# What does it mean the notation $\int{R\left( \cos{x}, \sin{x} \right)\mathrm{d}x}$

Sometimes I find this notation and I get confused: $$\int{R\left( \cos{x}, \sin{x} \right)\mathrm{d}x}$$

Does it mean a rational function or taking rational operations between $\cos{x}$ and $\sin{x}$ ?

Update: I think you did not understand the question well, Here is an example (maybe it is a lemma or a theorem):

All the integrals of the form $\int{R\left( \cos{x}, \sin{x} \right)\mathrm{d}x}$ can be evaluated using the substitution $u=\tan{\dfrac{x}{2}}$.

I think that $R$ here does not stand for a rational function but for taking rational operations(addition, subtraction, multiplication, division) between $\cos{x}$ and $\sin{x}$

Update : I did not noticed that $R$ is a rational function of two variables and that means exactly that we are taking rational operations.

• The assumption is exactly that R is a rational function. – Did May 4 '16 at 11:21
• What is, in your opinion, the difference between "a rational function of $\cos(x)$ and $\sin(x)$" and "taking rational operations between $\cos(x)$ and $\sin(x)$"? – Christian Blatter May 4 '16 at 11:22
• Please read the question before downvoting. the book says: The symbol R and does not say rational function – Navaro May 4 '16 at 11:50
• @Navaro No, $R(x,y)$ can be any rational function, so it could be $xy$ or $x+y$ or $\frac{x}{y}$ or $\frac{x^2+xy+x^3y^3}{x^2+y^2}$ or any number of things. It is not necessarily a ratio of one argument to the other. Perhaps that is why the term "rational operation" confused you. – Ian May 4 '16 at 12:28
• The point that Did and Christian Blatter and I have been trying to make is that you have not made it entirely clear what "taking rational operations" means exactly. By contrast "rational function" is a clearly defined term, whose definition can be looked up. I can tell you from the context (because you are talking about the half-angle substitution) that your $R$ is a rational function. Maybe it can be described the other way too, but that doesn't really matter. – Ian May 4 '16 at 12:42

Here $R$ is a function of two variables $s$ and $t$. For instance, if $$R(s,t) = \frac{s}{1+t}$$ then $$R(\cos x , \sin x) = \frac{\cos x}{1 + \sin x}.$$

• Comments are not for extended discussion; this conversation has been moved to chat. – Jyrki Lahtonen May 4 '16 at 13:01

A polynomial in two variables is an expression $$P(x, y) = \sum_{i, j=0}^{\infty} a_{ij} x^{i} y^{j}$$ with only finitely many non-zero coefficients $a_{ij}$. A rational function in two variables is a quotient of polynomials in two variables.

In your question, $R$ denotes an arbitrary rational function of two variables. (It's arguably reasonable to describe "evaluating a rational function of two variables $x$ and $y$" as "taking rational operations between $x$ and $y$", but the number of comments here and elsewhere suggests doing so is a recipe for ambiguity.)

We can be certain of this interpretation on grounds of mathematical culture: Thanks to the double-angle formulas for the circular functions and the chain rule, the substitution $u = \tan(x/2)$, or $x = 2\arctan u$, gives $$\cos x = \frac{1 - u^{2}}{1 + u^{2}},\qquad \sin x = \frac{2u}{1 + u^{2}},\qquad dx = \frac{2\, du}{1 + u^{2}}.$$ Consequently, $$\int R(\cos x, \sin x)\, dx = \int R\left(\frac{1 - u^{2}}{1 + u^{2}}, \frac{2u}{1 + u^{2}}\right) \frac{2\, du}{1 + u^{2}},$$ a rational function in $u$.

The significance is, every rational function in one variable has an elementary primitive (antiderivative).

• This is exactly what I expected the answer to be like – Navaro May 4 '16 at 14:06

In books with tables of integrals etc. (like Gradshteyn's book) $R(\cos x, \cos y)$ typically means a rational function of $(\cos x, \sin x)$. So take a rational function $R(a,b)=f^{1}(a,b)/f^{2}(a,b)$ and plug in $a=\cos x, b=\sin x$. Here, $f^{i}(a,b), i=1,2$ are polynomials in the two variables $(a,b)$, i.e., $$f^{i}(a,b)=c^i_{0,0}+c^i_{1,0}a+c^i_{0,1}b+c^i_{2,0}a^2+c^i_{0,2}b^2+c^i_{1,1}ab+\dots$$ with coefficients $c^{i}_{k,l}$.