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May 17

Integration problem

Hi everyone. I'm studying for my calc 2 final. I was integrating sin^4(x)cos^3(x)dx, and got as an answer 1/5 sin^5(x) - 1/7 sin^7(x) + C.

However, putting this into Wolfram Alpha gets me something really weird: http://www.wolframalpha.com/input/?i=integrate+(sinx)%5E4+*+(cosx)%5E3

So my question is, am I correct in my computation, and if so why is WA being strange? Or is that just another way of expressing the same thing? — Preceding unsigned comment added by 140.233.173.11 (talk) 16:32, 17 May 2016 (UTC)

Also, how the heck do I attack #2 on this page? https://math.dartmouth.edu/archive/m11f97/public_html/m12integrals.pdf — Preceding unsigned comment added by 140.233.173.11 (talk) 17:02, 17 May 2016 (UTC)

@140.23.173.11: We will not do your homework for you. The only things I will give you are:

1. Remember that $\cos(2\theta )=\cos ^{2}(\theta )-\sin ^{2}(\theta )$ and that consequently, $\cos(2\theta )=1-2\sin ^{2}(\theta )$. Given that, the proof of the equivalence of the integrals is left as an exercise for you.
2. You want $\int {\frac {dx}{\sqrt {6x-x^{2}}}}$? Have you tried the "completing the square" strategy cited there? What if I rewrote this as $\int {\frac {dx}{\sqrt {9-(x^{2}-6x+9)}}}$?--Jasper Deng (talk) 17:40, 17 May 2016 (UTC)

It's not homework. It's optional. (You can see that yourself on the sheet). I'm just studying. But thanks for your help!

Also, I'm sorry if this is a stupid question, how did you get from $\int {\frac {dx}{\sqrt {6x-x^{2}}}}$ to $\int {\frac {dx}{\sqrt {9-(x^{2}-6x+9)}}}$? — Preceding unsigned comment added by 140.233.173.11 (talk) 18:14, 17 May 2016 (UTC)

@140.233.173.11: How do we complete the square without changing the value of the expression?--Jasper Deng (talk) 18:27, 17 May 2016 (UTC)

um... well I know that x^2+b+c can be written as (x+(b/2))^2 + something ... but this seems different somehow — Preceding unsigned comment added by 140.233.173.11 (talk) 18:36, 17 May 2016 (UTC)

@140.23.173.11: It's no different. You're given an expression of the form $x^{2}+bx$. Do you agree that I can then write it as $x^{2}+bx+(b/2)^{2}-(b/2)^{2}=(x+(b/2))^{2}-(b/2)^{2}$? Do you also agree that the contents of that square root are just this, just with a negative sign in front of the whole expression? What is b equal to in that expression?--Jasper Deng (talk) 18:41, 17 May 2016 (UTC)

Ah. I get it now. Sorry, that was stupid. Appreciate the help — Preceding unsigned comment added by 140.233.173.11 (talk) 18:58, 17 May 2016 (UTC)

Mind if I ask one more? (Maybe for someone other than Jasper :) )

For number 10 on the sheet, I used tan = sin/cos and substituted t= cos x. and eventually got ln(abs(cos x)), but WA laughs at me, and says the answer is 1/2 sec^2(x). What gives? — Preceding unsigned comment added by 140.233.173.11 (talk) 19:04, 17 May 2016 (UTC)

@140.233.173.11: Great substitution! However, you forgot about the denominator, it would seem! What do you get after you make the substitution?--Jasper Deng (talk) 19:12, 17 May 2016 (UTC)

I get $\int {\frac {-tdt}{t^{3}}}$ — Preceding unsigned comment added by 140.233.173.11 (talk) 19:21, 17 May 2016 (UTC)

So that means, $-\int {\frac {dt}{t^{2}}}$ . — Preceding unsigned comment added by 140.233.173.11 (talk) 19:25, 17 May 2016 (UTC)

(edit conflict) @140.233.173.11: Not quite! Let's look at it in more detail. You are using $t=\cos(x)$, so $dt=-\sin(x)dx$. Shouldn't this leave us with just $\int {\frac {-dt}{t^{3}}}$?--Jasper Deng (talk) 19:28, 17 May 2016 (UTC)

doesn't the t in the numerator kill one of the t's in the denominator?— Preceding unsigned comment added by 140.233.173.11 (talk • contribs)

@140.233.173.11: Look closely. Is there a t in the numerator?--Jasper Deng (talk) 19:41, 17 May 2016 (UTC)

Well, it's (sinx/cosx) / (cos^2(x)). In other words, (sinx/cosx) * (1/ cos^2(x)), in other words ... oh. Damn. Sorry. — Preceding unsigned comment added by 140.233.173.11 (talk) 19:45, 17 May 2016 (UTC)

Any rational function of sine and cosine can be integrated explicitly via the tangent half-angle substitution $t=\tan(\theta /2)$, together with the method of residues. Sławomir Biały (talk) 23:01, 17 May 2016 (UTC)

