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    (Original post by APersonYo)
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    I can think of a few methods. Maybe the quickest is to write it like this

    \cos^3 x \left(1-\cos^2 x\right) \sin x

    Then if you substitute u=\cos x the \sin x will cancel.

    Or if you expand it then you can integrate each term by recognition.
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    (Original post by APersonYo)
    Please
    \cos^3x = (1-\sin^2x) \cos x

    Then use substitution.
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    (Original post by APersonYo)
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    Exactly the same as the method you'd use for cos^4 x sin ^3 x, for which you have already received two answers.

    Please stop posting question after question without taking the time to read and understand the answers to the questions you have already posted.
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    (Original post by APersonYo)
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    use  \displaystyle \cos ^2x + \sin ^2x = 1

    and get in the form  \displaystyle \cos x (\sin^3x - \sin^5x) or  \displaystyle \sin x (\cos ^3x - \cos ^5x) (you choose whichever one you want)

    from this get it in the form  \displaystyle k \int f'(x)f(x)^n dx which you can then easily integrate
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    (Original post by TeeEm)
    cos3xsin3x = 1/8(sin32x) = 1/8(sin22x)sin2x =1/8(1- cos22x)sin2x =1/8(sin2x) - 1/8(cos22x)sin2x = ....
    It's probablly quicker to use the method you gave in the cos^4 x sin^3 x thread isn't it?
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    (Original post by notnek)
    I can think of a few methods. Maybe the quickest is to write it like this

    \cos^3 x \left(1-\cos^2 x\right) \sin x

    Then if you substitute u=\cos x the \sin x will cancel.

    Or if you expand it then you can integrate each term by recognition.
    Absolute genius
 
 
 
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