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More Integration Problems

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Sorry for all the posts; I'm trying to study for a test tomorrow.

    1. Integrate

    2. Integrate

    2. Relevant equations

    3. The attempt at a solution

    [tex]u = x^2 + 8x + 65[/tex]
    [tex]du = 2x+8dx[/tex]

    [tex]\int2x^3(x^2+4)^{-\frac{1}{2}} dx[/tex]
    [tex] \frac{2x^4}{4}\frac{x^2+4}{\frac{1}{2}} + C[/tex]
    Not sure if this is right so far.

    Thanks for any help!!
  2. jcsd
  3. Apr 28, 2008 #2
    For the first one, a simple u-substitution won't work. Why not try completing the square?

    For the second one, think about trig substitution.

    Integrals that need either of these tricks tend to have pretty recognizable forms, so you should familiarize yourself with which forms need which trick.
    Last edited: Apr 28, 2008
  4. Apr 28, 2008 #3
    Also, with #1:

    I know it has something to do with the rule [tex]\int\frac{du}{a^2+u^2} = \frac{1}{a}\arctan \frac{u}{a} + C [/tex]
  5. Apr 28, 2008 #4
    For #1 I think you want to write x^2 + 8x + 65 in the form a + (x+b)^2, then you'll have [tex] \int \frac{1}{a+(x+b)^2} = \frac{1}{a} \int \frac{1}{1 + (x+b)^2/a} [/tex] which you can then use the substitution u^2 = (x+b)^2/a so that you'll end up with an inverse tangent.
  6. Apr 28, 2008 #5
    #1 - Complete the square
  7. Apr 28, 2008 #6
    Ok so how about this for #1:

    [tex] \int\frac{dx}{x^2+8x+16+49}[/tex]
    [tex] \int\frac{dx}{(x+4)^2+49}[/tex] ?

    So then

    [tex] u = x+4 [/tex]
    [tex]du = dx[/tex]
    [tex]\int\frac{du}{u^2+7^2} = \frac{1}{7}\arctan\frac{u}{7} + C = \frac{1}{7}\arctan\frac{x+4}{7} + C[/tex]
    Last edited: Apr 28, 2008
  8. Apr 28, 2008 #7
    Looks good
  9. Apr 28, 2008 #8
    I'm still stuck on the 2nd one tho..

    What sort of trig substitution?
  10. Apr 28, 2008 #9
    #2 - Trig sub.

  11. Apr 28, 2008 #10
    There's a standard form each trig substitution. The expression under the radical is of the form [tex]x^{2} + a^{2}[/tex]. Your book should tell you which trig function to substitute.
  12. Apr 28, 2008 #11


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    Too late, Roco did. But you don't really need the book to tell you. What sort a trig function can you square and add a constant and get something you can extract the square root of?
  13. Apr 28, 2008 #12
    No clue. =(
    I found [tex]\int\frac{du}{\sqrt{a^2-u^2}} = \arcsin \frac{u}{a} + C[/tex]
    Is this what I need?
  14. Apr 28, 2008 #13
    That has a negative sign.
  15. Apr 28, 2008 #14
    I don't know. I'm fixing to go to bed to get a good nights sleep though, is there any way you could give me another clue to what I should be looking for?
  16. Apr 28, 2008 #15


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    Sorry, I'm distracting you. tan(x)^2+1=sec(x)^2. Rocophysics already told you the correct substitution.
  17. Apr 28, 2008 #16
    haha, I thought someone might say that... I don't know how to use his substitution.
    The answer I was given to check myself is [tex]\frac{2}{3}(x^2+4)^{\frac{1}{2}}(x^2-8) + C[/tex]
  18. Apr 28, 2008 #17


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    That's it alright. Sleep on it and check it tomorrow.
  19. Apr 28, 2008 #18
    I can't. My test is tomorrow D=
    Not sure if this type is going to be on there, but I was hoping to get it down just in case.
  20. Apr 28, 2008 #19
    Let me try one more time. Can you tell me where I'm going wrong? From the beginning?

    [tex] u = 2x^3[/tex]
    [tex] du = 6x dx[/tex]
    [tex] \int\frac{u}{\sqrt{x^2+4}}\frac{du}{6x}[/tex]

    right so far?
  21. Apr 28, 2008 #20


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    Ok, you don't need a trig substitution for this one. Try u=x^2+4. That leaves you with an extra x^2 in the numerator. But that's ok. x^2=u-4.
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