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## Differential equations

### 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 at when 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 at
• 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 time
• 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 ].
• 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.
• you could search for euler's formula in khan academy, and the video is named euler's formula and identity :)
• at i still don't understand why there is still exist a boundaries from [0,infinity]. Can anyone explain this? thanks
• 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].
• why is -e^-s(0) not equal to -1
• -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.
• much easier to use complex, doesn't it?
• At about in 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?"
• 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 ] ) ]