Also, for # 18, I reduced it to $4\int {\frac {dx}{x^{2}+4}}+\int {\frac {dx}{x^{3}+4x}}$ but I am not sure how to continue. — Preceding unsigned comment added by 140.233.173.11 (talk • contribs)

@140.233.173.11: You know how to integrate the first term using trigonometric substitution. For the second, use partial fraction decomposition, viz. ${\frac {1}{x^{3}+4x}}={\frac {1/4}{x}}+{\frac {-x/4}{x^{2}+4}}$. For the last term, what if we let $u=x^{2}+4$?--Jasper Deng (talk) 23:03, 17 May 2016 (UTC)

All those problems are very simple:

$\int {\frac {\tan x}{\cos ^{2}x}}dx=\int \tan x\,d(\tan x)={\frac {1}{2}}\tan ^{2}x+C$

$\int {\frac {dx}{x^{3}+4x}}=-\int {\frac {(1/x)d(1/x)}{1+4(1/x)^{2}}}=-{\frac {1}{8}}\ln |1+{\frac {4}{x^{2}}}|+C$

Ruslik_Zero 13:41, 18 May 2016 (UTC)

At the risk of annoying people, I want to ask one more. How do I attack #27? Is it a partial fractions probleM? — Preceding unsigned comment added by 140.233.173.11 (talk) 19:32, 18 May 2016 (UTC)

If you know enough to ask that question then you know enough to try and answer it yourself. --JBL (talk) 02:15, 19 May 2016 (UTC)

As I said above all those problems are very simple:

$\int {\frac {xdx}{x^{4}-16}}={\frac {1}{2}}\int {\frac {d(x^{2})}{(x^{2})^{2}-4^{2}}}={\frac {1}{16}}\ln \left|{\frac {x^{2}-4}{x^{2}+4}}\right|+C$

Ruslik_Zero 20:28, 19 May 2016 (UTC)

Generalizing curl

I've been told that the generalization of curl to higher dimensions is a non-trivial affair, even in nice Cartesian coordinates. It bothers me that a change from three dimensions to even one dimension higher is really complicated.

The way I am trying to do it is to generalize the circulation-based definition ${\text{curl}}\ \mathbf {F} \cdot {\hat {\mathbf {n} }}=\lim _{A\to 0}{\frac {1}{|A|}}\oint _{\partial A}\mathbf {F} \cdot d\mathbf {x}$ as given in the article. In three dimensions, it is straightforward to show that, when differentiation under the integral sign is permitted, and the curl exists, then it is given in coordinates by the usual formula using partial derivatives. However, in higher dimensions, this runs into problems, namely that if I pick a given coordinate unit vector, there is not a single plane that is orthogonal to it.

I've concluded, and have read, that curl can't be a vector then. Does that invalidate the above definition of the curl?--Jasper Deng (talk) 20:17, 17 May 2016 (UTC)

And yes, I'm aware of differential forms, but the only treatment of them I can understand is that of Rudin, which I've been told is not the best.--Jasper Deng (talk) 20:20, 17 May 2016 (UTC)

There is curl in higher dimensions, just not of "vector fields". The reason that the above formula makes sense is because the dot product $\mathbf {F} \cdot d\mathbf {x}$ is well-defined. The differential $d\mathbf {x}$ is a one-form, which can be integrated against over a curve. In higher dimensions, you need a version of $d\mathbf {x}$ that can be integrated over a higher dimensional set. If A is n-dimensional, and $\partial A$ is $n-1$ dimensional, then we need would need an $n-1$ form to integrate over $\partial A$. In indices, here is such a form:

$dx^{i_{1}}\wedge dx^{i_{2}}\wedge \cdots \wedge dx^{i_{n-1}}.$

There are $n-1$ indices that need to be contracted away, so the field F needs to be an $(n-1)$-vector (that is, a skew-symmetric tensor with $n-1$ indices). The curl in higher dimensions, defined by your formula, turns out to be equivalent to the exterior derivative, which follows from Stokes' theorem. Sławomir Biały (talk) 20:49, 17 May 2016 (UTC)

The connoisseurs will undoubtedly point out that I have played a bit loose with the covariance and contravariance. Perhaps, if they are astute, they may realize that in this description a "vector field" is really not a vector field at all, but is in fact a differential form.

I know, you said you heard of these, but I would like to add some perspective, particularly in relation to your formula. The perspective I wish to advance is that any pair of objects "curl F" and "F" that behave in reasonable ways, if they are to make sense and define a "true formula", then they are actually just differential forms, and curl is actually the exterior derivative. This is what might be called an "abstract" differential form.

