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Integration - two sin functions

  1. Aug 22, 2009 #1
    1. The problem statement, all variables and given/known data

    I have the answer I just dont know how he got it! I want to know how he got from the second line to the third line in the image below.


    2. Relevant equations

    eq0010MP.gif

    3. The attempt at a solution

    I have used integration by part and the u substitution method.

    When substituting with u i am still left with the other terms in the other sine function.
    When integrating by parts, it just doesn't simplify..

    Is there another method i don't know about? There doesn't seem to be any trig identities either.
     
    Last edited: Aug 22, 2009
  2. jcsd
  3. Aug 22, 2009 #2
    Hi & welcome to PF!

    Use the product to sum formulae:

    img1.gif

    Regards.
     
  4. Aug 22, 2009 #3
    ok, that seems to be the way to go (i must remember these)
    but i am still having problems evaluating it...

    [tex]\frac{1}{2a^2}[/tex][tex]\int cos(at-2a\tau) d\tau - \int cos(at) d\tau[/tex]

    [tex]\frac{1}{2a^2}[\frac{sin(at-2a\tau)}{-2a}][/tex]from t to 0 [tex]- [cos(at)\tau] [/tex] from t to 0

    im going to stop there because something is wrong already....
     
  5. Aug 22, 2009 #4
    Are a and t constants?

    Obviously [itex]\int cos(x)dx = sin(x)[/itex].

    Regards.
     
    Last edited: Aug 23, 2009
  6. Aug 22, 2009 #5
    [tex]\sin(at-a\tau)\sin(a\tau) = ?[/tex]
     
  7. Aug 22, 2009 #6
    [tex]sin(at-a\tau)sin(a\tau) = \frac{1}{2}(cos(at-a\tau-a\tau) - cos(at-a\tau+a\tau)) = \frac{1}{2}(cos(at-2a\tau) - cos(at))[/tex]
    subbing this into the integral and taking the half outside:

    [tex]\frac{1}{2a^2}( \int cos(at-2a\tau) d\tau - \int cos(at) d\tau )[/tex]

    i treated cos(at) as a constant because we are integrating with respect to tau, not t....
    or is this where i am going wrong?
     
    Last edited: Aug 22, 2009
  8. Aug 23, 2009 #7
    Actually, you forgot the limits. So your integral would be:

    [tex]\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - \int_{0}^{t} cos(at) d\tau)[/tex]

    Now make a substitution,

    - for the first integral [itex]w=at-2a\tau[/itex]

    - for the second integral treat cos(at) as constant, so that:


    [tex]\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - cos(at)\int_{0}^{t} d\tau)[/tex]

    Now find dw but do not forget the limits of the integral. You need to substitute 0, t for tau, in w.

    Regards.
     
  9. Aug 23, 2009 #8
    Ok heres another attempt

    [tex]\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - cos(at)\int_{0}^{t} d\tau) [/tex]

    let [tex]u = at-2a\tau[/tex]
    [tex]\frac{du}{d\tau} = -2a[/tex]
    [tex]\frac{du}{-2a} = d\tau[/tex]

    [tex]\frac{1}{2a^2}( \int_{0}^{t} cos(u) \frac{du}{-2a} - cos(at)\int_{0}^{t} d\tau) [/tex]


    [tex]\frac{1}{2a^2}(\frac{1}{-2a} \int_{0}^{t} cos(u) du - cos(at)\int_{0}^{t} d\tau) [/tex]

    [tex]\frac{1}{2a^2}(\frac{1}{-2a} \int_{0}^{t} cos(u) du - cos(at)(t-0)) [/tex]

    [tex]\frac{1}{2a^2}(\frac{1}{-2a} \int_{0}^{t} cos(u) du - tcos(at)) [/tex]

    substituting u back in and evaluating

    [tex]\frac{1}{2a^2}(\frac{1}{-2a} (sin(at-2at) - sin(at)) - tcos(at)) [/tex]


    [tex]\frac{1}{2a^2}(\frac{sin(at-2at)}{-2a} - \frac{sin(at)}{-2a} - tcos(at)) [/tex]


    [tex]\frac{1}{2a^3}(\frac{sin(at-2at)}{-2} - \frac{sin(at)}{-2} - atcos(at)) [/tex]

    [tex]-\frac{1}{4a^3}(sin(at-2at) - sin(at)} + 2atcos(at)) [/tex]


    im stuck again, somehow i dont think that 4 is suppose to be in front there and the coefficient of 2 in front of the cos shouldn't be there either..... is there another trig identity i should know?
     
    Last edited: Aug 23, 2009
  10. Aug 23, 2009 #9
    As I said, again you missed the limit.

    [tex]\frac{1}{2a^2}( \int_{0}^{t} cos(at-2a\tau) d\tau - cos(at)\int_{0}^{t} d\tau) [/tex]

    let [tex]u = at-2a\tau[/tex]
    [tex]\frac{du}{d\tau} = -2a[/tex]
    [tex]\frac{du}{-2a} = d\tau[/tex]

    Now, for [itex]\tau=0[/itex], u=at and for [itex]\tau=t[/itex], u = at-2at=-at

    So your integral should look like:


    [tex]
    \frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - cos(at)\int_{0}^{t} d\tau)
    [/tex]

    Be careful with those limits!

    Regards.
     
    Last edited: Aug 23, 2009
  11. Aug 23, 2009 #10
    Oh, i now realize what you ment!
    i always forget about those when i substitute u!
    thanks will try again
     
  12. Aug 23, 2009 #11
    That help! Thankyou!! You dont know how long i have been trying to figure this out.

    you posted the limits the wrong way round, but you probably mixed them up when you typed it into the brackets {}{}..


    [tex]\frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - cos(at)\int_{0}^{t} d\tau) [/tex]

    [tex]\frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - cos(at)(t-0)) [/tex]

    [tex]\frac{1}{2a^2}(\frac{1}{-2a} \int_{at}^{-at} cos(u) du - tcos(at)) [/tex]

    [tex]\frac{1}{2a^2}(\frac{1}{-2a} (sin(-at) - sin(at)) - tcos(at)) [/tex]

    [tex]\frac{1}{2a^2}(\frac{1}{-2a} (- sin(at) - sin(at)) - tcos(at)) [/tex]

    [tex]\frac{1}{2a^2}(\frac{1}{a} (sin(at) - tcos(at)) [/tex]

    [tex]\frac{1}{2a^3}(sin(at) - atcos(at) [/tex]

    Thanks again. I can now sleep knowing how he does it now! ;)
     
  13. Aug 23, 2009 #12
    Yep, you did it right. :smile:

    Regards.
     
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