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Integration By Parts and Substitution

  1. Sep 18, 2007 #1
    [SOLVED] Integration By Parts and Substitution

    Short background; Took Calc 1 my senior year in highschool. Got As all 4 quarters and found it quite easy. Freshman year comes around and I sign up for Calc 2. Turns out the only teacher teaching Calculus 2 for my fall and spring semester is a horrible teacher who doesn't care about his students. So I decided to wait until my Sophomore year to take Calc 2.

    So fast forward; I am picking up on all the new techniques and what have you, but I keep forgetting all the old, little essential things because of that year off.

    So heres the problem;

    [tex]\int\pi^{\sqrt{x}}dx[/tex]

    Solve using integration by parts. At the bottom of the problem my professor gave a hint saying "You may like to do a substitution first."

    And I'm stumped. I haven't used substitution at all in the past 15 months.

    The only substitution I can think of is [tex]{\sqrt{x}}=m[/tex], but that is of no use.

    Thanks in advance. =P
     
  2. jcsd
  3. Sep 18, 2007 #2

    morphism

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    Well,

    [tex]\pi^{\sqrt{x}} = e^{\sqrt{x} \, \ln \pi}[/tex]

    And maybe the substitution u=\sqrt{x} from here?
     
  4. Sep 18, 2007 #3
    Played around with it for a while, didn't find anything. Not saying your wrong tho, I'm probably wrong.
     
  5. Sep 18, 2007 #4

    morphism

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    I decided to use the substitution [itex]u=\ln(\pi)\sqrt{x}[/itex] instead, but that doesn't really matter. This transforms the integral into

    [tex] \frac{2}{(\ln{\pi})^2} \int u e^u du. [/tex]

    And from here it's an easy application of integration by parts.
     
  6. Sep 18, 2007 #5
    lol thanks for the help. But this is the part I don't remember how to do. Can't figure out how to get from
    [tex]\int{\pi^\sqrt{x}}\ dx[/tex]
    to
    [tex] \frac{2}{(\ln{\pi})^2} \int u e^u du. [/tex]
     
  7. Sep 18, 2007 #6

    Dick

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    pi^(sqrt(x))=exp(ln(pi)*sqrt(x)). Then u=ln(pi)*sqrt(x), du=ln(pi)*dx/(2*sqrt(x)). Please check this and then try to continue.
     
  8. Sep 18, 2007 #7
    [tex]u=ln\pi\sqrt{x}[/tex]
    [tex]du=\frac{ln\pi dx}{2\sqrt{x}}[/tex]

    If I'm not wrong, that substitution would require;

    [tex]\int\frac{ln\pi}{2\sqrt{x}} e^\sqrt{x}ln\pi[/tex]

    I can account for the [tex]\frac{ln\pi}{2}[/tex] but I don't remember how to get the [tex]\frac{1}{\sqrt{x}}[/tex] before the e in the integral so that you can apply the du.
     
  9. Sep 18, 2007 #8

    Dick

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    You are wrong. The integral is exp(u)*dx. Would you please solve for dx in terms of u and du so we can get on with this? Sorry to be snappy, it's getting late.
     
  10. Sep 19, 2007 #9
    Still drawing a blank.

    And I've never seen the term exp(u). I was thinking of it as e^(u) but the more I play with the numbers the more I see I'm wrong.
     
  11. Sep 19, 2007 #10

    morphism

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    exp(u) is indeed just e^u.

    We have:
    du=ln(pi)*dx/(2*sqrt(x))

    and

    u=ln(pi)*sqrt(x).

    It shouldn't be hard to write dx in terms of u and du.
     
  12. Sep 20, 2007 #11
    Okay, pretty sure I got it. Like I said, I was just forgetting simple little substitution rules.

    Thanks.
     
  13. Sep 20, 2007 #12
    Last edited: Sep 20, 2007
  14. Sep 20, 2007 #13

    Dick

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    Ok, so u*e^u-e^u is ok. But you've lost all of the multiplicative constants like ln(pi) and 2. Where did they go?
     
  15. Sep 20, 2007 #14

    Ahhhh. Thank you very much.
     
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