If `I_(n)=int_(0)^(npi)max(|sinx|,|sin^(-1)(sinx)|)dx`, the `I_(2)+I_(4)` has the value `(lambdapi^(2))/(2)`, where `lambda` is

Answer

To solve the problem, we need to evaluate the integral \( I_n = \int_0^{n\pi} \max(|\sin x|, |\sin^{-1}(\sin x)|) \, dx \) for \( n = 2 \) and \( n = 4 \), and then find \( I_2 + I_4 \).

### Step 1: Understanding the Functions
First, we need to analyze the functions involved:
- \( |\sin x| \) oscillates between 0 and 1 with a period of \( 2\pi \).
- \( |\sin^{-1}(\sin x)| \) is a piecewise function that takes values in the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and is periodic with a period of \( 2\pi \).

### Step 2: Graphing the Functions
To find the maximum of these two functions, we can graph them over one period \( [0, 2\pi] \):
- The graph of \( |\sin x| \) reaches its maximum of 1 at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).
- The graph of \( |\sin^{-1}(\sin x)| \) reaches its maximum of \( \frac{\pi}{2} \) at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

### Step 3: Finding the Maximum
From the graphs, we can see:
- For \( x \in [0, \frac{\pi}{2}] \), \( |\sin x| \) is greater than \( |\sin^{-1}(\sin x)| \).
- For \( x \in [\frac{\pi}{2}, \frac{3\pi}{2}] \), \( |\sin^{-1}(\sin x)| \) is greater than \( |\sin x| \).
- For \( x \in [\frac{3\pi}{2}, 2\pi] \), \( |\sin x| \) is again greater than \( |\sin^{-1}(\sin x)| \).

Thus, we can express the integral as:
\[ I_2 = \int_0^{2\pi} \max(|\sin x|, |\sin^{-1}(\sin x)|) \, dx = \int_0^{\frac{\pi}{2}} |\sin x| \, dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\sin^{-1}(\sin x)| \, dx + \int_{\frac{3\pi}{2}}^{2\pi} |\sin x| \, dx \]

### Step 4: Calculating \( I_2 \)
Calculating each part:
1. \( \int_0^{\frac{\pi}{2}} |\sin x| \, dx = 1 \)
2. \( \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\sin^{-1}(\sin x)| \, dx = \frac{\pi^2}{2} \) (area of two triangles)
3. \( \int_{\frac{3\pi}{2}}^{2\pi} |\sin x| \, dx = 1 \)

Thus,
\[ I_2 = 1 + \frac{\pi^2}{2} + 1 = \frac{\pi^2}{2} + 2 \]

### Step 5: Calculating \( I_4 \)
Now for \( n = 4 \):
\[ I_4 = \int_0^{4\pi} \max(|\sin x|, |\sin^{-1}(\sin x)|) \, dx \]
This will consist of 4 identical segments of \( I_2 \):
\[ I_4 = 4 \cdot \left( 1 + \frac{\pi^2}{2} + 1 \right) = 4 \left( \frac{\pi^2}{2} + 2 \right) = 2\pi^2 + 8 \]

### Step 6: Summing \( I_2 + I_4 \)
Now we can find \( I_2 + I_4 \):
\[ I_2 + I_4 = \left( \frac{\pi^2}{2} + 2 \right) + \left( 2\pi^2 + 8 \right) = \frac{5\pi^2}{2} + 10 \]

### Step 7: Finding \( \lambda \)
According to the problem, \( I_2 + I_4 = \frac{\lambda \pi^2}{2} \):
\[ \frac{5\pi^2}{2} + 10 = \frac{\lambda \pi^2}{2} \]
Setting the coefficients equal gives:
\[ \lambda = 5 + \frac{20}{\pi^2} \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 5 \]