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### Course: Differential equations > Unit 3

Lesson 1: Laplace transform# Part 2 of the transform of the sin(at)

Part 2 of getting the Laplace transform of sin(at). Created by Sal Khan.

## Want to join the conversation?

- pretty sure a small mistake was made at2:04when you factored the -e^-st out and re-wrote the terms inside the brackets, you forgot the 'a' before the cos (the term was (a/s^2)e^(-st)cos(at)

if this is not a mistake please explain where that a went

edit: it was caught at7:30(47 votes) - I don't understand how to get from y+(a^2/s^2)y to (s^2+a^2/s^2)y on the left hand side of the equation. Around time2:22(6 votes)
- Inside the brackets: after =y [ 1 + (a^2/s^2) ] just replace the 1 with s^2/s^2 and you can see you now have a common denominator = y [ (s^2/s^2) + (a^2/s^2) ]. Now you can add the numerators and get =y [ (s^2 + a^2) / s^2 ].(13 votes)

- Actually, this long method is not needed. If you write sinat in euler form, ie. sinat= e^iat-e^-iat/ 2i, things turn out pretty simple and short.(9 votes)
- you could search for euler's formula in khan academy, and the video is named euler's formula and identity :)(3 votes)

- at3:11i still don't understand why there is still exist a boundaries from [0,infinity]. Can anyone explain this? thanks(7 votes)
- I also have the same question initially. I think it's because when applying int(u'v)=uv-int(uv'), u(t) and v(t) should have the boundaries from int(u'v), otherwise it does not make sense that the equation generates an expression with t but does not follow the original boundary. The final results should not have t, because t has a boundary of [0, infinity].(2 votes)

- why is -e^-s(0) not equal to -1(5 votes)
- -e^(-s(0)) is equal to -1 but when integrating, Sal evaluates the expression at t = infinity and then subtracts the expression evaluated at 0 so when he subtracts, it canceled with the -1.4:40(5 votes)

- much easier to use complex, doesn't it?(4 votes)
- At about2:00in the video Sal made a small mistake and as usual there was a corrective note added to the video. My question is, "Is there a way to make those corrective notes show up when watching the video in full screen mode or on YouTube?"(4 votes)
- Does anyone have any explanation of why this is?:

After watching some trig substitution videos, I realized that

L{sin(t)} = d/ds[arctan(s)]

Note: d/ds[arctan(s)] is the derivative with respect to s of arctan(s), in case having an s there instead of an x confused you.

After that, I figured out the more general case:

L{sin(at)} = d/ds[arctan(s/a)]

Why is this? Are there other properties like this for other trig functions?

Also, L{cos(at)} = d/ds[ ln( sqrt[ s^2 + a^2 ] ) ]

Note: sqrt[ s^2 + a^2 ] is the square root of s^2 + a^2.

Does this have any significance? How does it relate to L{sin(at)} = d/ds[arctan(s/a)]?

L{sin(at)} = d/ds[ arctan( s/a ) ]

L{cos(at)} = d/ds[ ln( sqrt[ s^2 + a^2 ] ) ](4 votes)- Good question. Coincidence maybe? What the Laplace transform actually does is compress an exponential function so that you can analyze it. I would need to learn more to really answer your question. Remind me in an year or 2 ;-)

But based on my current intuition, the transform doesn't seem to do anything useful to periodic functions, except remove the.. 'periodicity'(1 vote)

- Just a comment on this problem... I feel it is much easier to show and explain if when you use the partial differential formula you take uv|0 to infinity - differential vdu ....in that process it makes things much quicker and simpler without mistakes(3 votes)
- if we factor out -e^-st/s instead of -s^-st at the end, one of the s is lost and the answer becomes as/s^2 + a^2 instead, someone clear this up(2 votes)

## Video transcript

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.