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Integral Involving Trigonometric Functions with Varying Arguments

  • Thread starter Joshk80k
  • Start date
  • #1
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Homework Statement



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]


Homework Equations



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


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?
 

Answers and Replies

  • #2
ideasrule
Homework Helper
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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.
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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?
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]
 
  • #4
17
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So I am able to rewrite

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

as

[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 =)
 
  • #5
17
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Sorry for the redundant information - I posted at nearly the same time as you did.

Thanks very much for your help =).
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
6
So I am able to rewrite

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

as

[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 =)
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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