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Math Help - Integrate this!

  1. #1
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    Integrate this!

    Can you please show me how to do this...

    <br />
\int_{0}^{pie}{x}^2sin{x}\,\,\,dx<br />

    thanks.
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  2. #2
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    Quote Originally Posted by Unt0t View Post
    Can you please show me how to do this...

    <br />
\int_{0}^{pie}{x}^2sin{x}\,\,\,dx<br />

    thanks.
    Apply integration by parts. Twice.
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  3. #3
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    does it end up being

    <br />
{x}^2*-cos{x}-\int2{x}*-cos{x}<br />

    then


    <br />
{x}^2*-cos{x}+2{x}sin{x}-\int 2sin{x}<br />

    then

    <br />
{x}^2*-cos{x}+2{x}sin{x}+2cos{x}<br />

    i'm so confused...
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  4. #4
    Senior Member Pinkk's Avatar
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    This calls for integration by parts. In general, if you have an integral: \int u\,\,\,dv, where both u and dv are functions of x, the integral can be rewritten as:

    uv - \int v\,\,\,du

    Usually, you want to pick a u that is easy to differentiate and can usually differentiate down to 0. In this case, let's pick x^{2} as u. Differentiating, we get du=2x\,\,\,dx. When we let dv=sinxdx, we can integrate that to get v=-cosx. Having these, we can simply rewrite the original integral as:

    -x^{2}cosx + 2\int xcosx\,\,\,dx

    Notice how the integral there cannot be easily integrated, so for that integral we need to do integration by parts a second time. Same rules apply; make your u easy to differentiate. Let's pick x as u. Differentiating, we get du=dx. Picking cosxdx as your dv, when we integrate we will get v=sinx. We can therefore rewrite the previous expression as:

    -x^{2}cosx + 2xsinx - 2\int sinx\,\,\,dx

    The final integral is simply -cosx, we just simply write:

    -x^{2}cosx + 2xsinx + 2cosx. I'll leave it up to you to evaluate the integral from 0 to \pi.
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  5. #5
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    can I ask where you go the 2 in front of  \int sinx\,\,\,dx


    <br /> <br />
-x^{2}cosx + 2xsinx - 2\int sinx\,\,\,dx<br /> <br />
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  6. #6
    Senior Member Pinkk's Avatar
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    Notice how there was a 2 in front of the integral \int xcosx\,\,\,dx. That 2 is distributed to both the xsinx and \int sinx\,\,\,dx terms that resulted from the second time integrating by parts.

    Quote Originally Posted by Unt0t View Post
    does it end up being

    <br />
{x}^2*-cos{x}-\int2{x}*-cos{x}<br />

    then


    <br />
{x}^2*-cos{x}+2{x}sin{x}-\int 2sin{x}<br />

    then

    <br />
{x}^2*-cos{x}+2{x}sin{x}+2cos{x}<br />

    i'm so confused...
    You have the 2 in your evaluation right here. Same 2 as in mine.
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  7. #7
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    thanks heaps
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