frac{int_0^{pi / 2}(sin x)^{sqrt{2}+1} d x}{i | vedclass

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Question

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

Vedclass · Accepted Answer

$2-\sqrt{2}$

Vedclass · Answer

(D) ધારો કે $I = \frac{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} dx}{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} dx}$.ધારો કે $I_n = \int_0^{\pi / 2} (\sin x)^n dx$. તો $I = \frac{I_{\sqrt{2}+1}}{I_{\sqrt{2}-1}}$.$\int_0^{\pi / 2} (\sin x)^n dx = \frac{n-1}{n} I_{n-2}$ માટે રિડક્શન સૂત્રનો ઉપયોગ કરતા:$I_{\sqrt{2}+1} = \int_0^{\pi / 2} (\sin x)^{\sqrt{2}+1} dx = \frac{(\sqrt{2}+1)-1}{\sqrt{2}+1} \int_0^{\pi / 2} (\sin x)^{\sqrt{2}-1} dx$.$I_{\sqrt{2}+1} = \frac{\sqrt{2}}{\sqrt{2}+1} I_{\sqrt{2}-1}$.તેથી,$I = \frac{I_{\sqrt{2}+1}}{I_{\sqrt{2}-1}} = \frac{\sqrt{2}}{\sqrt{2}+1}$.છેદનું સંમેયીકરણ કરતા:$I = \frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{2-\sqrt{2}}{2-1} = 2-\sqrt{2}$.

$\frac{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} d x}{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x}$ નું મૂલ્ય $........$ છે.

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