   #5
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3, the answer is 3.
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Old 01-08-2011, 10:17 PM   #6
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Quote:
Originally Posted by android View Post
3, the answer is 3.
Wrong. It's banana.
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Old 01-09-2011, 01:24 AM   #7
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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.
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Old 01-09-2011, 01:25 AM   #8
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I always go with "all of the above"
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Old 01-10-2011, 01:48 PM   #9
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42
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Old 01-10-2011, 01:48 PM   #10
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Oops, sorry, wrong question!


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