Answer to Exercise 11.4-1 (alternative attack)

So you took the road less traveled, eh?

The problem is to use integration by parts to integrate

e^-ξωx cos(ωx) dx

Step 1: Choose your parts. The choices for parts to put into equation 11.4-6a appear to be

a)
u = e^-ξωx
and
dv = cos(ωx) dx

or
b)
u = cos(ωx)
and
dv = e^-ξωx dx

And you chose to go with b) -- if you chose to go with a), you can find its expansion by clicking here.

Step 2: What is v? Remember that you find v by integrating the expression you choose for dv/dx. So do that, then click here to see if you got the right expression.

You have dv/dx = cos(ωx). So using the table and equation 11.2-9

v =

e^-ξωx dx =

1 - e^-ξωx + C ξω

You remember that in this step you are allowed to drop the C in the above when you stick this in for v into equation 11.4-6a (where you keep 11.4-6a's C). So what you have so far is

1 - e^-ξωx cos(ωx) + C + ξω

1 ξω

e^-ξωx

du dx = dx

e^-ξωx cos(ωx) dx

Step 3: What is du/dx? You find that by taking the derivative of u = cos(ωx). So do that, then plug the result into the above and click here to move on.

Of course you have

   du
       =  - ω sin(ωx)
   dx

and when you put that into the integration by parts you have already developed, you get

1 - e^-ξωx cos(ωx) + C - ξω

ω ξω

e^-ξωx sin(ωx) dx =

e^-ξωx cos(ωx) dx

(go ahead and cancel those ω's in the numerator and denominator of that middle coefficient) which brings you to

Step 4: Do the remaining integral: It doesn't really look as though that integral to the left of the equal is any easier than the original integral, which is shown to the right of the equal. But bear with me. By applying integration by parts again to the left-hand integral, you'll see that a miracle will happen. To do this step, we have to do steps 1a) through 4a, just as we did in the example given in the main text.

Step 1a) Choose your parts. Here the choices are

a)
s = e^-ξωx
and
dt = sin(ωx) dx

or
b)
s = sin(ωx)
and
dt = e^-ξωx dx

It turns out that if you choose b) in the first round, you must choose b) again in the second round. Why? Try mixing strategies in that way and you'll see that just when you think you're getting near the end, everything in your equation will cancel (assuming no mistakes) and you'll be left with the universal truth that 0 = 0.

Step 2a: What is t? Just integrate the expression in b) for dt/dx. When you've done that, click here to move on.

You should have from the table and equation 11.2-9:

t =

e^-ξωx dx =

1 - e^-ωx + C ξω

At this point you should have dropped the C in the above, and combined the C that arises from 11.4-6a with the one you already have in the expression you have developed so far (into a single C). With all that, it ought to look like:

1 - e^-ξωx cos(ωx) + ξω

1 e^-ξωx sin(ωx) + C - ξ²ω

1 ξ²ω

e^-ξωx

ds dx = dx

e^-ξωx cos(ωx) dx

Step 3a: What is ds/dx? Just take the derivative of s = sin(ωx). Then substitute that expression into the above. When you're done, click here.

Your derivative is:

   ds
       =  ω cos(ωx)
   dx

Substituting and taking a cancellation of ω's, you will get

1 - e^-ξωx cos(ωx) + ξω

1 e^-ξωx sin(ωx) + C - ξ²ω

1 ξ²

e^-ξωx cos(ωx)

e^-ξωx cos(ωx) dx

Step 4a: Do the remaining integral -- Gotcha! Once again we have returned to the same integral that we started with. So use the same trick. Just move that integral over to the right-hand side of the equal. It's easier if you also multiply both sides by ξ² to boot. Do all that and then click here.

When you do, you will get

ξ − ω

e^-ξωx cos(ωx) +

1 ω

e^-ξωx sin(ωx) + C =

(1 + ξ²)

e^-ξωx cos(ωx) dx

Now divide out the (1 + ξ²) from both sides and factor everything so it looks pretty, swap the sin and cos terms according to the commutative law, and your final answer will come out looking like:

1 e^-ξωx ω(1 + ξ²)

sin(ωx) - ξ cos(ωx)

+ C =

e^-ξωx cos(ωx) dx

same as if you had taken choice a) at the beginning.

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a)	u = e^-ξωx	and	dv = cos(ωx) dx
or
b)	u = cos(ωx)	and	dv = e^-ξωx dx

a)	s = e^-ξωx	and	dt = sin(ωx) dx
or
b)	s = sin(ωx)	and	dt = e^-ξωx dx