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Integration by parts tanx help

  1. May 1, 2008 #1
    I actually have two here, so I will just list both:

    1. The problem statement, all variables and given/known data
    [tex]\int\frac{x}{x^{2}+4x+4}dx[/tex]


    2. Relevant equations
    None


    3. The attempt at a solution
    I tried this one twice. I honestly have no idea how to do it, and I used integration by parts. The first time, I reduced it down to:
    [tex]\int\frac{1}{x} + \frac{1}{4} + \frac{x}{4}dx[/tex]
    But, this is wrong.

    I tried it a second time by using integration by parts to obtain:
    [tex]\int\frac{x}{(x+2)(x+2)}dx[/tex], then I reduced that down, since integration by parts does not work. So, I was hoping to know what I am susposed to do.

    The second one is a bit different:
    1. The problem statement, all variables and given/known data
    [tex]/int[/tex] [tex](tan^{2}(x))dx[/tex]


    2. Relevant equations
    [tex]tan^{2}(x) + 1 = sec^{2}[/tex]


    3. The attempt at a solution
    I used the regular formula that I listed to get:
    [tex]\int(sec^{2}(x) - 1)dx[/tex].
    I just integrated to: [tex] tan^{2} - x + c [/tex]
    I wanted to see if this one is correct.

    Thankyou for your help.
     
    Last edited: May 1, 2008
  2. jcsd
  3. May 1, 2008 #2
    For the first problem you need to use integration by partial fractions to solve it.

    You were on the right track in the second problem changing (tanx)^2 to (secx)^2 - 1 now you just have to integrate each seperatly which you did right except for the integral of (secx)^2 is just tanx
     
  4. May 1, 2008 #3
    I did use that too for the first one, and I could not find an answer. When I did it, I got 2=0 which is not an answer.

    All right, thank you for the second one.
     
  5. May 1, 2008 #4

    jambaugh

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    Quicker solution to the first problem:

    Use variable substitution: [tex] u = x+2[/tex]
    You'll then get:
    [tex] \int \frac{something}{u^2}du[/tex]
    which you can then separate into powers of u and integrate.

    Your integration in the second problem is not quite correct.
    Recall that [tex]\frac{d}{dx} \tan (x) = \sec^2(x)[/tex] not tan squared!!!
     
  6. May 1, 2008 #5
    You got [tex]
    \int\frac{x}{(x+2)^2}dx[/tex]

    use the substitution u=x+2, so x=u-2 and dx=du and take it from there
     
  7. May 1, 2008 #6
    What both of you are saying will produce:
    [tex]\int\frac{x}{u^{2}}[/tex]
    Shouldn't the x in the numerator be apart of it too?
     
  8. May 1, 2008 #7

    jambaugh

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    You must finish the substitution.... rewrite x in terms of u.

    [edit: And of course when you're done integrating, rewrite your answer back in terms of x.]
     
  9. May 1, 2008 #8
    yes, you said that [tex]u = x+2[/tex]
    The derivative of that is [tex]du = 1dx[/tex]

    That would only get:
    [tex]\int\frac{x}{u^{2}}[/tex]

    Then you would solve x equal to u-2?

    That would get:

    [tex]\int\frac{u-2}{u^{2}}[/tex]

    That would make:
    [tex]\int\frac{1}{u} - \int\frac{2}{u^{2}}[/tex] = [tex]ln(x) - \frac{-2}{u}[/tex]
    Then, of course, you would put x+2 back in for all u's. Right?
     
  10. May 1, 2008 #9

    jambaugh

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    Right! (But don't forget the +C!!!) ;-)
     
  11. May 1, 2008 #10

    tiny-tim

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    Hi carbz! :smile:

    A partial fractions method (which avoids having to substitue and then substitute back again ) is:

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

    [tex]=\,\int\frac{x+2}{x^{2}+4x+4}dx
    \,-\,\int\frac{2}{x^{2}+4x+4}dx[/tex] :smile:
     
  12. May 1, 2008 #11

    jambaugh

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    You've not solved it yet. You'll still need to do a substitution to integrate the partial fractions. Now you must do it twice. Also since its a linear substitution you can do it in your head once you're practiced.
     
  13. May 1, 2008 #12
    his thing = (x+2)/(x+2)^2 - 2/(x+2)^2
    so, no *twice* substitution
     
    Last edited: May 1, 2008
  14. May 1, 2008 #13

    tiny-tim

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    No … jambaugh is right … you and I can do it in our heads, but without that knack it does require a substitution! :smile:
     
  15. May 1, 2008 #14
    actually couldnt you do it with partial fractions and integrating once?

    you can break up x/(x^2+4x+4) into x/(x+2)^2

    therefore by partial fractions you could break it into [x/(x+2)^2]= A/(x+2) + Bx+C/[(x+2)^2]

    from there you can solve for A, B, and C and plug that back into your original integral

    not sure if thats what you originally ment but idk.
     
  16. May 1, 2008 #15

    tiny-tim

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    Hi tron_2.0! :smile:

    Yes … and B will be 0 …

    … but that's exactly what we have done! :smile:
     
  17. May 1, 2008 #16
    ahh my mistake, i should have read the replies more closely =[
     
  18. May 1, 2008 #17
    Nice one carbz
     
  19. May 1, 2008 #18

    jambaugh

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    OK I take back the "twice", as it is the same substitution and you can write the two summed integrals as a single one.

    In effect it is the same two steps in different orders as when I say "you can then separate into powers of u and integrate" this is a trivial case of expansion into partial fractions. Ultimately the preference is one of which you can do more easily the first substitution or the first expansion by partial fractions.
     
  20. May 1, 2008 #19
    well, the C is always annoying, but thank you.
     
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