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Math Help - 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
    \int cos(2x)sin(3x)dx or \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; August 15th 2007 at 07:28 AM. Reason: typo
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  4. #4
    Forum Admin topsquark's Avatar
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    \int cos(2x)sin(3x)dx

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

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

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

    Then use
    \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
    \int cos(2x)sin(3x)dx

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

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

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

    Then use
    \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 sin^5(x) integral to a sin^3(x) integral, then to a 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
    \int cos(2x)sin(3x)dx

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

    Did you try to use \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:

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

    Shells:

    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; November 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 \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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