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Math Help - Integral Help

  1. #1
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    Integral Help



    Does anyone know how to do the above integral? I'm stumped on where to even start.
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  2. #2
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    \int -9 \cos^3 (x) \sin^2 (x) dx =

    -9 \int \cos (x) (1 - \sin^2 (x)) \sin^2 (x) dx =

    -9 \int \sin^2 (x) \cos (x) - \sin^4 (x) \cos (x) dx

    Can you do it from here?
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  3. #3
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    thanks for the help. From there you would just do u-sub with sinx, if I'm not mistaken
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  4. #4
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    Quote Originally Posted by cdlegendary View Post
    thanks for the help. From there you would just do u-sub with sinx, if I'm not mistaken
    Yes, that will work. Then you have

    u = \sin (x)

    du = \cos (x) dx

    -9 \int \sin^2(x) \cos (x) - \sin^4(x) \cos(x) dx =

    -9 \int (u^2 - u^4) du
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  5. #5
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    And then subbing back in after taking the integral would give you

    ((9sinx^5)/5)-(3sinx^3)+C

    thanks for the help
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