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Math Help - double integral

  1. #1
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    double integral

    Calculate the double integral where is the region:
    I dun know how to integrate it. Do i have to integrate with x first or y first? even though i tried both way, no clue to integrate.
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  2. #2
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    ok i think i figure out something..

    \int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(2x+y)dydx
    u= 2x+y
    du = dy

    \int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(u)dudx
    \int^{\pi/3}_{0} [xsin(u)]_{2x}^{2x+\pi/2}dx
    \int^{\pi/3}_{0} [xsin(2x+ \pi/2) - xsin(2x)]dx
    \int^{\pi/3}_{0} xsin(2x+ \pi/2) dx - \int^{\pi/3}_{0}xsin(2x)dx

    I'm stuck here. pls help
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  3. #3
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    Quote Originally Posted by zpwnchen View Post
    ok i think i figure out something..

    \int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(2x+y)dydx
    u= 2x+y
    du = dy

    \int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(u)dudx
    \int^{\pi/3}_{0} [xsin(u)]_{2x}^{2x+\pi/2}dx
    \int^{\pi/3}_{0} [xsin(2x+ \pi/2) - xsin(2x)]dx
    \int^{\pi/3}_{0} xsin(2x+ \pi/2) dx - \int^{\pi/3}_{0}xsin(2x)dx

    I'm stuck here. pls help
    Now use integration by parts.
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  4. #4
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    I did using by parts. But way too complicated.

    u = x
    du = dx

    dv = sin(2x+ \pi/2)dx
    v = -1/2 cos(2x+ \pi/2)

    If i substitude them in, it become complicated. isnt it right?
    Last edited by zpwnchen; November 1st 2009 at 08:47 PM.
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  5. #5
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    Quote Originally Posted by zpwnchen View Post
    I did using by parts. But way too complicated.

    u = x
    du = dx

    dv = sin(2x+ \pi/2)dx
    v = -1/2 cos(2x+ \pi/2)

    If i substitude them in, it become complicated. isnt it right?
    The whole point of integration by parts is that you end up with a simpler integral. In this case, you end up having to integrate -\frac{1}{2} \cos \left(2x+ \frac{\pi}{2} \right) - surely that's not too difficult a job (especially if you're studying double integrals) ....?
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