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# Laplace Transform 4: Part 2 of getting the Laplace transform of sin(at)

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- Welcome back.
- We were in the midst of figuring out the Laplace
- transform of sine of at when I was running out of time.
- This was the definition of the Laplace
- transform of sine of at.
- I said that also equals y.
- This is going to be useful for us, since we're going to be
- doing integration by parts twice.
- So I did integration by parts once, then I did integration
- by parts twice.
- I said, you know, don't worry about the boundaries of the
- integral right now.
- Let's just worry about the indefinite integral.
- And then after we solve for y-- let's just say y is the
- indefinite version of this-- then we can evaluate the
- boundaries.
- And we got to this point, and we made the realization, after
- doing two integration by parts and being very careful not to
- hopefully make any careless mistakes, we realized, wow,
- this is our original y.
- If I put the boundaries here, that's the same thing as the
- Laplace transform of sine of at, right?
- That's our original y.
- So now-- and I'll switch colors just avoid monotony--
- this is equal to, actually, let me just-- this is y.
- Right?
- That was our original definition.
- So let's add a squared over sine squared y to
- both sides of this.
- So this is equal to y plus-- I'm just adding this whole
- term to both sides of this equation-- plus a squared over
- s squared y is equal to-- so this term is now gone, so it's
- equal to this stuff.
- And let's see if we can simplify this.
- So let's factor out an e to the minus st. Actually, let's
- factor out a negative e to the minus st. So it's minus e to
- the minus st, times sine of-- well, let me just write 1 over
- s, sine of at, minus 1 over s squared, cosine of at.
- I really hope I haven't made any careless mistakes.
- And so this, we can add the coefficient.
- So we get 1 plus a squared, over s squared, times y.
- But that's the same thing as s squared over s squared, plus a
- squared over s squared.
- So it's s squared plus a squared, over s squared, y is
- equal to minus e to the minus st, times this whole thing,
- sine of at, minus 1 over s squared, cosine of at.
- And now, this right here, since we're doing everything
- with respect to dt, this is just a constant, right?
- So we can say a constant times the
- antiderivative is equal to this.
- This is as good a time as any to evaluate the boundaries.
- Right?
- If this had a t here, I would have to somehow get them back
- on the other side.
- Because the t's are involved in evaluating the boundaries,
- since we're doing our definite integral or improper integral.
- So let's evaluate the boundaries now.
- And we could've kept them along with us
- the whole time, right?
- And just factored out this term right here.
- But anyway.
- So let's evaluate this from 0 to infinity.
- And this should simplify things.
- So the right-hand side of this equation, when I evaluate it
- at infinity, what is e to the minus infinity?
- Well, that is 0.
- We've established that multiple times.
- And now it approaches 0 from the negative side, but it's
- still going to be 0, or it approaches 0.
- What's sine of infinity?
- Well, sine just keeps oscillating, between negative
- 1 and plus 1, and so does cosine.
- Right?
- So this is bounded.
- So this thing is going to overpower these.
- And if you're curious, you can graph it.
- This kind of forms an envelope around these oscillations.
- So the limit, as this approaches infinity, is going
- to be equal to 0.
- And that makes sense, right?
- These are bounded between 0 and negative 1.
- And this approaches 0 very quickly.
- So it's 0 times something bounded between 1
- and negative 1.
- Another way to view it is the largest value this could equal
- is 1 times whatever coefficient's on it, and then
- this is going to 0.
- So it's like 0 times 1.
- Anyway, I don't want to focus too much on that.
- You can play around with that if you like.
- Minus this whole thing evaluated at 0.
- So what's e to the minus 0?
- Well, e to the minus 0 is 1.
- Right?
- That's e to the 0.
- We have a minus 1, so it becomes plus 1 times-- now,
- sine of 0 is 0.
- Minus 1 over s squared, cosine of 0.
- Let's see.
- Cosine of 0 is 1, so we have minus 1 over s squared, minus
- 1 over s squared, times 1.
- So that is equal to minus 1 over s squared.
- And I think I made a mistake, because I shouldn't be having
- a negative number here.
- So let's backtrack.
- Maybe this isn't a negative number?
- Let's see, infinity, right?
- This whole thing is 0.
- When when you put 0 here, this becomes a minus 1.
- Yeah.
- So either this is a plus or this is a plus.
- Let's see where I made my mistake.
- e to the minus st-- oh, I see where my mistake is.
- Right up here.
- Where I factored out a minus e to the minus st, right?
- Fair enough.
- So that makes this 1 over s, sine of at.
- But if I factor out a minus e to the minus st, this becomes
- a plus, right?
- It was a minus here, but I'm factoring out of a minus e to
- the minus st.
- So that's a plus.
- This is a plus.
- Boy, I'm glad that was not too difficult to find.
- So then this becomes a plus.
- And then this becomes a plus.
- Thank God.
- It would have been sad if I wasted two videos and ended up
- with a careless, negative number.
- Anyway.
- So now we have s squared plus a squared, over s squared,
- times y is equal to this.
- Multiply both sides times s squared over-- s
- squared plus a squared.
- Divide both sides by this, and we get y is equal to 1 over s
- squared-- And actually, let me make sure that that is right.
- It's 1 over s squared.
- y is equal to 1 over s squared, times s squared, over
- s squared plus a squared.
- And then these cancel out.
- And let me make sure that I haven't made
- another careless mistake.
- Because I have a
- feeling I have. Yep.
- There.
- I see the careless mistake.
- And it was all in this term.
- And I hope you don't mind my careless mistakes, but I want
- you to see that I'm doing these things in real time and
- I am human, in case you haven't realized already.
- Anyway, so I made the same careless mistake.
- So I factor out an e to the minus st here, so it's plus.
- But it was a over s squared.
- So this is an a.
- That's an a.
- And so this is an a.
- And so this is an a.
- And so this is an a.
- Right?
- This was an a.
- And this is the correct answer.
- a over s squared plus a squared.
- So I hope those careless mistakes didn't
- throw you off too much.
- These things happen when you do integration by parts twice
- with a bunch of variables.
- But anyway, now we are ready to add a significant entry
- into our table of Laplace transforms. And that is that
- the Laplace transform-- I had an extra curl, there.
- That was unecessary.
- Let me do it again.
- The Laplace transform of sine of at is equal to a over s
- squared, plus a squared.
- And that's a significant entry.
- And maybe a good exercise for you, just to see how fun it is
- to do these integration by parts problems twice, is to
- figure out the Laplace transform of cosine of at.
- And I'll give you a hint.
- It's s over s squared over s squared plus a squared.
- And it's nice that there's that symmetry there.
- Anyway, I'm almost at my time limit.
- And I'm very tired working on this video.
- So I'll leave it there and I'll see you in the next one.

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