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Integration With Trig Substitution Calc II

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int \frac{\sqrt{196 x^2-144}}{x} dx[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I first rewrote the integral.....
    [tex]\int \frac{\sqrt{(14x)^2-12^2}}{x} dx[/tex]
    Then I let.......
    dx=6/7sec[tex] \theta tan \theta d \theta[/tex]

    My progression of this problem is as follows:

    [tex] \int \frac{\sqrt{(144(sec^2 \theta -1)}}{6/7sec \theta} d \theta[/tex]
    [tex]12 \int \frac{tan \theta}{6/7sec \theta}(6/7sec \theta tan \theta) d \theta[/tex]
    [tex]12 \int tan^2 \theta d \theta[/tex]
    [tex]12 \int sec^2 - 1 \theta d \theta[/tex]
    [tex]12 tan \theta [/tex]

    I am not sure how to finish the problem after that. I am also not sure I did that right since I am very new at this, but I have a feeling everything I did was right.
  2. jcsd
  3. Feb 18, 2010 #2


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    I didn't check all your steps, but didn't you forget to integrate the -θ in the last integral?

    Now draw a little right triangle in the first quadrant with hypotenuse 7x, x-side of 6 and y-side of [itex]\sqrt{(49x^2-36)}[/itex] and polar angle θ. You should be able to read everything off this triangle to get your answer back in terms of x.
  4. Feb 18, 2010 #3
    Sorry the last integral was a typo it should be....

    [tex]12 \int sec^2 \theta - 1 d \theta[/tex]

    Okay so after I fill this in the answer should be....

  5. Feb 18, 2010 #4
    That answer is wrong. I am not sure what mistake I made.
  6. Feb 19, 2010 #5
    I forgot to multiply it by 12 so I would have had...

    [tex]2 \sqrt{49x^2 - 36} + c [/tex]

    which is still wrong

    I checked Wolfram Alpha and they gave me the answer as.....

    [tex]2 \sqrt{49 x^2 - 36} + 12 tan^-1\left[\frac {6}{\sqrt{49 x^2 -36}}\right] + c[/tex]

    So I now know that I am close, but after checking it with the steps in Wolfram Alpha I am still not sure where I went wrong.
    Last edited: Feb 19, 2010
  7. Feb 19, 2010 #6
    What is ∫tan2θ dθ? There's really no need to change tan2θ to sec2θ - 1 that I can see.
    Once you find that integral, make sure you substitute for x correctly.
  8. Feb 19, 2010 #7


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    You forgot to integrate the 1 in the integrand.
  9. Feb 20, 2010 #8
    Wow, I did forget to do that! Thanks for noticing that. I got it now.
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