Answer to Exercise 11.4-1

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

It turns out that in this type of problem you can choose it either way and still arrive at the correct answer. So I will choose a) just to be contrary to those of you who chose b) and worked it all the way through that way. Seriously though, if you did work it through using b), you can click here to see that approach expanded. Whether you did it using a) or b), you would do well to familiarize yourself with both approaches.

To give you a vague answer to the obvious question: The reason that it works both ways with this example and not with others that we've tried is because both sine and cosine have secret lives as exponential functions. You will see how in a later section on complex numbers. Because both factors in the integrand, e^-ξωx cos(ωx), are, in reality, exponential functions, it doesn't matter in what order you assign u and dv/dx to them.

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 =

cos(ωx) dx =

1 sin(ωx) + C ω

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 sin(ωx) + C - ω

1 ω

sin(ωx)

du dx = dx

e^-ξωx cos(ωx) dx

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

Of course you have

   du
       =  -ξω e^-ξωx
   dx

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

1 e^-ξωx sin(ωx) + C + ω

ξω ω

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 a) in the first round, you must choose a) 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 a) 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 =

sin(ωx) dx =

1 - cos(ωx) + C ω

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 sin(ωx) − ω

ξ e^-ξωx cos(ωx) + C + ω

ξ ω

cos(ωx)

ds dx = dx

e^-ξωx cos(ωx) dx

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

Your derivative is (just as in the first round):

   ds
       =  -ξω e^-ξωx
   dx

Substituting you will get

1 e^-ξωx sin(ωx) - ω

ξ e^-ξωx cos(ωx) + C - ξ² ω

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. Do that and then click here.

When you do, you will get

1 e^-ξωx sin(ωx) - ω

ξ e^-ξωx cos(ωx) + C = ω

(1 + ξ²)

e^-ξωx cos(ωx) dx

Now divide out the (1 + ξ²) from both sides and factor everything so it looks pretty, 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

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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