Math Is Fun Forum

great_math

Member

Registered: 2008-10-07

Posts: 5

Integrate:

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Integrate:

According to wolfram's integrator; the answer to this is pretty darn complex (although still analytical)

http://integrals.wolfram.com/index.jsp?expr=1%2F((Sin[x])^2+%2B+Sin[x]%2B+1)&random=false

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All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Integrate:

According to wolfram's integrator; the answer to this is pretty darn complex (although still analytical)

http://integrals.wolfram.com/index.jsp?expr=1%2F((Sin[x])^2+%2B+Sin[x]%2B+1)&random=false

Interestingly, my Ti-89 returned a real (i.e. ∈ R) answer.

Last edited by All_Is_Number (2008-11-04 05:53:20)

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Integrate:

All_Is_Number, you probably mixed up the variable you're integrating with.  If I remember right (and it has been a while), you want:

Integrate(1/(sin(x)^2 + sin(x) + 1), x)

To solve the integral, remember that:

I haven't the time to work it out any further, but this looks like the trick you need.

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All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Integrate:

All_Is_Number, you probably mixed up the variable you're integrating with.

No, I integrated with respect to x. There was no i in the result, which is how I should have phrased my initial reply.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Integrate:

If you use the suggestion I made, you should get a result by using a mix of u-sub, trigonometric identities, and integration by parts.  The answer is:

http://integrals.wolfram.com/index.jsp?expr=(Sin[x]^3+-+1)%2F(Sin[x]+%2B+1)&random=false

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"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..."

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Integrate:

If you use the suggestion I made, you should get a result by using a mix of u-sub, trigonometric identities, and integration by parts.  The answer is:

http://integrals.wolfram.com/index.jsp?expr=(Sin[x]^3+-+1)%2F(Sin[x]+%2B+1)&random=false

Your observation that (1+sin(x)³)/(1+sin(x))=sin(x)²+sin(x)+1 does simplify the integration. My comment was only regarding the difference in the form of the answers given by my Ti-89 and Mathematica.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Integrate:

All, I'm not trying to argue with you...  I wasn't responding to your post, just clarifying my previous one.

And again, this comment is not directed to you, but to everyone in general, and specifically great_math:

What's really cool is that had I not taken a graduate course in abstract algebra, I would have been clueless on how to solve this integral.  Indeed, the identity I mentioned came from Gauss when he was studying cyclotomic polynomials.  He was doing so was to investigate the roots of unity and their algebraic properties.

Studying roots of polynomials transferred over directly over to being able to compute an integral.  That is beauty in mathematics, and also demonstrates how unexpected and useful results can come about.

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"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..."