Integral of trigonometric substitutions

In summary, the conversation discusses the use of trigonometric substitutions in integrating various expressions and the creation of a table for reference. The conversation also explores a problem with a specific integral and the potential use of trigonometric or other substitutions to solve it. The conversation ends with a discussion on the usefulness of tables for simplifying expressions and the importance of analytical math in finding answers to complex questions.
  • #1
Jhenrique
685
4

Homework Statement



Composing trigonometric functions, you realize that the main substitutions are related with the table below:
imagem2.png


So, I started to integrate each expression above and I created this other table:
imagem.png


But I had a problem with the integral circled in red, because I don't know how transform the arctan(...) in other expression that resembles with your adjacentes functions of upper and from below.

Homework Equations


The Attempt at a Solution


[tex]\int \frac{\sqrt{x^2-1}}{x}dx = \sqrt{x^2-1} + \frac{i}{2} \log(x^2 - 2i \sqrt{1-x^2}-2) -i \log(x)[/tex] You have some ideia that how make the integral of ##\frac{\sqrt{x^2-1}}{x}## be similar to integrals of ##\frac{\sqrt{x^2+1}}{x}## and ##\frac{\sqrt{1-x^2}}{x}## ?
 
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  • #2
Jhenrique said:

Homework Statement



Composing trigonometric functions, you realize that the main substitutions are related with the table below:
imagem2.png


So, I started to integrate each expression above and I created this other table:
imagem.png


But I had a problem with the integral circled in red, because I don't know how transform the arctan(...) in other expression that resembles with your adjacentes functions of upper and from below.

Homework Equations


The Attempt at a Solution


[tex]\int \frac{\sqrt{x^2-1}}{x}dx = \sqrt{x^2-1} + \frac{i}{2} \log(x^2 - 2i \sqrt{1-x^2}-2) -i \log(x)[/tex] You have some ideia that how make the integral of ##\frac{\sqrt{x^2-1}}{x}## be similar to integrals of ##\frac{\sqrt{x^2+1}}{x}## and ##\frac{\sqrt{1-x^2}}{x}## ?

Can't see the images clearly, and not entirely sure what you're attempting, but if you just want to get ##\int \frac{\sqrt{x^2-1}}{x}dx##, you're much better off with the sub ##x = \cosh y##.
 
  • #3
Or you could use a regular trig substitution, with θ = sec-1(x).

BTW, the index 2 in a square root is pretty much never shown. If there is no index, the radical is assumed to be a square root.

Also, what you have in the upper left corner of the first table is misleading/wrong. f(f-1(x)) = x, as long as x is in the domain of f-1
 
  • #4
Mark44 said:
Or you could use a regular trig substitution, with θ = sec-1(x).
No! I want to generate a formula!

BTW, the index 2 in a square root is pretty much never shown. If there is no index, the radical is assumed to be a square root.
Personal taste, because I think beautiful.

Also, what you have in the upper left corner of the first table is misleading/wrong. f(f-1(x)) = x, as long as x is in the domain of f-1
The idea is only explicit the order of composition.
 
  • #5
Mark44 said:
Or you could use a regular trig substitution, with θ = sec-1(x).
Jhenrique said:
No! I want to generate a formula!
For what purpose? Is it yet more formulas in your very long list of formulas? I have to ask why you are doing this, since this information is available in many other places. What I'm saying is that I don't see much value in your lists.

BTW, the index 2 in a square root is pretty much never shown. If there is no index, the radical is assumed to be a square root.
Jhenrique said:
Personal taste, because I think beautiful.
Then you're probably the only one. I have been looking at calculus books for more than 50 years, and I don't think I've seen a single one of them write a square root like this: ##\sqrt[2]{x^2 + 1}##. I.e., with an explicit index of 2.

Also, what you have in the upper left corner of the first table is misleading/wrong. f(f-1(x)) = x, as long as x is in the domain of f-1
Jhenrique said:
The idea is only explicit the order of composition.
It's wrong for almost all of the compositions, and it's silly for all of them. The notation f(f-1(x)) is the composition of a function f with its inverse.
 
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  • #6
While this looks like a completely futile exercise, it is interestingly enough quite important in some branches of mathematics. In particular, in the field of differential algebra, they question which functions exactly have an "elementary" primitive and which do not. Writing functions as logarithms/exponentials is actually the key to solving this question.

See: http://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra ) and http://en.wikipedia.org/wiki/Risch_algorithm

/end(random remark)
 
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  • #7
[tex]\int \frac{\sqrt{x^2-1}}{x} dx = \sqrt{x^2-1} - i \log(i+\sqrt{x^2-1})+i \log(x)+C[/tex]
 
  • #8
Jhenrique said:
[tex]\int \frac{\sqrt{x^2-1}}{x} dx = \sqrt{x^2-1} - i \log(i+\sqrt{x^2-1})+i \log(x)+C[/tex]
I'll leave the check of your antiderivative to you.

