To solve the integral \( \int_{\pi}^{2\pi} |\sin x| \, dx \), we can follow these steps:
### Step 1: Understand the behavior of \( |\sin x| \) from \( \pi \) to \( 2\pi \)
The sine function, \( \sin x \), oscillates between -1 and 1. In the interval \( [\pi, 2\pi] \):
- At \( x = \pi \), \( \sin \pi = 0 \)
- At \( x = \frac{3\pi}{2} \), \( \sin \frac{3\pi}{2} = -1 \)
- At \( x = 2\pi \), \( \sin 2\pi = 0 \)
Since \( \sin x \) is negative in the interval \( (\pi, 2\pi) \), we have:
\[
|\sin x| = -\sin x \quad \text{for } x \in [\pi, 2\pi]
\]
### Step 2: Rewrite the integral
Now we can rewrite the integral using the property of absolute value:
\[
\int_{\pi}^{2\pi} |\sin x| \, dx = \int_{\pi}^{2\pi} -\sin x \, dx
\]
### Step 3: Factor out the negative sign
We can factor out the negative sign from the integral:
\[
\int_{\pi}^{2\pi} -\sin x \, dx = -\int_{\pi}^{2\pi} \sin x \, dx
\]
### Step 4: Integrate \( \sin x \)
The integral of \( \sin x \) is:
\[
\int \sin x \, dx = -\cos x + C
\]
Thus, we evaluate:
\[
-\int_{\pi}^{2\pi} \sin x \, dx = -\left[-\cos x \bigg|_{\pi}^{2\pi}\right]
\]
### Step 5: Evaluate the limits
Now we evaluate the limits:
\[
-\left[-\cos(2\pi) + \cos(\pi)\right]
\]
Calculating \( \cos(2\pi) \) and \( \cos(\pi) \):
- \( \cos(2\pi) = 1 \)
- \( \cos(\pi) = -1 \)
Substituting these values:
\[
-\left[-1 - 1\right] = -\left[-1 + 1\right] = -(-2) = 2
\]
### Final Answer
Thus, the value of the integral is:
\[
\int_{\pi}^{2\pi} |\sin x| \, dx = 2
\]
