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Current time:0:00Total duration:10:44

L{sin(at)} - transform of sin(at)

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 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 we want to take the left actually to all of them we've done so far useful I'll tell you when the one we start doing not so useful ones let's say we want to take the Laplace transform of the sign of some constant times T well our definition of the Laplace transform that's that says that it's the improper integral remember the Laplace transform is just the definition it's just a tool that has turned out to be extremely useful and and will 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 a T DT and now we have to go back and and and find our integration by parts neuron and mine always disappears so I 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 if we have two functions U times V and if I were to 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 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 this integral the integral of U prime V is going to be equal to this u V minus the integral of U V 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 it 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 okay the integration by parts says if I take the derivative of the derivative of one thing and then a function and then just a regular function of another it equals the two functions times each other minus the integral of the reverse right here it when you take the subtraction you're taking the the one that had a derivative now doesn't and the one that didn't have a derivative now does but anyway let's apply that to our problem at hand to this one well let's make we could go either way about it let's make u prime is equal to so let's say u prime will 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 alright and actually just so this is going to be an integration by parts twice problems 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 the 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 in V if this is if this is U Prime right this is U Prime then this is V so V is equal to sine of a T and then what is V Prime well that's just a cosine of a T right the chain rule and now we're ready to to do our integration so the Laplace transform and I'll just say that's why Y is equal to right why is what we're trying to solve for the Laplace transform of sine of a T that is equal to u prime V we I defined u Prime and V right that's equal to that you probably the integral of u prime times V that equals UV so UV so that's minus 1 over s e to the minus s T times V sine sine of a t minus the integral and when you do what you do the integration by parts and you know this could be a an indefinite integral an improper integral or definitely whatever but the boundary stay so we can still say from 0 to infinity 0 to infinity of U V prime so U is minus 1 over s minus 1 over s e to the minus s T times V prime times a cosine of a t a cosine of a T fair enough DT well now we have another hairy integral we need to solve so this might involve another integration by parts which and it does so here let's see let's see if we can simplify it about let's take the constants out first so let me just rewrite this so we get Y is equal to minus e to the minus s T over s sine of ay T so you have a minus minus plus a over s plus a over s right a divided by s and there's two negative signs cancel out times the integral from 0 to infinity e to the minus s T cosine of a T DT let's do another integration by parts and I'll do this in a purple color just you know this is our second integration by parts over here well let's define let's define once again let's define u prime u prime is equal to e to the minus s T so this is u prime then U is equal to minus 1 over s e to the minus s T will make V equal to cosine of 80 d is equal to cosine of 80 the hardest part about this is just not making careless mistakes and then V Prime I just want it to be in the same row is equal to what - a - a sine of 80 right chain rule derivative of cosine is minus sine so let's substitute that back in and we get we get this is going to get hairy if unless if you actually already is hairy Y is equal to minus e to the minus s T over s sine of ay T plus a over s times x ok integration by parts UV so that's minus 1 over s e to the a minus s T times V times cosine a t - minus the integral minus the integral from 0 to infinity problems making me hungry it's it's taking so much blue 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 minus 1 over s e to the minus s T that's U and then V prime times minus a so let's put that - let's make that - cancel out with this one so that becomes a plus a sine of 80 sign of 80 DT I'm starting to see the light at the end of the tunnel so then let's let's simplify this thing and of course we're going to have to evaluate this whole thing right from infinity actually we're going to evaluate everything let's just focus on the indefinite integral for now we're going to take this whole thing and value let's just say that Y is the antiderivative and then evaluate it from infinity to zero from zero to infinity so this is equal to Y is equal to minus e to the minus s T over s sine of 80 now let's distribute this minus minus a over s squared e to the minus s T cosine of 80 right okay now I want to make sure I don't make a careless mistake okay now let's multiply this times this and take all the constants out so we have an a and an S a / s there's a minus sign we have a plus a to the S so we'll have a minus minus a squared a squared over s squared times the integral from zero well I said I'm just worrying about the improper I'm sorry the the indefinite integral right now and we'll take the value 8 at the boundaries later e to the minus s T sine of a T DT now this is the Parden we've done this before it's a bit 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've made a little even hair here so we can rewrite this integral as Y that was in our definition so we get 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