The value of frac{int_0^{pi / 2}(sin x)^{sqrt | vedclass

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(D) Let $I = \frac{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} dx}{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} dx}$.Let $I_n = \int_0^{\pi / 2} (\sin x)^n dx$. Th

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$2-\sqrt{2}$

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(D) Let $I = \frac{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} dx}{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} dx}$.Let $I_n = \int_0^{\pi / 2} (\sin x)^n dx$. Then $I = \frac{I_{\sqrt{2}+1}}{I_{\sqrt{2}-1}}$.Using the reduction formula for $\int_0^{\pi / 2} (\sin x)^n dx = \frac{n-1}{n} I_{n-2}$,we have:$I_{\sqrt{2}+1} = \int_0^{\pi / 2} (\sin x)^{\sqrt{2}+1} dx = \frac{(\sqrt{2}+1)-1}{\sqrt{2}+1} \int_0^{\pi / 2} (\sin x)^{\sqrt{2}-1} dx$.$I_{\sqrt{2}+1} = \frac{\sqrt{2}}{\sqrt{2}+1} I_{\sqrt{2}-1}$.Therefore,$I = \frac{I_{\sqrt{2}+1}}{I_{\sqrt{2}-1}} = \frac{\sqrt{2}}{\sqrt{2}+1}$.Rationalizing the denominator:$I = \frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{2-\sqrt{2}}{2-1} = 2-\sqrt{2}$.

The value of $\frac{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} d x}{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x}$ is $........$

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