<transcript><audio_3>sine of Omega T as 1 over 2i it's about I Omega t minus its power minus I Omega T because if you write it in this way you can it gets two Exponential's and integrating Exponential's is easy because you get here e to the power</transcript><transcript><audio_4>minus AZ plus I Omega T so if you integrate it you get the it is power I Omega minus s times T divided by minus s and for the other one you get a minus and we have e to the power minus st and</transcript><transcript><audio_5>e to the power minus I Omega T so you get e to the power minus I minus s times T its antiderivative I have to divide by minus I Omega minus s and then if you do it between the two boundaries well</transcript><transcript><audio_6>provided you take your as appropriately contributions to up about vanish only have to country we're lowering so you keep your own over 2i the lower boundary you get from this part 1 times</transcript><transcript><audio_7>1 over I Omega mines s missing dishes mines are further because you're on the lower boundary so this term over here now for the other term you get a minus the mines over here of course again one of two I and then we</transcript> sine of Omega T as 1 over 2i it's about I Omega t minus its power minus I Omega T because if you write it in this way you can it gets two Exponential's and integrating Exponential's is easy because you get here e to the power minus AZ plus I Omega T so if you integrate it you get the it is power I Omega minus s times T divided by minus s and for the other one you get a minus and we have e to the power minus st and e to the power minus I Omega T so you get e to the power minus I minus s times T its antiderivative I have to divide by minus I Omega minus s and then if you do it between the two boundaries well provided you take your as appropriately contributions to up about vanish only have to country we're lowering so you keep your own over 2i the lower boundary you get from this part 1 times 1 over I Omega mines s missing dishes mines are further because you're on the lower boundary so this term over here now for the other term you get a minus the mines over here of course again one of two I and then we
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