1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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:


  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].

    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.

  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:

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

    Be careful with those limits!

    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:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook