The value of the definite integral `int_(2pi)^(5pi//2)(sin^(-1)(cosx)+cos^(-1)(sinx))dx` is equal to

Options

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

Explanation

To solve the integral \( I = \int_{2\pi}^{\frac{5\pi}{2}} \left( \sin^{-1}(\cos x) + \cos^{-1}(\sin x) \right) dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the limits of the integral. We know that both \(\sin^{-1}(\cos x)\) and \(\cos^{-1}(\sin x)\) are periodic functions with a period of \(2\pi\). Therefore, we can express the integral as:
\[ I = \int_{2\pi}^{\frac{5\pi}{2}} \left( \sin^{-1}(\cos x) + \cos^{-1}(\sin x) \right) dx = \int_{0}^{\frac{\pi}{2}} \left( \sin^{-1}(\cos x) + \cos^{-1}(\sin x) \right) dx \]

### Step 2: Use the Property of Integrals
Using the property of integrals, we can transform the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( \sin^{-1}(\cos(\frac{\pi}{2} - x)) + \cos^{-1}(\sin(\frac{\pi}{2} - x)) \right) dx \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \left( \sin^{-1}(\sin x) + \cos^{-1}(\cos x) \right) dx \]

### Step 3: Simplify the Integral
Now, we know that:
\[ \sin^{-1}(\sin x) = x \quad \text{and} \quad \cos^{-1}(\cos x) = x \]
for \(x\) in the interval \([0, \frac{\pi}{2}]\). Therefore, we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} (x + x) dx = \int_{0}^{\frac{\pi}{2}} 2x \, dx \]

### Step 4: Evaluate the Integral
Now we can evaluate the integral:
\[ I = 2 \int_{0}^{\frac{\pi}{2}} x \, dx \]
Calculating the integral:
\[ \int x \, dx = \frac{x^2}{2} \quad \text{so} \quad \int_{0}^{\frac{\pi}{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{(\frac{\pi}{2})^2}{2} - 0 = \frac{\pi^2}{8} \]
Thus,
\[ I = 2 \cdot \frac{\pi^2}{8} = \frac{\pi^2}{4} \]

### Final Answer
The value of the definite integral is:
\[ \boxed{\frac{\pi^2}{4}} \]