Computing $\int_0^{2\pi}(1+2\cos t)^n\cos nt\ \mathrm{d}t$

Question

I'd like to calculate the following integral on the interval $[0,2\pi]$:

$$ I=\int_0^{2\pi}(1+2\cos t)^n\cos nt\ \mathrm{d}t = 2\pi. $$

Answer 1

It's standard to use the following substitution in these cases: $$ z=e^{it},\ \ \ \mathrm{d}t=\frac{\mathrm{d}z}{iz} $$ $$ \cos t=\frac{1}{2}\left(z+\frac{1}{z}\right),\ \ \ e^{int}=z^n. $$ Therefore if $C(0,1)=C$ is the unit circle centered at the origin, we have $$ I=\int_0^{2\pi}(1+2\cos t)^n\cos nt\ \mathrm{d}t= \Re e\int_0^{2\pi}(1+2\cos t)^ne^{int}\ \mathrm{d}t= \oint_C \left(1+z+\frac{1}{z}\right)^nz^n\frac{\mathrm{d}z}{iz} $$ and we now get a simplification as the singularity at the origin is reduced to a simple pole: $$ I=\Re e\oint_C\frac{(z^2+z+1)^n}{iz}\mathrm{d}z =\Re e \left(i2\pi\ \frac{1}{i}\right) = 2\pi $$ by Residue Theorem.

Observation: Hadn't we used Euler's formula to simplify $\cos nt$ we would have obtained a much higher pole and thus a much more confusing calculation to do!