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- Let's keep building our table of Laplace transforms. And now
- we'll do a fairly hairy problem, so I'm going to have
- to focus so that I don't make a careless mistake.
- But let's say we want to take the Laplace transform-- and
- this is a useful one.
- Actually, all of them we've done so far are useful.
- I'll tell you when we start doing not-so-useful ones.
- Let's say we want to take the Laplace transform of the sine
- of some constant times t.
- Well, our definition of the Laplace transform, that says
- that it's the improper integral.
- And remember, the Laplace transform is just a
- definition.
- It's just a tool that has turned out to
- be extremely useful.
- And we'll do more on that intuition later on.
- But anyway, it's the integral from 0 to infinity of e to the
- minus st, times-- whatever we're taking the Laplace
- transform of-- times sine of at, dt.
- And now, we have to go back and find our integration by
- parts neuron.
- And mine always disappears, so we have to reprove
- integration by parts.
- I don't recommend you do this all the time.
- If you have to do this on an exam, you might want to
- memorize it before the exam.
- But always remember, integration by parts is just
- the product rule in reverse.
- So I'll just do that in this corner.
- So the product rule tells us if we have two
- functions, u times v.
- And if I were take the derivative of u times v.
- Let's say that they're functions of t.
- These are both functions of t.
- I could have written u of x times u of x.
- Then that equals the derivative of the first times
- the second function, plus the first function times the
- derivative of the second.
- Now, if I were to integrate both sides, I get uv-- this
- should be review-- is equal to the integral of u prime v,
- with respect to dt-- but I'm just doing a little bit of
- shorthand now-- plus the integral of uv prime.
- I'm just trying to help myself remember this thing.
- And let's take this and subtract it from both sides.
- So we have this integral of u prime v is going to be equal
- to this, uv minus the integral of uv prime.
- And, of course, this is a function of t.
- There's a dt here and all of that.
- But I just have to do this in the corner of my page a lot,
- because I always forget this, and with the primes and the
- integrals and all that, I always forget it.
- One way, if you did want to memorize it, you said, OK, the
- integration by parts says if I take the integral of the
- derivative of one thing and then just a regular function
- of another, it equals the two functions times each other,
- minus the integral of the reverse.
- Right?
- Here, when you take the subtraction, you're taking the
- one that had a derivative, now it doesn't.
- And the one that didn't have a derivative, now it does.
- But anyway, let's apply that to our problem at
- hand, to this one.
- Well, we could go either way about it.
- Let's make u prime is equal to-- we'll do our definition--
- u prime is equal to e to the minus st, in which case you
- would be the antiderivative of that, which is equal to minus
- 1 over s e to the minus st, right?
- And actually, this is going to be an integration by parts
- twice problem, so I'm just actually going to define the
- Laplace transform as y.
- That'll come in useful later on.
- And I think I actually did a very similar example to this
- when we did integration by parts.
- But anyway, back to the integration by parts.
- So that's u.
- And let me do v in a different color.
- So when v-- if this is u prime, right?
- This is u prime, then this is v.
- So v is equal to sine of at.
- And then what is v prime?
- Well, that's just a cosine of at, right?
- The chain rule.
- And now, we're ready to do our integration.
- So the Laplace transform, and I'll just say that's y, y is
- equal to-- y is what we're trying to solve for, the
- Laplace transform of sine of at-- that is
- equal to u prime v.
- I defined u prime in v, right?
- That's equal to that.
- The integral of u prime times v.
- That equals uv.
- So that's minus 1 over s e to the minus st, times v, sine of
- at, minus the integral.
- And when you do the integration by parts, this
- could be an indefinite integral, an improper
- integral, a definite integral, whatever.
- But the boundary stays.
- And we can still say, from 0 to infinity of uv prime.
- So u is minus 1 over s e to the minus st, times v prime,
- times a cosine of at-- fair enough-- dt.
- Well, now we have another hairy
- integral we need to solve.
- So this might involve another integration by
- parts, and it does.
- Let's see if we can simplify it at [? all. ?]
- Let's take the constants out first. Let me
- just rewrite this.
- So we get y is equal to minus e to the minus st
- over s, sine of at.
- So you have a minus minus plus a over s-- a divided by s, and
- then these two negative signs cancel out-- times the
- integral from 0 to infinity, e to the minus st,
- cosine of at, dt.
- Let's do another integration by parts.
- And I'll do this in a purple color, just so you know this
- is our second integration by parts.
- Over here.
- Let's define once again, u prime is equal to e the minus
- st. So this is u prime.
- Then u is equal to minus 1 over s e to the minus st.
- We'll make v equal to cosine of at.
- The hardest part about this is not making careless mistakes.
- And then v prime-- I just want it to be on the same row-- is
- equal to minus a sine of at, right?
- The chain rule, derivative of cosine is minus sign.
- So let's substitute that back in, and we get-- this is going
- to get hairy; actually, it already is hairy-- y is equal
- to minus e to the minus st over s, sine of at, plus a
- over s, times-- OK.
- Integration by parts.
- uv.
- So that's minus 1 over s e to the minus st, times v, times
- cosine at, minus the integral from 0 to infinity.
- This problem is making me hungry.
- It's taking so much glucose from my bloodstream.
- I'm focusing so much not to make careless mistakes.
- Anyway, integral from 0 to infinity.
- And now, we have uv prime, so u is minus 1 over s e to the
- minus st. That's u.
- And then v prime times minus a.
- So let's make that minus cancel out with this one.
- So that becomes a plus.
- a sine of at, dt.
- I'm starting to see the light at the end of the tunnel.
- So then, let's simplify this thing.
- And, of course, we're going to have to evaluate this whole
- thing, right?
- Actually, we're going to have to evaluate everything.
- Let's just focus on the indefinite integral for now.
- We're going to have to take this whole thing and
- evaluate-- let's just say that y is the antiderivative and
- then evaluate it from infinity to 0.
- From 0 to infinity.
- So y is equal to minus e to the minus st
- over s, sine of at.
- Now let's distribute this.
- Minus a over s squared, e to the minus st, cosine of at.
- Right?
- OK, now I want to make sure I don't make a careless mistake.
- OK.
- Now, let's multiply this times this and take all of the
- constants out.
- So we have an a and an s.
- a over s.
- There's a minus sign.
- We have a plus a to the s.
- So we'll have a minus a squared over s squared, times
- the integral from 0-- well, I said I'm just worrying about
- the indefinite integral right now, and we'll evaluate the
- boundaries later.
- e to the minus st, sine of at, dt.
- Now, this is the part, and we've done this before, it's a
- little bit of a trick with integration by parts.
- But this expression, notice, is the same thing as our
- original y.
- Right?
- This is our original y.
- And we're assuming we're doing the indefinite integral, and
- we'll evaluate the boundaries later.
- Although we could have kept the boundaries the whole time,
- but it would have made it even hairier.
- So we can rewrite this integral as y.
- That was our definition.
- And actually, I just realized I'm running out of time, so
- I'll continue this hairy problem in the next video.
- See you soon.

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