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Thread: Intergration question

  1. #1
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    Intergration question

    intergrate:

    1) cos2xsin3x dx
    2) 1+x ((x+Inx))^0.5
    x

    3) got this question sitting in front of me and not sure how to get my head about it sounds simple but aint got a scooby doo:

    'Find the volume generated by rotating the area bounded by y=x^2 and y=4 around the line y=-1

    ty x
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    intergrate:

    1) cos2xsin3x dx
    $\displaystyle \int cos(2x)sin(3x)dx$ or $\displaystyle \int cos^2(x) sin^3(x) dx$?

    -Dan
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  3. #3
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    first one

    amanda x
    Last edited by Amanda-UK; Aug 15th 2007 at 07:28 AM. Reason: typo
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  4. #4
    Forum Admin topsquark's Avatar
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    $\displaystyle \int cos(2x)sin(3x)dx$

    The simplest attack I can think of is to use:
    $\displaystyle cos(2x) = 1 - 2sin^2(x)$
    $\displaystyle sin(3x) = -4sin^3(x) + 3sin^2(x)$

    So:
    $\displaystyle \int cos(2x)sin(3x)dx = \int (1 - 2sin^2(x))(-4sin^3(x) + 3sin^2(x))dx$

    $\displaystyle = \int (8sin^5(x) - 6sin^4(x) -4sin^3(x) + 3sin^2(x)) dx$

    Then use
    $\displaystyle \int sin^n(x) dx = -\frac{sin^{n-1}(x)cos(x)}{n} + \frac{n-1}{n}\int sin^{n - 2}(x) dx$

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    $\displaystyle \int cos(2x)sin(3x)dx$

    The simplest attack I can think of is to use:
    $\displaystyle cos(2x) = 1 - 2sin^2(x)$
    $\displaystyle sin(3x) = -4sin^3(x) + 3sin^2(x)$

    So:
    $\displaystyle \int cos(2x)sin(3x)dx = \int (1 - 2sin^2(x))(-4sin^3(x) + 3sin^2(x))dx$

    $\displaystyle = \int (8sin^5(x) - 6sin^4(x) -4sin^3(x) + 3sin^2(x)) dx$

    Then use
    $\displaystyle \int sin^n(x) dx = -\frac{sin^{n-1}(x)cos(x)}{n} + \frac{n-1}{n}\int sin^{n - 2}(x) dx$

    -Dan
    do i use the last bit to substitute the bit before? after i do this what will i need to do next? amanda xx
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    do i use the last bit to substitute the bit before? after i do this what will i need to do next? amanda xx
    Use the recursion relation to reduce the $\displaystyle sin^5(x)$ integral to a $\displaystyle sin^3(x)$ integral, then to a $\displaystyle sin(x)$ integral. etc.

    -Dan
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  7. #7
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    i have never used that method is it possible to do it by using intergration by parts? amanda x
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  8. #8
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    Quote Originally Posted by topsquark View Post
    $\displaystyle \int cos(2x)sin(3x)dx$

    The simplest attack I can think of is to use:
    $\displaystyle cos(2x) = 1 - 2sin^2(x)$
    $\displaystyle sin(3x) = -4sin^3(x) + 3sin^2(x)$
    Hello Dan

    Did you try to use $\displaystyle \sin\alpha\cos\beta=\frac{\sin(\alpha+\beta)+\sin( \alpha-\beta)}2$ ?
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  9. #9
    Eater of Worlds
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    For the solid of revolution, you can washers or shells.

    Washers:

    $\displaystyle {\pi}\int_{-2}^{2}[(-1-4)^{2}-(-1-x^{2})^{2}]dx$

    Shells:

    $\displaystyle 4{\pi}\int_{0}^{4}[(y+1)\sqrt{y}]dy$

    The diagram was supposed to be animated, but I had to resize and then it wouldn't work.
    Last edited by galactus; Nov 24th 2008 at 05:39 AM.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Hello Dan

    Did you try to use $\displaystyle \sin\alpha\cos\beta=\frac{\sin(\alpha+\beta)+\sin( \alpha-\beta)}2$ ?
    You know, I recently showed that one to another poster... Ah well!

    -Dan
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