Complex Integration using substitution

Question

Let n be a positive integer.
Compute the integral $$\int_{0}^{2\pi}(1+2\cos(t))^n\cos(nt)\mathrm dt$$
Hint : Use the substitution $z=e^{it}$
I starting by using the substitution but the integral didn't simplify. Any help is appreciated.

Answer 1

You can use residue theorem. We have

$$\int_{0}^{2\pi}(1+2\cos(t))^n\cos(nt)\mathrm dt =\Re \int_{0}^{2\pi}(1+2\cos(t))^ne^{int}\mathrm dt.$$ Since $2\cos(t) =e^{it}+1/e^{it},$ we have $$\int_{0}^{2\pi}(1+2\cos(t))^ne^{int}\mathrm dt =\frac{1}{i}\int_{C(0,1)}(1+z+1/z)^n z^{n-1}dz.$$ Now, I'm sure you can conclude.

Answer 2

An alternative solution. Since $2\cos(t)=e^{it}+e^{-it}$ and for any integers $m,n$ we have $\int_{0}^{2\pi}e^{im\theta}e^{ni\theta}\,d\theta=2\pi\delta(m,n)$, by the multinomial theorem the given integral equals:

$$ \frac{1}{2}\int_{0}^{2\pi}\sum_{a+b+c=n}\binom{n}{a,b,c}e^{ita}e^{-itb}(e^{nit}+e^{-nit})\,dt $$ or, by symmetry, $$ 2\pi \sum_{\substack{a+b+c=n\\a-b=n}}\binom{n}{a,b,c}=2\pi\binom{n}{n,0,0}=\color{red}{2\pi}.$$