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Math Help - a few more integrals from this year's MIT Integration Bee

  1. #1
    Super Member Random Variable's Avatar
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    a few more integrals from this year's MIT Integration Bee

    Since I don't know the solutions to all of these, I'm posting them here instead of in the Math Challenge Problems forum.


    None of these integrals should have very complicated solutions.

    EDIT: The fourth one might be too easy.


    1)  \int^{\pi/2}_{0} \sin (x) \sin (2x) \sin (3x) \ dx

    2)  \int x \ln \Big( 1 + \frac{1}{x} \Big) \ dx

    3)  \int^{1}_{0} \sin^{2} (\ln x) \ dx

    4)  \int x^{x} ( 1 + \ln x ) \ dx

    5)  \int^{2}_{1} (x-1)^{1/2} (2-x)^{1/2} \ dx
    Last edited by Random Variable; February 7th 2010 at 06:54 PM.
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  2. #2
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    I=\int x \ln(1+\frac{1}{x})dx=\int \left(\frac{x^2}{2}\right)' \ln\left(1+\frac{1}{x}\right)dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \ln\left(1+\frac{1}{x}\right) \right)'dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \dfrac{-1}{x(x+1)} \right)dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\int \left(\frac{x}{2(x+1)}\right)dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra  c{1}{2}\int \dfrac{x+1-1}{x+1}dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra  c{1}{2}\left(\int 1dx -\int\dfrac{1}{x+1}dx \right)
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra  c{1}{2}\left(x-\ln(1+x)\right) + C

    I think that it's ok
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  3. #3
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    The fourth:

    I=\int x^x (1+\ln(x))dx = \int e^{x\ln(x)}(1+\ln(x))dx

    Let be x\ln(x)=u, then

    I=e^u du = x^x+C
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by felper View Post
    I=\int x \ln(1+\frac{1}{x})dx=\int \left(\frac{x^2}{2}\right)' \ln\left(1+\frac{1}{x}\right)dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \ln\left(1+\frac{1}{x}\right) \right)'dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)-\int \left(\frac{x^2}{2}\right)\left( \dfrac{-1}{x(x+1)} \right)dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\int \left(\frac{x}{2(x+1)}\right)dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra  c{1}{2}\int \dfrac{x+1-1}{x+1}dx
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra  c{1}{2}\left(\int 1dx -\int\dfrac{1}{x+1}dx \right)
    I=\frac{x^2}{2}\ln\left(1+\frac{1}{x}\right)+\dfra  c{1}{2}\left(x-\ln(1+x)\right) + C

    I think that it's ok
    Wake up! You're correct.
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  5. #5
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    4)  \int x^{x} ( 1 + \ln x ) \ dx

    =

     \int e^{xlnx} ( 1 + \ln x ) \ dx

    let u = xlnx

     \int e^u du = x^x+C
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  6. #6
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    Quote Originally Posted by felper View Post
    The fourth:

    I=\int x^x (1+\ln(x))dx = \int e^{x\ln(x)}(1+\ln(x))dx

    Let be x\ln(x)=u, then

    I=e^u du = x^x+C
    LOL...this one was way too easy...not inclined to do any with bounds...no indefinites left. i would have posted this sooner, probably a minute after you, but i took a trip to the can.
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  7. #7
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    I=\int_0^1 \sin^2(\ln(x))dx=\int_0^1 \dfrac{1-\cos(2\ln(x))}{2}dx
    I=\dfrac{1}{2}-\int_0^1 \dfrac{\cos(2\ln(x))}{2}dx
    Let be \ln(x)=u \Leftrightarrow e^u=x \implies dx=e^u du
    \int_0^1 \dfrac{\cos(2\ln(x))}{2}dx=\dfrac{1}{2}\int_{-\infty}^0\cos(2u)e^u du
    Let be u=-t \implies du=-dt then:
    \int_0^1 \dfrac{\cos(2\ln(x))}{2}dx=-\dfrac{1}{2}\int_{\infty}^0\cos(-2u)e^{-u} du=\dfrac{1}{2}\int_{0}^{\infty}\cos(-2u)e^{-u} du

