Need help with a calculus II problem...

[Anthony_Lopez]

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:rockin:

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.

 

[remilard]

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

 

[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?


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

 

[remilard]

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).

 

[android]

3, the answer is 3.

 

[KayaBrew]

3, the answer is 3.

Wrong. It's banana.

 

[Anthony_Lopez]

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.

 

[Frankfurtvr4]

I always go with "all of the above"

 

[Homercidal]

42

 

[Homercidal]

Oops, sorry, wrong question!