# Integral of trigonometric substitutions

1. Mar 31, 2014

### Jhenrique

1. The problem statement, all variables and given/known data

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

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

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.

2. Relevant equations
3. The attempt at a solution
$$\int \frac{\sqrt{x^2-1}}{x}dx = \sqrt{x^2-1} + \frac{i}{2} \log(x^2 - 2i \sqrt{1-x^2}-2) -i \log(x)$$ 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 $\frac{\sqrt{1-x^2}}{x}$ ?

2. Mar 31, 2014

### Curious3141

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. Mar 31, 2014

### Staff: Mentor

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. Apr 2, 2014

### Jhenrique

No! I want to generate a formula!

Personal taste, because I think beautiful.

The idea is only explicit the order of composition.

5. Apr 2, 2014

### Staff: Mentor

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.

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.

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.

6. Apr 2, 2014

### micromass

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 [Broken]) and http://en.wikipedia.org/wiki/Risch_algorithm

/end(random remark)

Last edited by a moderator: May 6, 2017
7. Apr 2, 2014

### Jhenrique

$$\int \frac{\sqrt{x^2-1}}{x} dx = \sqrt{x^2-1} - i \log(i+\sqrt{x^2-1})+i \log(x)+C$$

8. Apr 2, 2014

### Staff: Mentor

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. Apr 2, 2014

### Jhenrique

Exist a table of composite functions for sn, cn, sc...?

10. Apr 2, 2014

### Staff: Mentor

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. Apr 2, 2014

### Jhenrique

Is extremely difficult find someone or some article or book that approach the elipitc functions of A to Z, answering all questions...

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. Apr 3, 2014

### Staff: Mentor

I don't see a connection between what you have put together in this table, and elliptic functions. See Elliptic functions.
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. Apr 3, 2014

### Jhenrique

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.