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## Differential equations

### Course: Differential equations > Unit 3

Lesson 1: Laplace transform# L{sin(at)} - transform of sin(at)

Laplace Transform of sin(at) (part 1). Created by Sal Khan.

## Want to join the conversation?

- @5:29shouldn't [(-1/s) e^(-st) sin(at)] also be evaluated from zero to infinity ?(7 votes)
- That's true. Later at8:53Sal notices it(4 votes)

- I have a way to understand the integration by parts graphically:

https://rubenayla.blogspot.com/p/visualizar-integrales-por-partes.html

Imagine a rectangle of sides u and v. Then the area of the rectangle would be u*v. Imagine an awkward function inside making a strange area that you can integrate u dv. That area would be the area of the entire rectangle minus the area of the other side, which graphically turns to be the integral of v du(3 votes) - I heard that sin, cos, and tan stood for sine cosine tangarine. Is this true?(2 votes)
- Yes...furthermore csc = cosecant, sec = secant, cot = cotangent.(2 votes)

- sir in laplace what is (D2+1)y=tcost with y(0)=0, y`(0)=0 sir pleases tell how to solve(2 votes)
- Wouldn't it be easier to integrate if we took u_prime = sin(at) and v = e^(-st). We can finish integrating in one step by this process. But the solution looks a bit compact as well. What is the significance of integrating in the way it is done in the video?(1 vote)
- There is no significance in integrating either way. They both take approximately the same amount of work, and neither have any advantages over the other.(3 votes)

- I'm not sure if I am correct, but shouldn't u' be e^-st dt not just e^-st ?(2 votes)
- u' would only have the dt applied at the end of it during integration. In other words, you'll see the integral of u'dt - the dt is independent of the u'.(1 vote)

- hey by the way, this "hairy problem" becomes a smoothly shaved, easy peasy integral if we use complex substitution (Euler's formula) to sub in for Sin at Or Cos at. It comes down to (s+ai)/(s^2+a^2). And so if we are wanting the

L (sin at), we note that the Sin is the imaginary part of the Euler formula, so we choose the imaginary part of the top... L (sin at) = a/(s^2+a^2)! Super easy. And we can use that same answer above for L (cos at). Since cos is the Real part of the Euler formula then its the Real part of the solution... Therefore,

L(cos at)= s/(s^2+a^2) ! Cool stuff. and MUCH easier integration to get these!(2 votes) - How did we come up with this definition of Laplace transform?? ie. The improper integral of e^(-st) * f(t) *dt.(1 vote)
- thats got to be the most confusing way i have ever saw integration by parts been done, i just can't get my head around what you're even doing(1 vote)
- Wouldn't this have been easier using Euler's formula, rewriting sin(ax) as 1/(2i)*(e^(i*ax) - e^(-i*ax)), converting the problem into the integration of two simple exponentials? I guess I'm not sure why parts is better...(1 vote)

## Video transcript

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 v 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 sine. 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.