The value of int_0^{1/2} frac{dx}{sqrt{1-x^{2 | vedclass

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(A) Let $I = \int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}}$.For $x \in (0, 1/2)$, we have $0  1$

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less than or equal to $\frac{\pi}{6}$

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(A) Let $I = \int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}}$.For $x \in (0, 1/2)$, we have $0  1$.If $n=1$, then $I = \int_0^{1/2} \frac{dx}{\sqrt{1-x^2}} = [\sin^{-1} x]_0^{1/2} = \frac{\pi}{6}$.If $n > 1$, then $x^{2n}  1 - x^2$.This implies $\sqrt{1 - x^{2n}} > \sqrt{1 - x^2}$.Therefore, $\frac{1}{\sqrt{1 - x^{2n}}} Integrating both sides from $0$ to $1/2$:$I Thus, for all $n \in N$, $I \leq \frac{\pi}{6}$.

The value of $\int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}}$ is $(n \in N)$

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