The value of the integral int_{pi / 6}^{pi / | vedclass

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(D) Let $I = \int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} \right) dx$ Using the identities $\sin 2x = 2\sin x \cos x$ and $

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$1$

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(D) Let $I = \int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} ight) dx$ Using the identities $\sin 2x = 2\sin x \cos x$ and $\cos 2x = 2\cos^2 x - 1$: $I = \int_{\pi / 6}^{\pi / 2} \left( \frac{1 + 2\sin x \cos x + 2\cos^2 x - 1}{\sin x + \cos x} ight) dx$ $I = \int_{\pi / 6}^{\pi / 2} \left( \frac{2\cos x(\sin x + \cos x)}{\sin x + \cos x} ight) dx$ $I = \int_{\pi / 6}^{\pi / 2} 2\cos x dx$ $I = 2[\sin x]_{\pi / 6}^{\pi / 2}$ $I = 2\left( \sin \frac{\pi}{2} - \sin \frac{\pi}{6} ight)$ $I = 2\left( 1 - \frac{1}{2} ight) = 2 	imes \frac{1}{2} = 1$

The value of the integral $\int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} \right) dx$ is equal to

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