How to integrate with functions of differentials?

In summary, the author falls and writes a wrting that integrates to the double of the tangeant, in series, but then gets confused about parentheses.
  • #1
jk22
729
24
I fell upon such a wrting :

$$du=tan(d\theta)$$

How to integrate this ?

I didn't try numerically but I thought of expanding the tangeant in series but then should for example $$\int d\theta^2$$ be understood as a double integration ?
 
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  • #2
No, ##d\theta^2 = 2\theta d\theta##.

You fell ? Did it hurt ? 😉
Was there a context that can help us make it clear to you or did it get lost in the fall ?
 
  • #3
BvU said:
No, ##d\theta^2 = 2\theta d\theta##.
it depends whether we mean ##d(\theta^2)## or ##(d\theta)^2##

And for me to answer the OP's question ##\int (d\theta)^n## is zero for all ##n\geq 2##
 
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  • #4
@Delta2 corrected me rightly: the Taylor series at ##\theta = 0## contains ##(d\theta)^n##, not ##d\theta^2##.

Still curious about the context ...
 
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  • #5
Yes missed parentheses from my part.

The context is easy :

Let be the parametrization of a unit sphere ##\vec{r}(\theta,\phi)=(cos\theta cos\phi,cos\theta sin\phi,sin\theta),\theta\in [-\pi/2,\pi/2],\phi\in [0,2\pi] ## and the associated basis vectors ##\vec{e}_\theta=\frac{\partial\vec{r}}{\partial\theta}## and ##\vec{e}_\phi##

The metric is ##g=diag(1,cos(\theta)^2)##

I wanted to orthonormalize the metric via the associated quadratic form : ##\underbrace{(du\vec{e}_\theta+dv\vec{e}_\phi)}_{\vec{x}}g\vec{x}^T##

The ##du,dv## are infinitesimals on the tangent plane and I express them as a function of ##\theta,\phi##: ##du=tan(d\theta)##
##dv'=cos(\theta)^2tan(d\phi)##

Instead of linearizing I did what I wrote in the OP with successive integrations (##\int(d\theta)^n=\theta^n/n!## by fixing all the successive constants of integration to 0) and found
##u=\sum_{n=0}^\infty(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}\frac{\theta^{2n+1}}{(2n+2)!(2n+1)!}##

I wanted to compute the shape of u,v which if you want should be a map of the sphere.

Is it periodic and what's the asymptotic behaviour ?

Could anyone help me or give me references in webpages, it should be found in common textbooks about differential geometry or map projections I suppose.

What I found weird is that if I consider a path on the sphere ##\theta(t),\phi(t)## then the endpoint coordinate on the map could depend on the path itself ?!🤔
Thanks.
 
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  • #6
I think ##du=\frac{d(\tan{x})}{dx}## at 0: ##d(\tan{x})=(\tan{(0+dx)}-\tan{0})dx##. So perhaps integrating this would be ##\int du \cdot dx=\int (1+\tan^2{0})dx##?
 
  • #7
##du## is an infinitesimal while ##\frac{d(\tan(x))}{dx}## is a finite quantity if ##|x|<\frac{\pi}{2}##. Did you mean ##du=d\tan(\theta=x)=(1+\tan(\theta)^2)d\theta##, so that we have not ##\tan(d\theta)##, which is meaningless ?

I cannot anymore link geometry with algebra, it is as if the same geometric case could lead to several formulas.
 
  • #8
jk22 said:
##du## is an infinitesimal while ##\frac{d(\tan(x))}{dx}## is a finite quantity if ##|x|<\frac{\pi}{2}##.
Yes, that is more-or-less correct. The expression ##\frac{d(\tan(x))}{dx}## is the derivative of the tangent function, with respect to its variable x.

The substitution was ##u = \tan(x)## from which the differentials of the two sides would be as you have below. Note that we would usually write ##\tan^2(\theta)## rather than ##\tan(\theta)^2##.

However, I'm not sure this is relevant to your problem where you had ##du = \tan(d\theta)##. I would be interested in seeing the work that led up to this equation.
jk22 said:
Did you mean ##du=d\tan(\theta=x)=(1+\tan(\theta)^2)d\theta##, so that we have not ##\tan(d\theta)##, which is meaningless ?

I cannot anymore link geometry with algebra, it is as if the same geometric case could lead to several formulas.
 
  • #9
Mark44 said:
However, I'm not sure this is relevant to your problem where you had ##du = \tan(d\theta)##. I would be interested in seeing the work that led up to this equation.

In fact there is no work : I suppose a circle and the tangent line at coordinate ##\theta##, then an infinitesimal change ##d\theta## leads to a change ##du=\tan(d\theta)## on this line, with just basic trigonometrical geometry.
 

Related to How to integrate with functions of differentials?

1. What is the purpose of integrating with functions of differentials?

Integrating with functions of differentials allows us to find the area under a curve or the total change in a function over a given interval. It is a fundamental tool in calculus and is used in many scientific fields to solve problems involving rates of change.

2. How do I integrate with functions of differentials?

To integrate with functions of differentials, we use a process called integration, which is the inverse operation of differentiation. This involves finding the antiderivative of the function and then evaluating it at the upper and lower limits of the integration.

3. What is the difference between indefinite and definite integration?

Indefinite integration involves finding the general antiderivative of a function, while definite integration involves finding the specific value of the integral over a given interval. Indefinite integration results in a function, while definite integration results in a numerical value.

4. What are some common techniques for integrating with functions of differentials?

Some common techniques for integrating with functions of differentials include substitution, integration by parts, trigonometric substitution, and partial fractions. These techniques are used to simplify the integrand and make it easier to evaluate the integral.

5. How can I check if my integration is correct?

You can check your integration by taking the derivative of the antiderivative you found. If the result is equal to the original function, then your integration is correct. You can also use online tools or graphing software to graph both the original function and its antiderivative to visually confirm the accuracy of your integration.

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