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    In the question I had to do  \displaystyle\frac{d}{dx}(4x-sin3x)^2=(8-6cos3x)(4x-sin3x) . Then later in it told me to  \displaystyle\int(cos3x)(4x-sin3x)\ dx suppose we let  v=(4x-sin3x)^2 can someone explain to me how this integral is equivalent to  \displaystyle-\frac{1}{6}\int\frac{dv}{dx}\ dx + \frac{4}{3}\int 4x-sin3x \ dx . Thanks.
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    (Original post by JBKProductions)
    In the question I had to do  \displaystyle\frac{d}{dx}(4x-sin3x)^2=(8-6cos3x)(4x-sin3x) . Then later in it told me to  \displaystyle\int(cos3x)(4x-sin3x)\ dx suppose we let  v=(4x-sin3x)^2 can someone explain to me how this integral is equivalent to  \displaystyle-\frac{1}{6}\int\frac{dv}{dx}\ dx + \frac{4}{3}\int 4x-sin3x \ dx . Thanks.
    Note that \displaystyle\int (\cos 3x)(4x-\sin 3x) dx

     = -\dfrac{1}{6}\displaystyle\int (-6\cos 3x)(4x-\sin 3x)

     = -\dfrac{1}{6} \displaystyle\int ((8-6\cos 3x)-8)(4x-\sin 3x)  dx

    Now try expanding the brackets in a way that takes advantage of the distributive property and then attempt to use your substitution.

    Spoiler:
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    -\dfrac{1}{6} \displaystyle\int (8-6\cos 3x)(4x -\sin 3x) dx +\frac{1}{6} \displaystyle\int 8(4x-\sin 3x)dx

    Then use your substitution of the left hand integral.


    EDIT: I feel as though this is too close to a full solution so I'll spoilerise some of it. :p:
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    (Original post by Farhan.Hanif93)
    Note that \displaystyle\int (\cos 3x)(4x-\sin 3x) dx

     = -\dfrac{1}{6}\displaystyle\int (-6\cos 3x)(4x-\sin 3x)

     = -\dfrac{1}{6} \displaystyle\int ((8-6\cos 3x)-8)(4x-\sin 3x)  dx

    Now try expanding the brackets in a way that takes advantage of the distributive property and then attempt to use your substitution.

    Spoiler:
    Show
    -\dfrac{1}{6} \displaystyle\int (8-6\cos 3x)(4x -\sin 3x) dx +\frac{1}{6} \8(4x-\sin 3x)dx

    Then use your substitution of the left hand integral.


    EDIT: I feel as though this is too close to a full solution so I'll spoilerise some of it. :p:
    Thanks!
 
 
 
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