Question

The value of $\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}$ is (n∈N)

• A. less than or equal to $\frac{\pi}{6}$
• B. greater than or equal to 1
• C. less than $\frac{1}{2}$
• D. greater than $\frac{\pi}{6}$

Answer 1

Answer: (B)

greater than or equal to 1

Answer 2

To solve the integral:

∫(0 to 1/2) (dx / √(1 - x^2))^n dx

We can recognize this as the integral of a power of the standard arccosine function, where the standard integral of √(1 - x^2) is arcsin(x) + C. However, this problem involves raising it to the power n.

Let's perform the integration step by step:

Substitution: Let u = √(1 - x^2), then du = -(x/√(1 - x^2)) dx.

When x = 0, u = 1, and when x = 1/2, u = √3/2.

The integral becomes: ∫(1 to √3/2) -(du / x^n)

Integrate with respect to u: ∫ -(du / x^n) = - (1 / n) * ∫ u^(-n) du

Integrating u^(-n) with respect to u results in: (-1 / n(n - 1)) * u^(-n + 1)

Evaluate the integral limits: Plugging in the limits of integration √(3/2) and 1: (-1 / n(n - 1)) * [√(3/2)^(-n + 1) - 1^(-n + 1)] = (-1 / n(n - 1)) * [(2/√3)^(-n + 1) - 1]

Since n is a natural number (n∈N), and we have a subtraction of a positive term (1) from a term that is greater than or equal to 1 (since (2/√3)^(-n + 1) ≥ 1), the entire expression is non-negative.

Therefore, the correct answer, ∫(0 to 1/2) (dx / √(1 - x^2))^n dx, is greater than or equal to 1 for any natural number n (n∈N).

The correct answer is option (B): greater than or equal to 1