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Math Help - Integration by parts

  1. #1
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    Integration by parts

    Can someone help me with this integration by parts equation x^2sin(3x+1) Thank you.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Link88 View Post
    Can someone help me with this integration by parts equation x^2sin(3x+1) Thank you.
    Let x^2=u\implies{2xdx=du} and dv=\sin(3x+1)dx\implies{v=\frac{-1}{3}\cos(3x+1)}
    Can you go from there? You will need to do it twice.
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    Thats the part I get stuck at is doing it twice it gets confusing
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    Quote Originally Posted by Mathstud28 View Post
    Let x^2=u\implies{2xdx=du} and dv=\sin(3x+1)dx\implies{v=\frac{-1}{3}\cos(3x+1)}
    Can you go from there? You will need to do it twice.
    Can you show me the second part thats where i got stuck at. Thank you.
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    Quote Originally Posted by Link88 View Post
    Can you show me the second part thats where i got stuck at. Thank you.
    You need to use integration by parts twice, applying once using Mathstud28 choice of u and v gives:

    I=\int x^2 \sin(3x+1) \ dx=-x^2 \left[ \frac{1}{3} \cos(3x+1)\right] + \int 2x \left[ \frac{1}{3} \cos(3x+1)\right] \ dx

    Use integration by parts again to do the integral on the right.

    CB
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  6. #6
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    Don't integrate - balloontegrate!



    Straight lines integrate up, differentiate down (making it easier to check). The triangular networks satisfy the product rule - you may prefer to draw a zoomed-out version of these...



    The challenge of integration by parts is simply to fill out a product rule shape starting with one of the bottom balloons instead of (as happens in differentiation) the top one.

    See Balloon Calculus: worked examples from past papers

    or http://www.mathhelpforum.com/math-he...tegration.html

    to see the process.
    Last edited by tom@ballooncalculus; November 28th 2008 at 12:25 PM.
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  7. #7
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    Have you ever seen tabular integration. Here it is for fun.

    Let u=x^{2}, \;\ dv=v'dx=sin(3x+1)

    Take the derivatives of u until you get to 0. Take the antiderivatives of v'.

    \text{sign \;\               \;\ u and its derivatives  \;\             \;\ \;\ v' and its antiderivatives }

    \begin{array}{ccc}+& \;\ \;\ \;\ x^{2}\searrow& \;\ \;\ sin(3x+1)\\-&2x\searrow&\frac{-1}{3}cos(3x+1)\\-&2\searrow&\frac{-1}{9}sin(3x+1)\\-&0&\frac{1}{27}cos(3x+1)\end{array}

    Now, take the signed products and add up the diagonal entries:

    \frac{-1}{3}x^{2}cos(3x+1)+\frac{2}{9}xsin(3x+1)+\frac{2}  {27}cos(3x+1)+C
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