Let `f(x) = int_(0)^(pi)(sinx)^(n) dx, n in N` then

Options

A. `I_(n)` is rational if n is odd
B. `I_(n)` is irrational if n is even
C. `I_(n)` is an increasing sequence
D. `I_(n)` is a decreasing sequence

Answer

To solve the problem, we need to analyze the function \( f(x) = \int_0^{\pi} (\sin x)^n \, dx \) for \( n \in \mathbb{N} \). We will evaluate the integral for both odd and even values of \( n \) and determine the nature of the function \( f(n) \).

### Step 1: Evaluate \( f(n) \) for odd \( n \)

1. **Let \( n = 1 \)**:
\[ f(1) = \int_0^{\pi} \sin x \, dx \]
The integral of \( \sin x \) is:
\[ -\cos x \bigg|_0^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]
Thus, \( f(1) = 2 \), which is rational.

2. **For any odd \( n \)** (1, 3, 5, ...):
The integral can be computed using the reduction formula:
\[ f(n) = \frac{n-1}{n} f(n-2) \]
Since \( f(1) \) is rational, it follows that \( f(n) \) will also be rational for all odd \( n \).

### Conclusion for odd \( n \):
- \( f(n) \) is rational if \( n \) is odd.

### Step 2: Evaluate \( f(n) \) for even \( n \)

1. **Let \( n = 2 \)**:
\[ f(2) = \int_0^{\pi} \sin^2 x \, dx \]
Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \):
\[ f(2) = \int_0^{\pi} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \left( x - \frac{\sin(2x)}{2} \right) \bigg|_0^{\pi} \]
Evaluating this gives:
\[ = \frac{1}{2} \left( \pi - 0 \right) = \frac{\pi}{2} \]
Since \( \frac{\pi}{2} \) is irrational, \( f(2) \) is irrational.

2. **For any even \( n \)** (2, 4, 6, ...):
By using the reduction formula, we find that:
\[ f(n) = \frac{n-1}{n} f(n-2) \]
Since \( f(2) \) is irrational, it follows that \( f(n) \) will also be irrational for all even \( n \).

### Conclusion for even \( n \):
- \( f(n) \) is irrational if \( n \) is even.

### Step 3: Determine if \( f(n) \) is increasing or decreasing

- As \( n \) increases, \( (\sin x)^n \) becomes smaller for \( x \in (0, \pi) \) because \( \sin x \) is always less than or equal to 1. Thus, the integral \( f(n) \) decreases as \( n \) increases.

### Final Conclusion:
- \( f(n) \) is rational if \( n \) is odd.
- \( f(n) \) is irrational if \( n \) is even.
- \( f(n) \) is a decreasing function as \( n \) increases.