How to evaluate the integral $ \int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt $? How to evaluate the integral $$\int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt $$
I have attempted it replacing $e^{it}$ with $\cos t +i\sin t$ but it doesn't seem to be working. How would I go about this?

$\int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it}dt=\\
\int_{0}^{2\pi} |2\cos(t)+i\sin(t)-1|^2 2e^{it}dt=\\
\int_{0}^{2\pi} ((2\cos(t)-1)^2+(2\sin(t))^2) 2ie^{it}dt=\\
\int_{0}^{2\pi} (4\cos(t)^2+4\sin(t)^2+1-4\cos(t)) 2ie^{it}dt=\\
\int_{0}^{2\pi} (5-4\cos(t))2i(\cos(t)+i\sin(t)) dt=...$
