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How to do it faster?

  1. Apr 2, 2005 #1
    (excuse my poor LaTex...i don't know it very well yet :redface: )
    [tex] 2\pi \int {x^3 \sin 2x\,dx} \Rightarrow \left\{ \begin{array}{l}
    u = 2x \\
    du = 2dx \\
    \end{array} \right\} \Rightarrow \frac{\pi }{8}\int {u^3 \sin u\,du} = \frac{\pi }{8}\left( { - u^3 \cos u + \int {u^2 \cos u\,du} } \right) [/tex]
    [tex] = \frac{\pi }{8}\left( { - u^3 \cos u + u^2 \sin u + u\cos u - \sin u} \right) [/tex]
    [tex] = \frac{{\pi \left[ { - 8x^3 \cos \left( {2x} \right) + 4x^2 \sin \left( {2x} \right) + 2x\cos {2x} - \sin {2x}} \right]}}{8} [/tex]

    How can I do this faster? Are there things I can skip or connect--etc--?
    Last edited: Apr 2, 2005
  2. jcsd
  3. Apr 3, 2005 #2
    The substitution u = 2x seems unnecessary.
  4. Apr 3, 2005 #3
    just use tabular integration. that would be the fastest way to do it.
  5. Apr 3, 2005 #4
    Thanx--I tried it and it took much less time
  6. Apr 3, 2005 #5
    no problem. glad i could help. :smile:
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