Simplying $ie^{it}(1+e^{-it})^n$

Question

Need some help simplifying $$ie^{it}(1+e^{-it})^n$$ where n is an integer, so I can integrate it between $0$ and $2\pi$

I tried using De Moivres Theorem but the 1+ didn't allow me too

Answer 1

Alternatively, expand $(1+e^{-it})^n$ using the binomial theorem and integrate term by term. Most of the integrals will be $0$.

Answer 2

$$ie^{it}\left(1+e^{it}\right)^n=ie^{it\left(1+\frac{n}2\right)}\left(e^{-\frac12it}+e^{\frac12it}\right)^n=ie^{it\left(1+\frac{n}2\right)}2^n\cos^n\left(\frac{t}2\right)$$