n in N के लिए,int_{0}^{npi + V} sqrt{frac{1 + | vedclass

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(C) हम जानते हैं कि $\sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} = |\cos x|$।अतः,समाकलन $I = \int_{0}^{n\pi + V} |\cos x| dx$ हो जाता ह

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$2n + 2 - \sin V$

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(C) हम जानते हैं कि $\sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} = |\cos x|$।अतः,समाकलन $I = \int_{0}^{n\pi + V} |\cos x| dx$ हो जाता है।चूंकि $|\cos x|$ का आवर्तकाल $\pi$ है,हम लिख सकते हैं $I = \int_{0}^{n\pi} |\cos x| dx + \int_{n\pi}^{n\pi + V} |\cos x| dx$।$I = n \int_{0}^{\pi} |\cos x| dx + \int_{0}^{V} |\cos x| dx$।यहाँ $\int_{0}^{\pi} |\cos x| dx = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi} = 1 - (0 - 1) = 2$।दूसरे भाग के लिए,चूंकि $\frac{\pi}{2} अतः,$\int_{0}^{V} |\cos x| dx = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{V} (-\cos x) dx = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{V} = 1 - (\sin V - 1) = 2 - \sin V$।इसलिए,$I = n(2) + 2 - \sin V = 2n + 2 - \sin V$।

$n \in N$ के लिए,$\int_{0}^{n\pi + V} \sqrt{\frac{1 + \cos 2x}{2}} dx$ का मान . . . है (जहाँ $\frac{\pi}{2} < V < \pi$)

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