Integration of Rational Functions with Trigonometric Substitutions

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Discussion Overview

The discussion revolves around the integration of rational functions, specifically the integral of the form ∫[1/(a-bx^2)] dx. Participants explore various methods for solving this integral, including trigonometric substitutions and partial fractions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about the procedure for integrating the function and suggests that a trigonometric identity might be applicable.
  • Another participant proposes using partial fractions to rewrite the integral, provided that a and b are non-negative.
  • It is noted that if a and b have opposite signs, the integral may involve an arctangent function instead.
  • A suggestion is made to factor out b from the integral, allowing for the use of an inverse sine formula.
  • Some participants discuss the implications of using different methods, noting that certain approaches may lead to more complex solutions, particularly when asked to solve using partial fractions.
  • There is a debate about the applicability of the inverse sine formula, with some participants asserting that it requires square roots, while others argue for its relevance through trigonometric substitutions.

Areas of Agreement / Disagreement

Participants express differing opinions on the best approach to integrate the function, with no consensus reached on a single method. Some advocate for trigonometric substitutions while others prefer partial fractions, leading to an unresolved discussion.

Contextual Notes

Participants highlight that the integration method may depend on the signs of a and b, and that different approaches can yield varying complexities in the final answer.

hms.tech
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What is the procedure to integrate this kind of a fraction, i am guessing some trigonometric
identity will apply but i am not quite sure how


∫[1/(a-bx^2)] dx

can someone provide me a link which would help me learn about this kind of integration.

Any help is Appreciated
 
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hms.tech said:
What is the procedure to integrate this kind of a fraction, i am guessing some trigonometric
identity will apply but i am not quite sure how


∫[1/(a-bx^2)] dx

can someone provide me a link which would help me learn about this kind of integration.

Any help is Appreciated




Google "partial fractions integration". In your case,
$$\frac{1}{a-bx^2}=\frac{1}{(\sqrt a-\sqrt bx)(\sqrt a+\sqrt bx)}$$
iff \,a,b\geq 0\,

DonAntonio
 
The above works as long as a and b have the same sign - if both are negative, replace by minus absolute value. If they are of opposite sign, then you end up with something involving an arctan.
 
Can't you just pull out a {b}, giving \frac{1}{{b}(\frac{a}{{b}}-x^{2})}

Then use a simple inverse sin formula?
 
You can, but that does not mean you have to. It depends on what solution is being asked from you, because for example you are asked to solve the problem with partial fractions and you try a sine substitution, you will have to break up your answer into two natural logarithms later on. For definite integrals though, this often is not a problem.
 
johnqwertyful said:
Can't you just pull out a {b}, giving \frac{1}{{b}(\frac{a}{{b}}-x^{2})}

Then use a simple inverse sin formula?



Inverse sine formula? For that one needs \,(\sqrt{1-x^2})^{-1}\, and we have no square roots here.

DonAntonio
 
DonAntonio, it is still applicable. Let \sin(t)=x. Then we have that \displaystyle \int \frac{dt}{\cos(t)}=\int \sec(t)dt which has a closed-form solution.
 
DonAntonio said:
Inverse sine formula? For that one needs \,(\sqrt{1-x^2})^{-1}\, and we have no square roots here.

DonAntonio

Yeah, you need a square root for inverse sin, I knew that. :redface:
But still, just do a trig sub.
 
Millennial said:
DonAntonio, it is still applicable. Let \sin(t)=x. Then we have that \displaystyle \int \frac{dt}{\cos(t)}=\int \sec(t)dt which has a closed-form solution.


Yes...so? I was referring to other poster's idea of using something about the inverse sine, or \,\arcsin\, , which I

can't see how it can appear here more or less "naturally". Of course, one can alwasy make up substitutions as to make

almost any function pop up, but this is not what I meant.

DonAntonio
 

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