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Finding integral

  1. Jun 26, 2007 #1
    Hi,

    1. The problem statement, all variables and given/known data
    find value of
    [tex]
    \int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx
    [/tex]

    2. Relevant equations

    3. The attempt at a solution

    I first used subsitution u = lnx, that gave me:

    [tex]
    \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
    [/tex]
    then letting
    [tex]
    I = \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
    [/tex]
    integrating by parts twice, gave me:
    [tex]
    I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du
    [/tex]
    [tex]
    I = - e^{u} \cos u + e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
    [/tex]

    therefore

    [tex]
    I = - e^{u} \cos u + e^{u} \cos u + I
    [/tex]
    which of course cancels to zero which is not very useful

    Any ideas on finding the value of this integral, or have i made an error in my working?

    Many Thanks
     
    Last edited: Jun 26, 2007
  2. jcsd
  3. Jun 26, 2007 #2
    I think you need to re-check the 'by-parts'.
     
    Last edited: Jun 26, 2007
  4. Jun 26, 2007 #3

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Yep. In particular, in the second "by parts", think where that sine inside the integral comes from :)
     
  5. Jun 26, 2007 #4
    Taking the result of the first integration by parts:

    [tex]
    I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du
    [/tex]

    then working with the integral part:
    [tex]
    +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du
    [/tex]
    let u= cos u
    therefore
    [tex]
    \frac{du}{dx}= - sinu
    [/tex]
    let
    [tex]
    \frac{dv}{dx}= u e^{u}
    [/tex]
    therefore
    [tex]
    v = e^{u}
    [/tex]

    then using identity that:

    [tex]
    \int u \frac{dv}{dx} \equiv uv - \int v \frac{du}{dx}
    [/tex]

    I got
    [tex]
    e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
    [/tex]

    Note- i chose u=cos u and dv/dx =ue^u [rather than u=ue^u]since i thought otherwise I would never be able to achieve an equivalent expression without an integral involved as I would get greater and greater powers of u in front of e^u as I kept going.

    Sorry if i'm not seeing something obvious, but i cant find the error there

    thanks for the replies
     
    Last edited: Jun 26, 2007
  6. Jun 26, 2007 #5
    How did you get a 'u' within the integral at the result of the first integration by parts? (And it's now the derivative with respect to u, not x)

    [I'm using t instead of u for the sake of clarity]

    The first integral, after substitution [tex]= \int_{0}^{\pi/2}{e^t\sin{t}}dt[/tex]

    Let u = cos(t) and dv/dt = e^t. Therefore, du/dt = -sin(t) and v = e^t.
     
    Last edited: Jun 26, 2007
  7. Jun 26, 2007 #6
    Thanks, I think i've got it now; but just to check here my full working:

    let
    [tex]
    I = \int_{0}^{\frac{\pi}{2}} e^{t}\sin t dt
    [/tex]

    u = sin t
    dv/dt= e^t

    therefore
    du/dt = cos t
    v = e^t

    so:
    [tex]
    I = e^{t} sin t - \int_{0}^{\frac{\pi}{2}} e^{t} cos t dt
    [/tex]

    and integrating by parts again:
    u = cos t
    dv/dt = e^t

    so
    du/dt = -sin t
    v = e^t

    [tex]
    I = e^{t} sin t - e^{t} cos t - \int_{0}^{\frac{\pi}{2}} e^{t} sin t dt
    [/tex]

    [tex]
    I = e^{t} sin t - e^{t} cos t - I
    [/tex]

    [tex]
    2I = [e^{t} ( sin t - cos t )]_{0}^{\frac{\pi}{2}}
    [/tex]
    [tex]
    I = \frac{1}{2} [(e^{\frac{\pi}{2}}( 1 - 0 )) - ( -1)]
    [/tex]

    and finally, giving
    [tex]
    I = \frac{1}{2} [e^{\frac{\pi}{2}} + 1 ]
    [/tex]

    Hopefully its right this time?

    Thanks again
     
    Last edited: Jun 26, 2007
  8. Jun 26, 2007 #7
    Perfect! ...
     
  9. Jun 26, 2007 #8
    Great! Thank you very much for the advice
     
  10. Jun 26, 2007 #9
    You're welcome. :)
     
  11. Jun 27, 2007 #10

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Hmm, I don't quite follow what you're doing in your subsequent posts (it looks correct though), but your first try was already correct if it weren't for one minus sign:

    so
    [tex]2I = - e^{u} \cos u + e^{u} \cos u[/tex]
    and the answer follows.
     
  12. Jun 27, 2007 #11
    And you end up with 0, which is incorrect.
     
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