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?
Motto: Complex exponentials are often simpler to manipulate than their underlying cosine and sine parts.


Several answers are already posted, which rush to cosine-sine decompositions at the expense of overly long computations, so let us explain in details how to apply the general principle above to the present case.

1. We start with the identities $$|z|^2=z\cdot\bar z\qquad \overline{e^{it}}=e^{-it}$$ They yield at once $$|2e^{it}-1|^2e^{it}=(2e^{it}-1)(2e^{-it}-1)e^{it}=5e^{it}\color{red}{-2}-2e^{2it}\tag{$\ast$}$$
The good news is that, basically, the computation part is already over... To see why, let us continue.

2. Our second ingredient is that, for every nonzero integer $n$, $$\int_0^{2\pi}e^{int}dt=0$$ which can be proven in several different, elementary, ways, and that the integral for $n=0$ is simply $2\pi$. Thus, once integrated over $(0,2\pi)$, the only remaining term in the expansion $(\ast)$ is $\color{red}{-2}$.

3. Finally, using 1. and 2., $$\int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt=2i\int_0^{2\pi}(\color{red}{-2})\,dt=2i\cdot(\color{red}{-2})\cdot2\pi=-8i\pi$$
Expanding the squared modulus, 

$$((2\cos t-1)^2+\sin^2t)(\cos t+i\sin t)=(4\cos^2t-4\cos t+1+\sin^2t)(\cos t+i\sin t)=(3\cos^2t-4\cos t+2)(\cos t+i\sin t).$$

Now the integrals are over a full period and the functions can be replaced by their average (odd functions cancel out):

$$\cos^3t\to0,$$
$$\cos^2t=\frac{\cos 2t+1}2\to\frac12,$$
$$\cos t\to0,$$
$$\cos^2t\sin t\to\sin t-\sin^3t\to0,$$
$$\cos t\sin t\to\frac12\sin 2t\to 0,$$
$$\sin t\to0.$$

Hence
$$-4\cdot2i\cdot2\pi\cdot\frac12=-8i\pi.$$



A faster way:

$$2i|2e^{it}-1|^2e^{it}=2i(2e^{it}-1)(2e^{-it}-1)e^{it}=2i(2e^{it}-1)(2-e^{it}).$$

Then over a full period, only the constant term yields a nonzero contribution,

$$2i\cdot 2\pi\cdot(-2).$$