For n in N,the value of int_{0}^{npi + V} sqr | vedclass

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(C) We know that $\sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} = |\cos x|$.Thus,the integral becomes $I = \int_{0}^{n\pi + V} |\cos x| d

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$2n + 2 - \sin V$

Vedclass · Answer

(C) We know that $\sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} = |\cos x|$.Thus,the integral becomes $I = \int_{0}^{n\pi + V} |\cos x| dx$.Since the period of $|\cos x|$ is $\pi$,we can write $I = \int_{0}^{n\pi} |\cos x| dx + \int_{n\pi}^{n\pi + V} |\cos x| dx$.$I = n \int_{0}^{\pi} |\cos x| dx + \int_{0}^{V} |\cos x| dx$.Since $\int_{0}^{\pi} |\cos x| dx = \int_{0}^{\pi/2} \cos x dx + \int_{\pi/2}^{\pi} (-\cos x) dx = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi} = 1 - (0 - 1) = 2$.For the second part,since $\frac{\pi}{2} So,$\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$.Therefore,$I = n(2) + 2 - \sin V = 2n + 2 - \sin V$.

For $n \in N$,the value of $\int_{0}^{n\pi + V} \sqrt{\frac{1 + \cos 2x}{2}} dx$ is . . . (where $\frac{\pi}{2} < V < \pi$)

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