    Which is the Laplace Transform for cos(2t), so, the integral is

    I=\frac{1}{2}-\frac{1}{2}\cdot \frac{1}{5}=\frac{2}{5}

    End.
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  8. #8
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    I=\int^{\pi/2}_{0} \sin (x) \sin (2x) \sin (3x) \ dx=\dfrac{1}{2}\int_0^{\frac{\pi}{2}}(\cos(2x)-\cos(4x))\sin(2x)dx<br />
    I=\frac{1}{2}\left (\int_0^{\frac{\pi}{2}}\cos(2x)\sin(2x)dx-\int_0^{\frac{\pi}{2}}\cos(4x)\sin(2x)dx \right)
    I=\frac{1}{2}\left( \frac{1}{2}\int_0^{\frac{\pi}{2}}\sin(4x)dx - \frac{1}{2}\int_0^{\frac{\pi}{2}}(\sin(6x)-\sin(2x))dx \right)
    I=\frac{1}{4}\left(\int_0^{\frac{\pi}{2}}\sin(4x)d  x+\int_0^{\frac{\pi}{2}}\sin(2x)dx-\int_0^{\frac{\pi}{2}}\sin(6x)dx \right)
    I=\frac{1}{4}\left( 0+1-\frac{1}{3}\right)=\frac{1}{6}
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  9. #9
    Super Member Random Variable's Avatar
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    Quote Originally Posted by felper View Post
    I=\int^{\pi/2}_{0} \sin (x) \sin (2x) \sin (3x) \ dx=\dfrac{1}{2}\int_0^{\frac{\pi}{2}}(\cos(2x)-\cos(4x))\sin(2x)dx<br />
    I=\frac{1}{2}\left (\int_0^{\frac{\pi}{2}}\cos(2x)\sin(2x)dx-\int_0^{\frac{\pi}{2}}\cos(4x)\sin(2x)dx \right)
    I=\frac{1}{2}\left( \frac{1}{2}\int_0^{\frac{\pi}{2}}\sin(4x)dx - \frac{1}{2}\int_0^{\frac{\pi}{2}}(\sin(6x)-\sin(2x))dx \right)
    I=\frac{1}{4}\left(\int_0^{\frac{\pi}{2}}\sin(4x)d  x+\int_0^{\frac{\pi}{2}}\sin(2x)dx-\int_0^{\frac{\pi}{2}}\sin(6x)dx \right)
    I=\frac{1}{4}\left( 0+1-\frac{1}{3}\right)=\frac{1}{6}
    Correct again. These are too easy for you.
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  10. #10
    Super Member General's Avatar
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    Do you have a website for thier problems?
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  11. #11
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    <br />
\int^{2}_{1} (x-1)^{1/2} (2-x)^{1/2} \ dx<br />




    Choose

    u=x-1



    \Rightarrow



    \int^{2}_{1} (x-1)^{1/2} (2-x)^{1/2} \ dx

    =

    \int^{1}_{0} (u)^{1/2} (1-u)^{1/2} \ du

    Choose

    sin^2v = u

    du = 2sinvcosv dv = sin2v dv

    \Rightarrow

    \int^{1}_{0} (u)^{1/2} (1-u)^{1/2} \ du

    =

    \int^{\pi/2}_{0} (sinv)(cos^2v)^{1/2} \ sin2v \ dv

    =

    \int^{\pi/2}_{0} sinv \ sin2v \ cosv \ dv

    =

    \int^{\pi/2}_{0}cosv\times\frac{1}{2}(cosv-cos3v)dv

    =

    \frac{1}{2}\int^{\pi/2}_{0}cos^2v-\frac{1}{2}(cos2v+cos4v) \ dv



    The last 2 terms evaluate to 0.

    \int^{\pi/2}_{0} \frac{1}{2} \times\cos^2v \ dv

    =

    \pi/8





    All in all, quite a yawner of a set. The one where Felper recognized the Laplace was the best one, but even that one was str8tforward. This must be for the total newbie undergrads at mit.
    Last edited by vince; February 8th 2010 at 05:59 AM. Reason: added an extra equality for clarity
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