To solve the problem, we need to find the value of the integral \( \int_{\pi}^{2\pi} f^{-1}(x) \, dx \) where \( f(x) = x + \sin x \).
### Step-by-step Solution:
1. **Understanding the Function**:
We have \( f(x) = x + \sin x \). We need to find \( f^{-1}(x) \), which is the inverse function of \( f(x) \).
2. **Finding the Derivative**:
The derivative of \( f(x) \) is:
\[
f'(x) = 1 + \cos x
\]
Since \( \cos x \) oscillates between -1 and 1, \( f'(x) \) is always positive for \( x \in [\pi, 2\pi] \). This means \( f(x) \) is strictly increasing in this interval, and thus \( f(x) \) is invertible.
3. **Setting Up the Integral**:
By the property of inverse functions, we have:
\[
\int_a^b f^{-1}(x) \, dx = b f^{-1}(b) - a f^{-1}(a) - \int_{f^{-1}(a)}^{f^{-1}(b)} f(t) \, dt
\]
Here, \( a = \pi \) and \( b = 2\pi \).
4. **Finding \( f^{-1}(\pi) \) and \( f^{-1}(2\pi) \)**:
- For \( f^{-1}(\pi) \):
\[
f(t) = \pi \implies t + \sin t = \pi
\]
Checking \( t = \pi \):
\[
f(\pi) = \pi + \sin(\pi) = \pi
\]
So, \( f^{-1}(\pi) = \pi \).
- For \( f^{-1}(2\pi) \):
\[
f(t) = 2\pi \implies t + \sin t = 2\pi
\]
Checking \( t = 2\pi \):
\[
f(2\pi) = 2\pi + \sin(2\pi) = 2\pi
\]
So, \( f^{-1}(2\pi) = 2\pi \).
5. **Evaluating the Integral**:
Now we can substitute into the integral formula:
\[
\int_{\pi}^{2\pi} f^{-1}(x) \, dx = 2\pi \cdot 2\pi - \pi \cdot \pi - \int_{\pi}^{2\pi} f(t) \, dt
\]
This simplifies to:
\[
= 4\pi^2 - \pi^2 - \int_{\pi}^{2\pi} (t + \sin t) \, dt
\]
\[
= 3\pi^2 - \int_{\pi}^{2\pi} t \, dt - \int_{\pi}^{2\pi} \sin t \, dt
\]
6. **Calculating the Integrals**:
- For \( \int_{\pi}^{2\pi} t \, dt \):
\[
= \left[ \frac{t^2}{2} \right]_{\pi}^{2\pi} = \frac{(2\pi)^2}{2} - \frac{\pi^2}{2} = \frac{4\pi^2}{2} - \frac{\pi^2}{2} = \frac{3\pi^2}{2}
\]
- For \( \int_{\pi}^{2\pi} \sin t \, dt \):
\[
= \left[ -\cos t \right]_{\pi}^{2\pi} = -\cos(2\pi) + \cos(\pi) = -1 + 1 = 0
\]
7. **Final Calculation**:
Putting it all together:
\[
\int_{\pi}^{2\pi} f^{-1}(x) \, dx = 3\pi^2 - \frac{3\pi^2}{2} - 0 = 3\pi^2 - \frac{3\pi^2}{2} = \frac{6\pi^2}{2} - \frac{3\pi^2}{2} = \frac{3\pi^2}{2}
\]
Thus, the final answer is:
\[
\frac{3\pi^2}{2} + 2
\]
### Final Answer:
\[
\int_{\pi}^{2\pi} f^{-1}(x) \, dx = \frac{3\pi^2}{2} + 2
\]
