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More integration by parts

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    ∫ x2 sin x

    2. Relevant equations

    uv - ∫ v du

    3. The attempt at a solution

    u = x2

    du = 2x

    dv = sin x

    v = -cos x

    step 1. x2 - cos x - ∫ -cos x 2x

    I think -cos x * 2x becomes -2x cos x
    so now we have

    step 2. x2 - cos x - ∫ -2x cos x

    which means I have to integrate by parts again. Here, concentrating on just the right hand side

    u = 2x
    du = 2
    dv = cos x
    v = sin x

    step 3. 2x sin - ∫ sin x * 2

    [after 10 minutes of research I've decided that I have to move that 2 to the left of the integral. That sort of helps. previously I took the antiderivative of 2.]

    step 4. 2x sin + 2 -cosx

    now add the left hand side part from above

    step 5. x2 - cos x

    step 6. x2 - cos x + 2x sin x + - 2 cosx + C

    the book says the answer is

    -x2 cos x + 2x sin x + 2 cos x + C

    So I'm almost correct, I just don't understand how they got the negative on x2, Also my right cos x is negative and their's is positive.
    Last edited: Feb 9, 2012
  2. jcsd
  3. Feb 9, 2012 #2


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    Homework Helper

    You missed a pair of parentheses.

  4. Feb 9, 2012 #3
    There's your answer for the first part of your question. For the second part, remember that ∫ Sin[x] = -Cos[x]
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