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**Krizalid****Member**- Registered: 2007-03-09
- Posts: 51

Hi y'all!!

I wanna start an

Rules are easy: person which solves an integral, may post the next one but he must have the approval of the person which proposed the previous integral if the answer is correct or not.

Let's start with an easy one

Compute

Best regards!

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

sec(x) - tan(x) + x + C?

A logarithm is just a misspelled algorithm.

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**Krizalid****Member**- Registered: 2007-03-09
- Posts: 51

That's correct.

You can post the next one.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

*Last edited by mikau (2007-07-11 19:51:38)*

A logarithm is just a misspelled algorithm.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Krizalid wrote:

but

hemust have the approval of the person which proposed the previous integral if the answer is correct or not

Why not just differentiate back and check? Also, is there any need to use blatantly sexist language?

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**Krizalid****Member**- Registered: 2007-03-09
- Posts: 51

mikau wrote:

sec(x) - tan(x) + x + C?

Sorry, you have to show us your solution (I dunno what I was thinking that I didn't ask you that).

JaneFairfax wrote:

Show your solution too.

JaneFairfax wrote:

Why not just differentiate back and check? Also, is there any need to use blatantly sexist language?

Well, actually when I was writing I didn't realize of that.

If you don't mind, please limit to make all type of disagreeable commentary. There're forms and forms to request the things.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Thats right, Jane. Good job! Looks like you're next as soon as you show your solution.

and here's my solution to the first problem.

∫sin x /(1 +sin x) dx = ∫[-sin^2(x) + sin x ]/(1 - sin^2(x)) dx

= ∫ -sin^2(x)/cos^2(x) - sin(x)/cos^2(x) dx

= ∫ -tan^2(x) + sec(x)tan(x) dx

= ∫ 1 -sec^2(x) + sec(x)tan(x)dx = x - tan(x) + sec(x) + C QED.

*Last edited by mikau (2007-07-12 08:26:04)*

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

JaneFairfax wrote:

Also, is there any need to use blatantly sexist language?

Which reminds me:

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

That last comic was superb!

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

i'll say!

We're still waiting for you to post the next one, Jane!

A logarithm is just a misspelled algorithm.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

Doesn't this sort of thing also belong in Puzzles and Games? It is *kind of* a game, after all, it does follow the rule of most forum-based games.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Maybe it used to, but now it belongs in the "Jane is mean " section.

We don't have one of those yet, so it might as well stay here.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

mathsyperson wrote:

Jane is mean

Why?

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

is that meant to be interpreted as:

or

?

The Beginning Of All Things To End.

The End Of All Things To Come.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oops, typo. That should read "mathsy is dumb "

For some reason I read that as e^-x², and thought you were giving an impossible one. The didn't help either.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

*Last edited by JaneFairfax (2007-07-16 00:22:43)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

mathsyperson wrote:

is continuous for all real For some reason I read that as e^-x², and thought you were giving an impossible one. The didn't help either.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Now Jane, go again.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Yes.

Ive also seen

I believe you can integrate an absolutely convergent series term by term within its radius of convergence, and the result would be another absolutely convergent series?

Rickys turn.

*Last edited by JaneFairfax (2007-07-16 11:02:54)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Pass.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Krizalid****Member**- Registered: 2007-03-09
- Posts: 51

I forgot to mention that don't post integrals which don't have a primitive function in elementary terms.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Pass? :-( Who's next?

how about

this was in one of my books, i could never figure out how to do it. i think its non elementary.

*Last edited by mikau (2007-07-16 12:36:11)*

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Not impossible but it involves complex analysis. I would suggest picking a different integral.

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