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Math Help - Integration by parts problem

  1. #1
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    Integration by parts problem

    Hi, Can someone tell me how to do this integration problem by parts please, its driving me mental .

    integrate 4tsin(4t-2)dt

    Thanks
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    Quote Originally Posted by wolfhound View Post
    Hi, Can someone tell me how to do this integration problem by parts please, its driving me mental .

    integrate 4tsin(4t-2)dt

    Thanks
    u = 4t \implies u' = 4

    v' = \sin(4t-2) \implies v = -0.25\cos(4t-2)

    \int udv = uv - \int vdu = -t\cos(4t-2) + \int \cos(4t-2)

    The integral of \cos(4t-2) is a standard trig integral


    edit: please check my signs because I'm tend to get an embarrassing number of them wrong
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  3. #3
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    Thank you very much for that!
    while we are on the subject, one more problem please

    This integrate : -336x(4x+4)^(13) dx

    u=-336x du=-336
    dv=(4x+4)^(13)
    v= (4x+4)^(14)
    ...... 14

    and I got -24x(4x+4)^(14) +24(4x + 4)^(15) + C
    .... ................ .................... 60

    I think my answer is wrong ?
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  4. #4
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    e^(i*pi)'s Avatar
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    You could always use the binomial theorem to expand

    It looks like you haven't divided by 4 when finding v.

    I would be inclined to say that (4x+4)^{13} = [4(x+1)]^{13} = 4^{13}(x+1)^{13}

    \int -336 \cdot 4^{13} x (x+1)^{13}

    u = -336 \cdot 4^{13}x \implies du = -336 \cdot 4^{13}

    dv = (x+1)^{13} \implies v = \dfrac{1}{14}(x+1)^{14}

    -24\cdot 4^{13}x(x+1)^{14} + \int 24 \cdot 4^{13} (x+1)^{14}


    4^{13} is a massive number but it is, and thus acts like, a constant
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  5. #5
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    Excellent, thanks
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  6. #6
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    Quote Originally Posted by wolfhound View Post
    Thank you very much for that!
    while we are on the subject, one more problem please

    This integrate : -336x(4x+4)^(13) dx

    u=-336x du=-336
    dv=(4x+4)^(13)
    v= (4x+4)^(14)
    ...... 14

    and I got -24x(4x+4)^(14) +24(4x + 4)^(15) + C
    .... ................ .................... 60

    I think my answer is wrong ?
    Looks like you made calculation errors:

    Let \; u = -336x \;\;\;\;\;du=-336\;dx

     dv = (4x+4)^{13}\;dx \;\;\;\;\; v = \dfrac{(4x+4)^{14}}{56}

    integrate by parts:

     \displaystyle \int u\;dv= uv - \int v\;du

    = \displaystyle -336x \cdot   \dfrac{(4x+4)^{14}}{56} \;-\;\int \dfrac{(4x+4)^{14}}{56}\;(-336)\;dx

    =\displaystyle -6x(4x+4)^{14}+6 \int {(4x+4)^{14}} \;dx

    =\displaystyle -6x(4x+4)^{14}+ 6 \cdot \dfrac{(4x+4)^{15}}{60}+C

    =-6x(4x+4)^{14}+ \dfrac{(4x+4)^{15}}{10}+C
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    Thanks
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  8. #8
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by wolfhound View Post
    Thanks, but where did 56 come from when you integrate dv should it not be that v= ((4x+4)^14)/14
    \displaystyle \int (4x+4)^{13}\;dx

    use substution rule:

    Let\;\; a = 4x+4 \; \rightarrow da = 4 dx \rightarrow dx = \dfrac{da}{4}

    therefore your integral becomes:

     \displaystyle \int a^{13} \dfrac{da}{4} = \dfrac{1}{4} \int a^{13}\;da = \dfrac{1}{4} \times \dfrac{a^{14}}{14} = \dfrac{a^{14}}{56}=\dfrac{(4x+4)^{14}}{56}
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  9. #9
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    Thanks again people ..
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