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.

1 answer

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)
Similar Questions
  1. find the trigonometric form of -12-12(square root)3ia. 24(cos2pi/3 + isin 2pi/3) b. 24(cos4pi/3 + isin 4pi/3) c. 12(cos 4pi/3 +
    1. answers icon 1 answer
  2. Find the fourth roots of− 1/2 + (square root)3/2 i Write the roots in trigonometric form. A - w 1=cos(35°)+isin(35°) w2
    1. answers icon 1 answer
    1. answers icon 1 answer
  3. 1. Prove using mathematical induction that 1+2+3+...+n=[ n(n+1)]/22. Find the derivative of 2^x^2+log2(2x^2-1) 3. Use
    1. answers icon 0 answers
more similar questions