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Homework Help: Integral Involving Trigonometric Functions with Varying Arguments

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm in an Intermediate Mechanics course right now, and while the Physics itself isn't giving me too much trouble, I am lagging behind in the Math department. I am trying to solve the integral:

    [tex]\int cos(\omega t) sin(\omega t - \delta) dt[/tex]

    2. Relevant equations

    [tex]sin(A-B) = sin(A)cos(B) - sin(B)cos(A)[/tex]

    3. The attempt at a solution

    The first thing I recognized is that the trig functions had the same argument, plus a value, so I figured I could apply the above equation to the integral. However, that really just made things look more complicated.

    [tex]\int cos(\omega t)sin(\omega t)cos(\delta) -sin(\delta)cos^2(\omega t) dt[/tex]

    I stared at this for a while, but I couldn't find any substitutions (Which is what I was expecting.) I then thought that maybe I should try an integral table, to see if this was listed somewhere, but I couldn't find any functions that might have made sense. The added value in the argument of the "Sin" function is what's tripping me up.

    Can anyone give me a push in the right direction?
  2. jcsd
  3. Mar 15, 2010 #2


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    sin(wt)cos(wt) is easy enough to integrate; just use u=sin(wt). For cos^2(wt), the standard way of integrating this is to use the identity cos(2x)=2cos^2(x) - 1.
  4. Mar 15, 2010 #3


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    Looks good so far, now just split the integral into two and pull the constants out front:

    [tex]\int \left[\cos(\omega t)\sin(\omega t)\cos(\delta) -\sin(\delta)\cos^2(\omega t)\right]dt= \cos(\delta)\int\sin(\omega t)\cos(\omega t)dt-\sin\delta\int\cos^2(\omega t)dt[/tex]

    The first integral can be easily done by substituting [itex]u=\sin(\omega t)[/itex] the second integral can be evaluated by using another trigonometric identity, [itex]\cos^2(x)=\frac{1}{2}\left(\cos(2x)+1\right)[/itex]
  5. Mar 15, 2010 #4
    So I am able to rewrite

    [tex] \int cos(\omega t)sin(\omega t)cos(\\delta) -sin(\delta)cos^2(\omega t) dt [/tex]


    [tex] \int cos(\omega t)sin(\omega t)cos(\\delta)dt - \int sin(\delta)cos^2(\omega t) dt [/tex]


    I guess that does make it really easy - thanks =)
  6. Mar 15, 2010 #5
    Sorry for the redundant information - I posted at nearly the same time as you did.

    Thanks very much for your help =).
  7. Mar 15, 2010 #6


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    Sure, one of the fundamental rules of calculus is that [itex]\int\left[f(x)+g(x)\right]dx=\int f(x)dx+\int g(x)dx[/itex].
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