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Difficult integral by sub., parts & table

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int xe^{2x^2} cos(3x^2) dx[/tex]
    This is the hardest integral I've attempted so far. I've come up with an answer that fits a table in my book but I'm not sure if I arrived there correctly. Thanks for reading!

    2. Relevant equations
    [tex]\int e^{au} cos\ bu\ du\ = \frac{e^{au}}{a^2+b^2}(a\ cos\ bu\ + b\ sin\ bu)+C[/tex]

    3. The attempt at a solution
  2. jcsd
  3. Mar 25, 2008 #2


    [tex]du=2x dx[/tex]

    [tex]I=\int xe^{2x^2}\cos(3x^2){\rm dx}\,=\, {1\over 2} \int e^{2u}\cos(3u) {\rm du}[/tex]

    I have done some shortcut here, with use of the formula, your number two:

    [tex]I={1\over 26} e^{2u} (2 \cos(3u)\,+\,3\sin(3u))\,+\,C [/tex]

    [tex]I={1\over 26} e^{2x^2} (2 \cos(3x^2)\,+\,3\sin(3x^2))\,+\,C [/tex]

    this is the same as integrator...


    check it out
  4. Mar 25, 2008 #3
    Here's the solution, I didn't simplify it any more like I should have, but I hope you understand the basic process. I checked my answer with CAS and it's correct, so don't worry about that.
  5. Mar 26, 2008 #4
    How do you submit scanned work onto here?
  6. Mar 26, 2008 #5
    Assuming the scanned image is in a format the forum recognises, jpeg,mpeg and so on either click on the icon in yellow with a black mountain or two and a black sun, visible in the new reply or go advanced: edit windows, not quick post. Or type [noparse][​IMG][/noparse] around the files url.

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    Last edited: Mar 26, 2008
  7. Mar 26, 2008 #6
    Thanks alot!
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