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Math Help - Integration by parts.

  1. #1
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    Integration by parts.

    I almost understand all of this question, except in the last step where we have "128". I will show my work up to that step so you can follow.

    \int{x^2sin\frac{x}{4}}dx

    -4x^2cos\frac{1}{4}x--8\int{xcos\frac{1}{4}}xdx

    -4x^2cos\frac{1}{4}x+8[{4xsin\frac{1}{4}x-4\int{sin\frac{1}{4}xdx}}]

    -4x^2cos\frac{1}{4}x+32xsin\frac{1}{4}x+128cos\frac  {1}{4}x+c

    I thought that distributing the 8 in step 3 would give us "+32cos...x+c?"
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  2. #2
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    Quote Originally Posted by gammaman View Post
    I almost understand all of this question, except in the last step where we have "128". I will show my work up to that step so you can follow.

    \int{x^2sin\frac{x}{4}}dx

    -4x^2cos\frac{1}{4}x--8\int{xcos\frac{1}{4}}xdx

    -4x^2cos\frac{1}{4}x+8[{4xsin\frac{1}{4}x-4\int{sin\frac{1}{4}xdx}}]

    -4x^2cos\frac{1}{4}x+32xsin\frac{1}{4}x+128cos\frac  {1}{4}x+c

    I thought that distributing the 8 in step 3 would give us "+32cos...x+c?"
    It comes from dividing the 32 by the 1/4 in sin(x/4) to give 128
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  3. #3
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    Ok I see it now. The anti-derivative of sin[(1/4)] is -4cos so -4 x -4 = +16
    16 x the 8 in front gives us 128.
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