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Math Help - Integrate

  1. #1
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    Integrate

    (cosx)^3 * (sinx)^2 dx

    Can someone tell me how to start this, how to get at it? Thanks
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Morgan82 View Post
    (cosx)^3 * (sinx)^2 dx

    Can someone tell me how to start this, how to get at it? Thanks
    =\int(1-\sin^2x)\sin^2x\cos{x}dx

    Let u=\sin{x}...
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    =\int(1-\sin^2x)\sin^2x\cos{x}dx

    Let u=\sin{x}...
     = \int{(\sin^2{x} - \sin^4{x})\cos{x}\,dx}.

    NOW let u = \sin{x}.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Prove It View Post
     = \int{(\sin^2{x} - \sin^4{x})\cos{x}\,dx}.

    NOW let u = \sin{x}.
    I don't see anything wrong with

    \int(1-u^2)u^2du=\int(u^2-u^4)du

    The distribution can come after or before.
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  5. #5
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    True, there's nothing wrong with it.

    It's just that if a student is trying to get a function to be of the form f\left(u(x)\right)\cdot\frac{du}{dx}, it's hard enough to find the correct product as it is, let alone having a second product in there.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Prove It View Post
    True, there's nothing wrong with it.

    It's just that if a student is trying to get a function to be of the form f\left(u(x)\right)\cdot\frac{du}{dx}, it's hard enough to find the correct product as it is, let alone having a second product in there.
    Fair enough.
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  7. #7
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    I got it. Thanks
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