A different approach that doesn't involve imaginary numbers follows.

$$ \int \frac{\sqrt{x^2 - 1}dx}{x}$$
Let x = sec(θ), so dx = sec(θ)tan(θ)dθ, and tan(θ) = ##\sqrt{x^2 - 1}##

Using this trig substitution, the integral above becomes
$$ \int \frac{tan(θ)sec(θ)tan(θ)dθ}{sec(θ)}$$
$$ =\int tan^2(θ)dθ = \int (sec^2(θ) - 1)dθ$$
$$ = \int sec^2(θ)dθ - \int 1 dθ$$
$$ = tan(θ) - θ + C$$
Undoing the substitution, the above is equal to
$$\sqrt{x^2 - 1} - sec^{-1}(x) + C$$

If you differentiate the expression above, you get ##\frac{\sqrt{x^2 - 1}}{x}##, the original integrand.
 
  • #9
Exist a table of composite functions for sn, cn, sc...?
 
  • #10
Jhenrique said:
Exist a table of composite functions for sn, cn, sc...?
Does there exist a table like the one you have here? Is that what you're asking? You could probably find one online somewhere if you looked hard enough.

Speaking for myself, I don't have any use for such a table. If I want to simplify, say arcsin(cos(x)), I just draw a right triangle and label the sides and angles appropriately, and work it out.
 
  • #11
Mark44 said:
Does there exist a table like the one you have here? Is that what you're asking? You could probably find one online somewhere if you looked hard enough.
Is extremely difficult find someone or some article or book that approach the elipitc functions of A to Z, answering all questions...

Speaking for myself, I don't have any use for such a table. If I want to simplify, say arcsin(cos(x)), I just draw a right triangle and label the sides and angles appropriately, and work it out.

But this is the essence of the analytical math, you find for non-inuitive answers (not necessarily non-inuitive) that answers mechanically the questions. IMO.
 
  • #12
Jhenrique said:
Is extremely difficult find someone or some article or book that approach the elipitc functions of A to Z, answering all questions...
I don't see a connection between what you have put together in this table, and elliptic functions. See Elliptic functions.
Jhenrique said:
But this is the essence of the analytical math, you find for non-inuitive answers (not necessarily non-inuitive) that answers mechanically the questions. IMO.
Nonintuitive answers that are not necessarily nonintuitive? Huh?

That mechanically answer the questions? Do you mean with no thinking necessary?

I don't have any idea what you're saying here.
 
  • #13
Nonintuitive answers that are not necessarily nonintuitive? Huh?

That mechanically answer the questions? Do you mean with no thinking necessary?

I don't have any idea what you're saying here.

Pardon me, is little difficult express myself in english, but I'm trying to say that in certain case, like when you'll create a program of computer (like a program of integration, for example), how much more mechanical is your program, less it needs of AI for resolve the questions. So, I think that the analytical formulas are useful for this, because they not need of a interpretation inteligent, need just work.

BTW, I found this table in the wiki:
https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Related_identities
Very didactic for understand the tradicional trigonometric substitution.
 

What is the purpose of using trigonometric substitutions in integrals?

Trigonometric substitutions are used to simplify integrals involving trigonometric functions. They can help to transform complex expressions into simpler ones, making it easier to solve the integral.

How do I know when to use a trigonometric substitution?

Trigonometric substitutions are typically used when the integral contains a square root of a quadratic expression or a sum/difference of squares. It can also be used when there are sine, cosine, or tangent functions raised to an even power.

What are the most commonly used trigonometric substitutions?

The three most commonly used trigonometric substitutions are:
1) u = sin(x)
2) u = cos(x)
3) u = tan(x)
Other substitutions such as u = sec(x) and u = cot(x) can also be used in certain cases.

How do I solve an integral using a trigonometric substitution?

The first step is to identify the type of substitution needed based on the expression in the integral. Then, substitute the appropriate trigonometric function and its derivative into the integral. This will help to simplify the expression and make it easier to integrate. After integrating, don't forget to back-substitute and simplify the expression back to its original form.

What are some common mistakes to avoid when using trigonometric substitutions?

One common mistake is forgetting to back-substitute and simplify the expression after integrating. Also, it's important to make sure the limits of integration are changed accordingly when substituting in a new variable. Another mistake is using the wrong trigonometric substitution for a given integral, so it's important to carefully analyze the expression before making a substitution.

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