Integration by Parts
integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be pi.
**I pull i out because it is a constant.
My work:
let u=e^(it) du=ie^(it)dt dv=sin(t) v=cos(t)
i integral sin(t)e^(it)dt= e^(it)cos(t)+i*integral cost(t)e^(it)dt
do integration by parts again, then I get
=e^(it)cos(t)+i[e^(it)sin(t)i*integral e^(it)sin(t)dt.
then i add i*integral e^(it)sin(t)dt to both sides
2i*integral e^(it)sin(t)dt= e^(it)(cos(t)i(sin(t)))
divide by 2 and because e^(it)=cos(t)isin(t)
i*integral e^(it)sin(t)dt=1/2
So my question is how do I get pi from this? I have not plugged in 0 to 2pi interval and I should have t/2 as a function to plug the interval in.

since sin(t) = (e^t  e^it)/2,
isin(t) = i/2 (e^it  e^it)
i sint e^it = i/2 (e^2it  1)
so, the integral is just
i/2 ((1/2i) e^2it  t)
= 1/4 e^2it  t/2
do that from 0 to 2π and you have
[1/4 e^4πi  π][1/4]
= 1/4  π  1/4
= π
I haven't checked your integration by parts, but I get
t/2 + 1/4 (sin 2t  i cos 2t)posted by Steve
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