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integration of rational function of sine and cosine

Defines: 
universal hyperboloc substitution
Synonym: 
universal trigonometric substitution
Type of Math Object: 
Topic
Major Section: 
Reference

Mathematics Subject Classification

26A36 no label found

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Stevecheng's correction question (if I understood it):
R(sin{x}, cos{x})
is supposed continuous in odd multiples of pi. If we integrate through such a point using t = tan(x/2), can we obtain right results?
If anybody knows the thing, please supplement the entry.
Jussi

I have a calculus book that states that the Weierstrass substitution formulas (t=tan(x/2), sin x=(2t)/(1+t^2), etc.) are only valid for -\pi<x<\pi. I am hesitant to add this though, as I do not know *why* x cannot take other values. In any case, for definite integrals, one could throw in a disclaimer such as, "this method works provided that the function is integrable on the specified domain."

I think that

$\lim_{x\to (-\pi,\pi)}\tan{x/2}=(-\infty,\infty)$,

covering all the real line.
I'm missing something?

Just to clarify, the issue does not seem to be the covering of the real line, but rather if the Weierstrass substitution method is valid if x is allowed to take values outside of this interval. This issue extends to improper integrals: Are the formulas for improper integrals obtained in this manner only valid for -\pi<x<\pi?

I see. Thanks Warren.
perucho

Does the following reasoning work?

Suppose we restrict $-\pi < x < \pi$.
In any small interval within this such that the integrand
is continuous (i.e. it has no poles), it is also holomorphic.
Then its indefinite integral is also a holomorphic function.
If one obtains an analytic expression on the right-hand side
after integrating, then, by rigidity, that expression and the indefinite integral has to agree on the whole real line... ?

// Steve

1/(5-3cos{x}) is continuous in the whole R. Using t = tan(x/2) the integration gives the result
(1/2)arctan(2tan(x/2))+C
which isn't defined in odd multiples of pi. The right antiderivative is, by E. Lindel\"of,
x/4+(1/2)arctan(sin{x}/(3-cos{x})).

Jussi

1/(5-3cos{x}) is continuous in the whole R. Using t = tan(x/2) the integration gives the result
(1/2)arctan(2tan(x/2))+C
which isn't defined in odd multiples of pi. The right antiderivative is, by E. Lindel\"of,
x/4+(1/2)arctan(sin{x}/(3-cos{x})).

Jussi

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