The basic idea is that from the essential Stokes formula:

$\int _{A}{\text{curl}}\ \mathbf {F} \cdot d\mathbf {A} =\oint _{\partial A}\mathbf {F} \cdot d\mathbf {x}$

one can infer that F and curl F are dual to currents. Currents encompass the usual thing one may be familiar with from physics: a vector field, for example. But they also include other things like vector fields with delta-function profiles. Included in the menagerie of currents, there are objects like $(A,d\mathbf {A} )$ and $(\partial A,d\mathbf {x} )$ that represent a concept of integration over submanifolds. Currents form a natural homology theory (that is, there is a concept of "boundary" rather than "coboundary"). Anything dual to currents must behave functorially like a differential form rather than a vector field.

Stokes' theorem is a statement about duality between the exterior derivative on the (smooth) differential forms, and the boundary operator in a well-defined homology theory. Sławomir Biały (talk) 00:04, 18 May 2016 (UTC)Sławomir Biały (talk) 21:24, 17 May 2016 (UTC)

Perhaps, to clarify from the "Rudin" perspective, differential forms are things that are dual to chains. So if you ever encounter any kind of "reasonable" object that happens to be dual to chains, then that is actually a differential form. Sławomir Biały (talk) 00:33, 18 May 2016 (UTC)

In any number of dimensions, even three, the curl of a vector field is a bivector field. Bivectors aren't complicated, just unfamiliar.

You can understand the generalized curl via differential forms as Sławomir Biały suggested above, or via geometric algebra (aka Clifford algebra). The geometric algebra of a normed vector space is an algebra of scalars and vectors in which s+s' and v+v' sums, ss' and sv and vs products, and v² squared norm have their usual meanings from vector calculus, and there are no other constraints. Surprisingly, this generates a consistent and nontrivial mathematical structure. In it you can define a total derivative $\partial \!\!\!/$ (see Feynman slash notation), which when applied to a scalar field gives you the gradient, and when applied to a vector field gives you the sum of the divergence and the curl. In two real dimensions, the divergence+curl is a complex number (a value in the even part of the Clifford algebra, which is isomorphic to the complex numbers) with the divergence in the real part and the curl in the imaginary part. In three real dimensions it's a quaternion, and in 3+1 dimensions it's a biquaternion or Dirac spinor. In any number of dimensions it can be understood as an element of the Lie algebra of rotation and scaling, representing the orientation and volume change of a collection of dust particles under the flow defined by the vector field. -- BenRG (talk) 23:58, 17 May 2016 (UTC)

But isn't a bivector just the "ordinary curl"? What if A is not two-dimensional? Sławomir Biały (talk) 00:06, 18 May 2016 (UTC)

You're right, I only generalized to 2D surfaces in arbitrarily many dimensions. The original question was how you generalize the vector normal ${\hat {\mathbf {n} }}$ to more dimensions where there isn't a unique plane perpendicular to a vector. For 2D surfaces in general, the answer is that you use a bivector (a wedge product of vectors that span the tangent plane). The 3D vector normal is the Hodge dual of that bivector. For k-dimensional A you'd use a k-vector. (Or a k-form. I also didn't distinguish between k-vectors and k-forms because Clifford algebra doesn't.)

I think that $\partial \!\!\!/=\mathrm {d} +\delta$ where $\delta$ is the codifferential, so the Clifford version of the generalized Stokes theorem (for arbitrary-dimensional A in arbitrary-dimensional space) would combine the usual Stokes theorem and its Hodge dual. But I don't know how to write it. -- BenRG (talk) 02:55, 18 May 2016 (UTC)

I think the natural k-vector in this business is the dA in

$\int _{A}\operatorname {curl} F\cdot dA.$

For example, in two dimensions, we parameterize a surface by u and v using a vector valued function r, say. Then

$dA={\frac {\partial \mathbf {r} }{\partial u}}du\wedge {\frac {\partial \mathbf {r} }{\partial u}}dv$

which is a 2-vector. Sławomir Biały (talk) 10:19, 18 May 2016 (UTC)

Well as BenRG says the curl would normally be considered the Hodge dual of the exterior derivative of a 1-form and so be a k-2 form rather than a 2-form. I suppose that is just a bit of historical cruft though and it would really be better to just forget about the Hodge dual business when talking about curl and things like that. That has k(k-1)/2 dimension and in 2 dimensions the curl is a scalar and for 3 dimensions it is 3 dimensional vector. For 4 dimensions it has 6 dimensions and so can't be mapped into 4-space. I think it takes a bit of imagination to think of the space of 2-dimensional plane areas in 3-space as forming a 3 dimensional vector space never mind the six dimensions needed for 4-space areas. However you can get a 4-dimensional vector space of 3-dimensional volume in 4-space,, that would form the basis for a non-standard generalization. Dmcq (talk) 12:59, 18 May 2016 (UTC)

Yes, and I'm saying that it's more natural to regard the curl as a 2-form, if we are taking Jasper Deng's formula as a definition. 2-forms, not 2-vectors, are dual to cycles. Of course, once you start throwing in Hodge duality, metric tensors, etc, there are identifications that can be made. But the "natural duality", as I see it, is between cycles and forms, not cycles and k-vectors. Sławomir
Biały 13:43, 18 May 2016 (UTC)