Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    and...

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

    Can we just do it by parts that way?

    So...

    [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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

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

    ssd

    User Avatar

    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:

    [tex]x^2=x^2+4-4[/tex]

    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

    ssd

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: One last integration problem
  1. One tough integral (Replies: 2)

Loading...