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Thread: Integrate this!

  1. #1
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    Integrate this!

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

    $\displaystyle
    \int_{0}^{pie}{x}^2sin{x}\,\,\,dx
    $

    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...

    $\displaystyle
    \int_{0}^{pie}{x}^2sin{x}\,\,\,dx
    $

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

    $\displaystyle
    {x}^2*-cos{x}-\int2{x}*-cos{x}
    $

    then


    $\displaystyle
    {x}^2*-cos{x}+2{x}sin{x}-\int 2sin{x}
    $

    then

    $\displaystyle
    {x}^2*-cos{x}+2{x}sin{x}+2cos{x}
    $

    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: $\displaystyle \int u\,\,\,dv$, where both $\displaystyle u$ and $\displaystyle dv$ are functions of $\displaystyle x$, the integral can be rewritten as:

    $\displaystyle uv - \int v\,\,\,du$

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

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

    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$\displaystyle u$ easy to differentiate. Let's pick $\displaystyle x$ as $\displaystyle u$. Differentiating, we get $\displaystyle du=dx$. Picking $\displaystyle cosxdx$ as your $\displaystyle dv$, when we integrate we will get $\displaystyle v=sinx$. We can therefore rewrite the previous expression as:

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

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

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


    $\displaystyle

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

    $
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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 $\displaystyle \int xcosx\,\,\,dx$. That 2 is distributed to both the $\displaystyle xsinx$ and $\displaystyle \int sinx\,\,\,dx$ terms that resulted from the second time integrating by parts.

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

    $\displaystyle
    {x}^2*-cos{x}-\int2{x}*-cos{x}
    $

    then


    $\displaystyle
    {x}^2*-cos{x}+2{x}sin{x}-\int 2sin{x}
    $

    then

    $\displaystyle
    {x}^2*-cos{x}+2{x}sin{x}+2cos{x}
    $

    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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