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One last integration problem

  1. Feb 20, 2008 #1
    This is what I have so far:

    [tex]\int (\frac{x^2}{\sqrt{4 + x^2}}) dx [/tex]

    [tex]x = 2 tan u[/tex]
    [tex]dx = (2 sec^2u) du[/tex]

    [tex]x^2 = 4 tan^2 u[/tex]

    [tex]\sqrt{4 + x^2} = \sqrt{4 + 4 tan^2 u}[/tex]

    [tex]\sqrt{4 + x^2} = \sqrt{4 (1 + tan^2 u}[/tex]

    [tex]\sqrt{4 + x^2} = \sqrt{4 sec^2 u}[/tex]

    [tex]\sqrt{4 + x^2} = 2 sec u[/tex]

    Then I have:

    [tex]\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = \int \frac{4 tan^2 u}{2 sec u} 2 sec^2 u du[/tex]

    [tex]\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = \int (4 tan^2 u)(sec u) du[/tex]

    I keep getting stuck on this part, any ideas?
  2. jcsd
  3. Feb 20, 2008 #2
    Assuming your work is correct ...

    [tex]4\int\tan^{2}\theta\sec \theta d\theta[/tex]

    [tex]4\int(\sec \theta \tan \theta) \tan \theta d\theta[/tex]

    Or go straight to this ...

    [tex]4\int(\sec^{3}\theta-\sec \theta) d\theta[/tex]

    Either way, integration by parts!
    Last edited: Feb 20, 2008
  4. Feb 20, 2008 #3
    Taking that...

    [tex]4\int(\sec u \tan u) \tan u du[/tex]

    And knowing that:

    [tex]\int(\sec u \tan u) du = \sec u + C[/tex]


    [tex]\int\sec u du = ln |\sec u + \tan u|)+ C[/tex]

    Can we just do it by parts that way?


    [tex]\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = 4(\sec u + ln |\sec u + \tan u|) + C[/tex]

    (crosses figures) :blushing:
  5. Feb 20, 2008 #4
    You must be drunk!

    [tex]4\int(\sec \theta \tan \theta) \tan \theta d\theta[/tex]

    Usually, you want to reserve "u" so I'm using theta in it's place.

    [tex]u=\tan \theta[/tex]
    [tex]du=\sec^{2} \theta d\theta[/tex]

    [tex]dV=\sec \theta \tan \theta d\theta[/tex]
    [tex]V=\sec \theta[/tex]

    [tex]4\int(\sec \theta \tan \theta) \tan \theta d\theta=4(\sec \theta \tan \theta-\int\sec^{3}\theta d\theta)[/tex]

    Do parts again, you will actually have to do it 3x, I think. Take [tex]\int\sec^{3}\theta d\theta[/tex] to the side and you will notice that you get it back. Just move it to the other side and divide by the constant, and it's solved.
    Last edited: Feb 20, 2008
  6. Feb 20, 2008 #5
    No I'm not drunk, I'm just new to this material. However I think were just about done:

    I understand how you did everything, so now...

    [tex]4\int(\sec \theta \tan \theta) \tan \theta d\theta=4(\sec \theta \tan \theta-\int\sec^{3}\theta d\theta)[/tex]
    Where... [tex]\int\sec^{3}\theta d\theta) = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} ln|sec \theta + tan \theta| + C[/tex]

    This gives us:
    [tex]\int (\frac{x^2}{\sqrt{4 + x^2}}) dx = 4 (\sec x \tan x)-\frac{1}{2} \sec x \tan x + \frac{1}{2} ln|sec x + tan x| + C[/tex]

  7. Feb 20, 2008 #6
    To be honest, I did not work it out myself. Looks good though! Btw, hopefully you weren't angry with my comment. I didn't mean anything by it in terms of you just having learned this :p But um, your answer could be right, all depends if your teacher wants you to re-substitute ... this would be the most painful part, lol.
  8. Feb 20, 2008 #7
    Haha no offense taken and thanks for your help!
  9. Feb 20, 2008 #8


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    Actually, he meant it helps to be drunk when doing calculus!
  10. Feb 20, 2008 #9


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    A much faster way to learn these type of integrals is to go through a standard text book step by step, and use paper and pen.
  11. Feb 20, 2008 #10
    Really? Mr. Obvious.
  12. Feb 20, 2008 #11
    Another way of doing this integral is by using the substitution:

    [tex]x=2\cdot sinh(t)[/tex]

    after setting:


    in the numerator. You will end up with:

    [tex]I=\frac{x}{2}\sqrt{4+x^2}-2\cdot argsinh\left( \frac{x}{2}\right)+C[/tex]

    Btw: I'm sober :wink:
  13. Feb 21, 2008 #12


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    Some times people has to be more obvious than needed for some people who are reluctent to see in text books the standard result like

    integral [sqrt(a^2 + x^2)] dx = (x/2) sqrt(a^2 + x^2)+ [(a^2)/2] ln|(x+sqrt(a^2+x^2)|+c
    (I leave the derivation of this result for obvious reasons)

    and cannot see x^2 = x^2 +a^2-a^2, as shown in the previous post by COOMAST. Then the remaining term is easily (obviously?) undone by substituting x=2tan(u).
    After COOMAST's method, I would prefer this method as it will be easier to go back in terms of the original veriable. Is it obv....?
    Last edited: Feb 21, 2008
  14. Feb 21, 2008 #13
    Don't drink and derive.
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