Uploader Comments (khanacademy)
Top Comments
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I saw an interesting and much faster way to derive this by using the backwards Euler Formula: sin(at)=(exp(iat)-exp(-iat))/2
i or by taking the imaginary part of the Laplace transform of exp(iat)
All Comments (40)
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is this right?: i let u = e^-st and dv = sin at
then differentiate u for du = -1/s e^-st
then integrate dv for v = -a cos at
i dont get how you got (a cos at) as positive.
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haha hate it too but look to this simple method=
we all know--> cos t=1/2[e^jt+e^-jt] ,we can use that to solve
so it will be F(s)= (integration of cos t .e^-st)=
integration of (1/2[e^jt+e^-jt].e^-st)=
i/2[(1/s-j)=(1/s+j)]...=s/s^2+
1^2 its much easier than all of that ^_^
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that's y i hate integration by parts... it make me confius... huh... =.="
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I FREAKIN hate integration by parts!!
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9:25 "...cos at, now i wanna make sure i dont make any careless mistakes *checks intently* ... ok so now there's a minus sign *scratches out t thinking its a + sign*" lol
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wow ! great videos ! pure concept building
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Whoa infinite loop!!!... going to watch the next video to see how you get out of this one xD
Depends on what you pick for u and v. I rederive the formula from the product rule everytime (it's the only way you'll remember it 15 years after taking calculus)
khanacademy 3 years ago 12