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Trig Integral

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex] 59x(cos(x))2 dx

    2. Relevant equations

    3. The attempt at a solution

    I tried doing integration by parts with u= (cos(x))2 and dv= xdx
    v= [tex]\frac{x^2}{2}[/tex]
    However this didnt get me very far can some one tell me what the first step or two are.
  2. jcsd
  3. Oct 20, 2009 #2
    I got the the answer to be 59([tex]\frac{1}{4}[/tex]x sin(2x) + [tex]\frac{1}{8}[/tex]cos(2x) + [tex]\frac{x^2}{4}[/tex])
    But this was from my calculator I still dont know how to do it.
  4. Oct 20, 2009 #3


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    [tex]\cos^2(x) = \frac {1 + \cos{(2x)}}{2}[/tex]

    and integration by parts.
  5. Oct 20, 2009 #4
    ok Ill check it out thanks
  6. Oct 20, 2009 #5
    OK I tried it and it just got more complicated. Whats the next step?
  7. Oct 20, 2009 #6


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    what LCKurtz suggested shoudl lead to a pretty simple integral, maybe show what you did
  8. Oct 20, 2009 #7
    just did x(1+co(2x)/(2))
    Do i distribute or use by parts now?
  9. Oct 20, 2009 #8


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    yep multiply out and use parts on the (x.cos(2x)) part
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