Integral Involving Trigonometric Functions with Varying Arguments

[Joshk80k]

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:

\int cos(\omega t) sin(\omega t - \delta) dt

Homework Equations

sin(A-B) = sin(A)cos(B) - sin(B)cos(A)

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.

\int cos(\omega t)sin(\omega t)cos(\delta) -sin(\delta)cos^2(\omega t) dt

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?

 

[ideasrule]

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.

 

[gabbagabbahey]

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.

\int cos(\omega t)sin(\omega t)cos(\delta) -sin(\delta)cos^2(\omega t) dt

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:

\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

The first integral can be easily done by substituting u=\sin(\omega t) the second integral can be evaluated by using another trigonometric identity, \cos^2(x)=\frac{1}{2}\left(\cos(2x)+1\right)

 

[Joshk80k]

So I am able to rewrite

\int cos(\omega t)sin(\omega t)cos(\\delta) -sin(\delta)cos^2(\omega t) dt

as

\int cos(\omega t)sin(\omega t)cos(\\delta)dt - \int sin(\delta)cos^2(\omega t) dt

?

I guess that does make it really easy - thanks =)

 

[Joshk80k]

Sorry for the redundant information - I posted at nearly the same time as you did.

Thanks very much for your help =).

 

[gabbagabbahey]

So I am able to rewrite

\int cos(\omega t)sin(\omega t)cos(\\delta) -sin(\delta)cos^2(\omega t) dt

as

\int cos(\omega t)sin(\omega t)cos(\\delta)dt - \int sin(\delta)cos^2(\omega t) dt

?

I guess that does make it really easy - thanks =)

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