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.
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
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)
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
I haven't checked your integration by parts, but I get
-t/2 + 1/4 (sin 2t - i cos 2t)posted by Steve