1. Let's start by stating the problem clearly: We need to evaluate the integral $$\int_{\pi}^{2\pi} \cos(2x)\, dx.$$\n\n2. Recall the integral formula for cosine: $$\int \cos(ax)\, dx = \frac{1}{a} \sin(ax) + C.$$\n\n3. Applying this formula where $a=2$, we get $$\int \cos(2x)\, dx = \frac{1}{2} \sin(2x) + C.$$\n\n4. Now, apply the definite integral limits from $x=\pi$ to $x=2\pi$:\n$$\int_{\pi}^{2\pi} \cos(2x)\, dx = \left[ \frac{1}{2} \sin(2x) \right]_{\pi}^{2\pi} = \frac{1}{2} \sin(4\pi) - \frac{1}{2} \sin(2\pi).$$\n\n5. Evaluate the sine values:\n$$\sin(4\pi) = 0 \quad \text{and} \quad \sin(2\pi) = 0.$$\n\n6. Therefore, the value of the integral is $$\frac{1}{2} \times 0 - \frac{1}{2} \times 0 = 0.$$\n\nFinal answer: $$\int_{\pi}^{2\pi} \cos(2x)\, dx = 0.$$