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Evaluating the integral, correct?

  1. Jan 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral

    [tex]\int x^2 \cos mx dx[/tex]


    2. Relevant equations


    Evaluating the integral, correct?

    3. The attempt at a solution

    [tex]u = x^2[/tex]
    [tex]du = 2x[/tex]
    [tex]dv = \cos mx[/tex]
    [tex]v= \frac {\sin mx }{m}[/tex]

    (x^2)(sin mx / m) - [integral] (sin mx / m)(2x)

    (x^2)(sin mx / m) - 2 [integral] (sin mx / m) (x)

    [My final answer:] (x^2)(sin mx / m) + 2 (cos mx / m) + c
     
    Last edited: Jan 18, 2008
  2. jcsd
  3. Jan 18, 2008 #2
  4. Jan 18, 2008 #3

    dynamicsolo

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    Homework Helper

    I believe you lost a term in evaluating the second term integral in this:

    (x^2)(sin mx / m) - 2 [integral] (sin mx / m) (x) .

    If you differentiate your final result,

    (x^2)(sin mx / m) + 2 (cos mx / m) + c ,

    you don't cancel out the additional terms beyond the original integrand...
     
  5. Jan 18, 2008 #4
    i heard that i should do parts with xsinmx / m, should i do that?
     
  6. Jan 18, 2008 #5

    dynamicsolo

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    Homework Helper

    Yes, you'll need to do that. In general, integration of functions defined by polynomials times sin kx, cos kx, or e^kx need multiple stages of integration by parts.
     
  7. Jan 18, 2008 #6
    ok with doing parts again confuses me since i have xsinmx /m. The u should it be m? but then what is the derivative of m?
     
  8. Jan 18, 2008 #7

    dynamicsolo

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    Homework Helper

    The 'm' is a constant, so that will not be involved in the integration. Make the integral
    (1/m) · integral[ x sin(mx) ] dx . Now u = x and dv = sin(mx) dx . You can append the multiplicative constant (1/m) afterwards...
     
  9. Jan 18, 2008 #8
    ok so now I have

    x^2/m sinmx + 2/m [integral] xsinmx

    u = x
    du = dx

    dv = sin mx
    v = 1/m cos mx

    and now im lost
     
  10. Jan 18, 2008 #9
    Ok let's start from scratch.

    [tex]I=\int x^2\cos{mx}dx[/tex]

    [tex]u=x^2[/tex]
    [tex]du=2xdx[/tex]

    [tex]dV=\cos{mx}dx[/tex]
    [tex]V=\frac{1}{m}\sin{mx}[/tex]

    [tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\int x\sin{mx}dx[/tex]

    Now we have to do Parts again.

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

    [tex]dV=\sin{mx}dx[/tex]
    [tex]V=\frac{-1}{m}\cos{mx}[/tex]

    [tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)[/tex]

    Now you can easily evaluate this Integral!!!
     
    Last edited: Jan 18, 2008
  11. Jan 18, 2008 #10
    Ok, I'm officially done typing! Sorry about that, had too many typographical errors.
     
  12. Jan 18, 2008 #11
    [tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)[/tex]
    ok where did the last [tex]\int\cos{mx}dx\right)[/tex] come from
     
  13. Jan 18, 2008 #12
    By doing Parts again to evaluate ...

    [tex]\int x\sin{mx}dx[/tex]
     
  14. Jan 18, 2008 #13
    ok i see now, thank you, sad part is im not done with this question and i have one more just like it :(
     
    Last edited: Jan 18, 2008
  15. Jan 18, 2008 #14
    Post it and we'll work on it step by step.
     
  16. Jan 18, 2008 #15
    [integral] e^-x cos 2x dx

    u = e^-x
    du = -e^-x

    dv = cos 2x
    v = sin 2x / 2
     
  17. Jan 18, 2008 #16
  18. Jan 18, 2008 #17
    [tex]I=\int e^{-x}\cos{2x}dx[/tex]

    [tex]u=e^{-x}[/tex]
    [tex]du=-e^{-x}dx[/tex]

    [tex]dV=\cos{2x}dx[/tex]
    [tex]V=\frac{1}{2}\sin{2x}[/tex]

    [tex]I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\int e^{-x}\sin{2x}dx[/tex]

    Ok done typing. Evaluate your current Integral, notice that your Original Integral reappears? Just move it to the left then divide by the constant and you're done!
     
    Last edited: Jan 18, 2008
  19. Jan 18, 2008 #18
    ok so [tex]\frac{1}{2}\int e^{-x}\sin{2x}dx[/tex] turns into -1/2e^-xcos2x + c?
     
    Last edited: Jan 18, 2008
  20. Jan 18, 2008 #19
    You're supposed to have "2" Parts just like any Integration by Parts.

    What is the other term?
     
  21. Jan 18, 2008 #20
    so i dont take the integral just yet? i do another integration by parts?
     
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