 Now let's do something even more difficult. It's like the Laplace transform of this f of t. So the Laplace transform of the sine of 2t and there's a reason why probably although it's a pain we keep on writing this on this side so that's going to be the improper integral again from 0 to infinity of e to the power negative of s t sine of 2t dt. Let me not keep on writing it but you'll see it becomes important later. So two functions both of t product tool I've better decide on a u and a v prime which one's u which one's v prime let's choose this one when it is like this we're going to just iterate between sine and cosine always choose this as your u unless this is like t which will make it 1 which will make it easier as we go so u prime is going to equal 2 cosine 2t that's equal to e to the power if this is equal to e to the power negative s t that means v has got to equal negative 1 over s e to the power negative s t if I take the first derivative of that I land up with that so this has got to equal uv so I've got a negative 1 over s I've got 1 over e to the power s t and again I'm suggesting that s must now be larger than zero otherwise this does not remain a negative and I have the sine of 2t and all of this must go from 0 to infinity as far as t is concerned minus the integral of uv prime but there's a 2 which is a constant there's a negative 1 over s which is a constant integral 0 to infinity of I'm still left with e to the power negative s t cosine of 2t dt okay now let's see what we can do here it's going to be a negative 1 over s so let's plug infinity into this quickly and then zero so it's going to be 1 over infinity so that's going to be 0 1 1 multiplied by we've got to put infinity in there now of course the sine of it can only go from positive 1 to negative 1 to positive 1 but at a positive infinity I suppose it would be at the positive side but there's a proper way of doing this but let's just suggest that it is going to be a 1 minus if I put a zero in there that is going to be 1 if I put a zero in there it's going to be a zero so what are we left with here and I always have to check that I that I don't make a mistake I think I did in as much as if I plug infinity in there this is a zero if I say e to the power s times infinity this becomes a zero so I have zero minus zero so it was actually a naughty of me to suggest that this equals one I mean there's a way to look at it but there's zero in another zero so this all cancels out here I'm left with a positive 2 over s improper integral zero to infinity of e to the power negative s t cosine of 2t dt so I'm almost where I started but instead of the sine I now have a cosine you can see I'm going to get back to that so this term is all zero apologies for not for doing that incorrectly so we have 2 over s now again I'm going to have to let u equals something let's make it the cos of 2t and v prime still a product of two functions of t e to the power negative s t that makes u prime equal to negative 2 sine of 2t and that makes v equal to negative 1 over s e to the power negative s t so it's 2 over s still now we're going to have uv so there's a negative 1 over s there's a 1 over e to the power s t still with s larger than zero to remain to make this there and a cosine of 2t and all of this has got to go from zero to infinity then minus but the product of these two there is a negative two of constant that I can bring out another negative one over s that I can bring out integral of going from zero to infinity of e to the power negative s t sine of 2t guess what I'm back where I started I'm back where I started we only need to do this now that's going to be 2 over s so we're going to be 2 over s let's see what happens here be an interesting one we're going to have the negative 1 over s let's see what we have if we plug infinity in there one divided by e to the power s infinity well that's one that is definitely going to be a one times if I plug infinity in there now what would be what would be the cosine of infinity indeed what will be the cosine of infinity and that is something that we will have to that we will have to to ponder let's just leave it for a little bit let's just leave it for a little bit and now if we plug in zeros in here it's going to be one and the cosine of zero well that's so that's going to be minus and we have to get these two so let's leave that one the cosine of two times infinity out for just a little bit so these two brackets go together so what are we going to have if we plug in zero there that's one over one that's just a one cosine of zero is one so we're definitely going to have a negative a negative one there and what do we have left on this side well now see what I did see what I did made exactly the same mistake didn't I so if I plugged in infinity there I have zero okay if I plug in infinity there I have zero I have zero and anything times this so it's this zero times whatever that would still remain a zero and this would remain a one there's a negative a negative a negative so it'll definitely be a negative and we have two over let's see two over s and here we are back to the Laplace transform of the sine of two t we are just back there so this is a zero a minus one so that's negative one so here we're definitely going to have a two s and here we're going to have a one over s there we're going to have one over s we're going to have a one over s and here we're going to have a negative two over s times the Laplace transform of the sine of two t what we're going to have we've got to multiply this two s in so we're going to have two over s squared minus four over s squared times the Laplace transform of the sine of two t but remember that equals the Laplace transform of the sine of two t I still have that on my left hand side so let's bring this over to this side so I have four over s squared times the Laplace transform of the sine of two t plus the Laplace transform of the sine of two t equals two over s squared so I have to get s squared as a common denominator here that'll leave me with four times the Laplace transform of the sine of two t let's just do that minus what do we have on this side s squared times the Laplace transform of the sine of two t that equals two over s squared the two s squareds will go same in the denominator I can take out the Laplace transform of the sine of two t out as a common factor and that's going to equal two and whatever I have left on this side I'm just doing it all in one go it's four minus s squared if I took out the Laplace transform there the Laplace transform there out I'm left with four minus s squared I'm just taking it dividing both sides by it so I get that so the Laplace transform of the sine of two t equals two over two squared minus minus s squared now I must say just looking at it quickly I don't know where this suddenly became from a positive into a negative apologies for that that's that and we usually write it as two over s squared plus four and so what we basically what we do see is that this s squared is there the two is there and two squared is there so it's actually what if this was sine of eight t that would have been a over s squared plus a squared it'll always work out like this so sorry for the hickson stutters and etc for this problem you've got to think about it fortunately for us fortunately for me you don't have to consider the cosine of two t at infinity because you are multiplying it by zero so zero is left the same with a one over e to the power s times s times infinity is going to be zero zero times whatever this the whatever the sine of two times infinity is it's always multiplied by zero so zero remains so that falls away for us unfortunately yeah we this falls away and we just have the negative one times the negative one over s leaving us with one over s okay so the few hiccups there but we still get to the answer and you have to practice these these are plus transforms initially they do take some getting used to