If `l=int_(0)^(1//2)(dx)/(sqrt(1-x^(2n))), n in N` then
(Where, `[.]` denotes G.I.F)

Options

A. `[l]=1`
B. `[l]=0`
C. `l le (pi)/(2)`
D. `l le (pi)/(6)`

Answer

To solve the integral \( L = \int_{0}^{\frac{1}{2}} \frac{dx}{\sqrt{1 - x^{2n}}} \), where \( n \in \mathbb{N} \), we will follow these steps:

### Step 1: Understanding the Integral
We need to evaluate the integral \( L \). The integrand \( \frac{1}{\sqrt{1 - x^{2n}}} \) suggests a trigonometric substitution might be useful, as it resembles the derivative of the arcsine function.

**Hint:** Look for a substitution that simplifies the square root in the denominator.

### Step 2: Trigonometric Substitution
Let \( x = \sin(\theta) \). Then, \( dx = \cos(\theta) d\theta \). The limits of integration change as follows:
- When \( x = 0 \), \( \theta = 0 \).
- When \( x = \frac{1}{2} \), \( \theta = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \).

Now, substituting into the integral, we have:
\[ L = \int_{0}^{\frac{\pi}{6}} \frac{\cos(\theta) d\theta}{\sqrt{1 - \sin^{2n}(\theta)}} \]

**Hint:** Remember that \( \sqrt{1 - \sin^{2n}(\theta)} = \cos^{n}(\theta) \) for \( n \in \mathbb{N} \).

### Step 3: Simplifying the Integral
Using the identity \( \sqrt{1 - \sin^{2n}(\theta)} = \cos^{n}(\theta) \), we can rewrite the integral:
\[ L = \int_{0}^{\frac{\pi}{6}} \frac{\cos(\theta)}{\cos^{n}(\theta)} d\theta = \int_{0}^{\frac{\pi}{6}} \cos^{1-n}(\theta) d\theta \]

**Hint:** Consider the case when \( n = 1 \) and how it affects the integral.

### Step 4: Evaluating the Integral
For \( n = 1 \):
\[ L = \int_{0}^{\frac{\pi}{6}} \cos^{0}(\theta) d\theta = \int_{0}^{\frac{\pi}{6}} 1 d\theta = \frac{\pi}{6} \]

For \( n > 1 \):
The integral \( \int \cos^{1-n}(\theta) d\theta \) can be evaluated, but it will yield values less than \( \frac{\pi}{6} \) as \( n \) increases.

**Hint:** The value of the integral decreases as \( n \) increases.

### Step 5: Conclusion
Thus, we conclude that:
\[ L \leq \frac{\pi}{6} \]
The maximum value of \( L \) occurs when \( n = 1 \), giving us \( L = \frac{\pi}{6} \).

**Final Answer:** The greatest integer function value is \( \lfloor L \rfloor = 0 \).