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Techniques for Integrating Radical Functions?

  1. May 14, 2009 #1
    What are some techniques for integrating functions that are or contain radicals? I am familiar with trigonometric substitution. Are there any other "common" techniques or special functions that can be used to integrate these functions? or does trigonometric substitution basically take care of everthing? This is a general question, I don't have a specific function in mind.

  2. jcsd
  3. May 15, 2009 #2


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    Most common is to use trigonometric (or hyperbolic) substitutions.

    [itex]sin^2(t)+ cos^2(t)= 1[/itex] so [itex]\sqrt{1- sin^2(t)}= cos(t)[/itex]

    For example, to integrate [itex]\int \sqrt{1- x^2}dx[/itex], let x= sin(t). Then dx= cos(t) dt and [itex]\sqrt{1- x^2}= \sqrt{1- sin^2(t)}= cos(t)[/itex] so the integral becomes [itex]\int cos^2(t) dt[/itex] which can be integrated using the identity [itex]cos^2(t)= (1/2)(1+ cos(2t))[/itex].

    For something like [itex]\sqrt{1+ x^2}[/itex] you can divide [itex]sin^2(t)+ cos^2(t)= 1[/itex] by cos(t) to get [itex]tan^2(t)+ 1= sec^2(t)][/itex]

    To integrate [itex]\int \sqrt{9+ x^2}dx[/itex], let x= 3tan(t). Then [itex]dx= 3 sec^2(t) dt[/itex] and [itex]\sqrt{9+ x^2}= \sqrt{9+ 9tan^2(t)}= 3\sqrt{1+ tan^2(t)}= 3sec(t)[/itex] so the integral becomes [itex]\int (3 sec(t))(3sec^2(t)dt)= 9\int sec^3(t)dt[/itex]. That can be integrated by writing it as [itex]9\int dt/cos^3(t)= 9\int cos(t)dt/cos^4(t)= 9\int cos(t)dt/(1- sin^2(t))^2[/itex] and using the substitution u= sin(t).

    But it is also true that [itex]cosh^2(t)- sinh^2(t)= 1[/itex] or [itex]sinh^2(t)+ 1= cosh^2(t)[/itex] so you could also use the substitution x= 3 sinh(x). Then dx= 3 cosh(x)dx and [itex]\int \sqrt{9+ x^2}dx= 9\int cosh^2(x) dx[/itex]

    Just about any calculus text will devote at least a section, if not a chapter, to trig substitutions though hyperbolic substitution are less commonly covered.
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