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Homework Help: Integral question

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data

    I have a question dealing with this integral: [tex]\int\frac{x}{\sqrt{x^2-4}}[/tex]. I did the trig substitution to check my other method of substitution and got two different answers: [tex]\sqrt{x^2-4}[/tex] and [tex]\sqrt{x^2-4}/2[/tex]. The latter is from trig substitution and the former from regular substitution.

    3. The attempt at a solution

    Ok, here is my work for the trig substitution

    (x^2-4) = 4tan^2@

    I finally get the integral of sec^2@ which gives me tan@ and looking from the triangle I drew its answer is [tex]\sqrt{x^2-4}/2[/tex]. Why don't these two answers agree? Is there some rule suggesting not to use trig substitution or one way over the other?

    Thank You
  2. jcsd
  3. Sep 6, 2008 #2
    Oh! I made a mistake with a coefficient, they agree with each other! Sorry
  4. Sep 6, 2008 #3


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    You skipped over the critical step! Yes, if you let [itex]x= 2 sec(\theta)[/itex], then dx= 2sec(\theta)tan(\theta)d\theta[/itex] (you dropped the "[itex]d\theta[/itex]" just as you dropped the "dx" in the original integral- bad habit.) and [itex]x^2- 4= 4tan^2(\theta)[/itex] so [itex]\sqrt{x^2- 4}= 2 tan^2(\theta)[/itex]. Did you forget that the"4" in [itex]4 tan^2(\theta)[/itex] became "2" when you took the square root?

    Now, the part you skipped- putting all that into the integral.
    [tex]\int\frac{x dx}{\sqrt{x^2- 4}}= \int \frac{[2 sec(\theta)][2sec(\theta)tan(\theta)d\theta}{2 tan(\theta)}[/tex]
    [tex]= 2\int sec^2(\theta)d\theta= 2 tan(\theta)+ C= 2tan(sec^{-1}(\frac{x}{2})+ C[/tex]
    Notice the "2" still in there?

    Now, if [itex]sec(\theta)= x/2[/itex], then we can represent that as a triangle with angle [itex]\theta[/itex], near side= 2, and hypotenuse= x. By the Pythagorean theorem, the opposite side has length [itex]\sqrt{x^2- 4}[/itex] and so [itex]tan(\theta)= \sqrt{x^2- 4}/2[/itex]. But because of the "2" multiplying the integral,
    [tex]\int \frac{x dx}{\sqrt{x^2- 4}}= \sqrt{x^2- 4}+ C[/tex]
    exactly as you would get if you make the substitution u= x2- 4.
  5. Sep 6, 2008 #4
    Excellent Halls,

    I found my mistake but thanks for showing me the extended steps, :D
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