The value of the definite integral `int_(0)^(2npi) max (sinx,sin^(-1)( sinx)) dx` equals to (where, n in l)

Options

A. `(n(pi^(2)-4))/(2)`
B. `(n(pi^(2)-4))/(4)`
C. `(n(pi^(2)-8))/(4)`
D. `(n(pi^(2)-2))/(4)`

Answer

To solve the definite integral \( \int_{0}^{2n\pi} \max(\sin x, \sin^{-1}(\sin x)) \, dx \), we will analyze the functions involved and calculate the integral step by step.

### Step 1: Understand the Functions
First, we need to understand the behavior of the functions \( \sin x \) and \( \sin^{-1}(\sin x) \).

- The function \( \sin x \) oscillates between -1 and 1 with a period of \( 2\pi \).
- The function \( \sin^{-1}(\sin x) \) is defined as follows:
- For \( x \in [0, \pi] \), \( \sin^{-1}(\sin x) = x \).
- For \( x \in [\pi, 2\pi] \), \( \sin^{-1}(\sin x) = 2\pi - x \).

### Step 2: Identify the Interval
The integral is from \( 0 \) to \( 2n\pi \). We can break this into segments of \( [0, 2\pi] \) and then multiply the result by \( n \) since the functions are periodic with a period of \( 2\pi \).

### Step 3: Calculate the Integral from \( 0 \) to \( 2\pi \)
We will evaluate the integral over one period \( [0, 2\pi] \):

1. **From \( 0 \) to \( \pi \)**:
- Here, \( \max(\sin x, \sin^{-1}(\sin x)) = \max(\sin x, x) \).
- The function \( \sin x \) starts at 0, reaches 1 at \( \frac{\pi}{2} \), and returns to 0 at \( \pi \).
- The function \( x \) is a straight line from 0 to \( \pi \).
- The intersection point occurs at \( x = \frac{\pi}{2} \) where both functions equal \( \frac{\pi}{2} \).
- Thus, \( \max(\sin x, \sin^{-1}(\sin x)) = x \) for \( x \in [0, \frac{\pi}{2}] \) and \( \max(\sin x, \sin^{-1}(\sin x)) = \sin x \) for \( x \in [\frac{\pi}{2}, \pi] \).

2. **Calculate the integral**:
\[ \int_{0}^{\pi} \max(\sin x, \sin^{-1}(\sin x)) \, dx = \int_{0}^{\frac{\pi}{2}} x \, dx + \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx \]

- For the first integral:
\[ \int_{0}^{\frac{\pi}{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{(\frac{\pi}{2})^2}{2} = \frac{\pi^2}{8} \]

- For the second integral:
\[ \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx = \left[-\cos x\right]_{\frac{\pi}{2}}^{\pi} = -\cos(\pi) - (-\cos(\frac{\pi}{2})) = 1 - 0 = 1 \]

- Therefore, the total integral from \( 0 \) to \( \pi \) is:
\[ \frac{\pi^2}{8} + 1 \]

### Step 4: Calculate the Integral from \( 0 \) to \( 2n\pi \)
Since the integral is periodic with period \( 2\pi \):
\[ \int_{0}^{2n\pi} \max(\sin x, \sin^{-1}(\sin x)) \, dx = n \left( \frac{\pi^2}{8} + 1 \right) \]

### Final Result
Thus, the value of the definite integral is:
\[ \int_{0}^{2n\pi} \max(\sin x, \sin^{-1}(\sin x)) \, dx = n \left( \frac{\pi^2}{8} + 1 \right) \]