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Homework Help: What's the indefinite integral formula?

  1. Aug 19, 2008 #1
    1. The problem statement, all variables and given/known data
    evaluate the integral
    ∫x^2 sinpi x dx

    2. Relevant equations
    ∫u dv= uv - ∫v du
    integration by parts formula

    3. The attempt at a solution
    u=x^2 dv= sin pi x dx
    du = 2x v = -cos pi x dx ?????? the pi is giving me trouble
  2. jcsd
  3. Aug 19, 2008 #2

    Would using 3.14 instead of pi would help?

    Sorry, I am not used to this type of Int by parts. I make
    d/dx (f(x)) = .. equation and integrate

    But, seems like you having trouble integrating cos(pi*x)?
  4. Aug 19, 2008 #3
    i figured it out already
  5. Aug 19, 2008 #4
    my answer is -1/pi cos pi x + 1/pi 2x sin pi x - 2(1/pi) -cos pi x +c
    but in the back of the book the answer is - 1/pi cos pi x + 2/pi^2 sin pi x + 2/pi^3 cos pi x +c
  6. Aug 19, 2008 #5
    possible that you can write out the solution?

    I hate to simplify and compare but this is what I got with matlab if you want confirm it with book:

    >> int('x^2*sin(pi*x)','x')

    ans =

    Last edited: Aug 19, 2008
  7. Aug 19, 2008 #6
    here's me work

    u=x^2 dv= sin pi x
    du = 2x v = 1/pi -cos pi x dx

    x^2 (1/pi) (-cos pi x) -∫2x (1/pi) -cos pi x dx or - (1/pi) x^2 (cos pi x) + (1/pi) ∫2x cos pi x dx

    u=2x dv= (1/pi) cos pi x
    du=2 v= (1/pi) sin pi x

    (1/pi) ∫2x cos pi x dx= 2x (1/pi) sin pi x - ∫ 2(1/pi) sin pi x dx
    plug this equation to the other one and you get

    (1/pi) x^2 (cos pi x) + (2/pi) x sin pi x + 2/pi cos pi x + c
  8. Aug 19, 2008 #7
    You are forgetting to divide by pi when you substitute (you are multiplying by 2 but forgetting about pi as it is 2/pi?)
    and second problem is you should have cos(pi*x)/pi^2 but you have cos(pi*x)/pi

    Only seems to be coefficient problem, everything else looks good!
  9. Aug 19, 2008 #8


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    Science Advisor
    Homework Helper

    Hi afcwestwarrior! :smile:

    (have a pi: π :smile:)

    The only difference is that you have 1/π everywhere, but the answer has 1/π2 or 1/π3

    that's because each time you integrate a function of (πx), you must divide by π …

    two integrations π2, three integrations π3. :wink:
  10. Aug 19, 2008 #9
    Ok, so your saying that i have to divide by pi each time i integrate a function of nx.
  11. Aug 19, 2008 #10
    so if that pi were a 5, then i'd have to divide by 5 each time right
  12. Aug 19, 2008 #11
    I get it now thanks.
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