Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How to do it faster?
  1. How to do it faster? (Replies: 1)

  2. How did they do this? (Replies: 4)

Loading...