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(C) We evaluate the integral $I = \int_0^\pi \frac{dx}{1 + \sin x}$.Multiply the numerator and denominator by $(1 - \sin x)$:$I = \int_0^\pi \frac{1 -

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$2$

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(C) We evaluate the integral $I = \int_0^\pi \frac{dx}{1 + \sin x}$.Multiply the numerator and denominator by $(1 - \sin x)$:$I = \int_0^\pi \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx = \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x} dx = \int_0^\pi \frac{1 - \sin x}{\cos^2 x} dx$.Split the integral into two parts:$I = \int_0^\pi (\sec^2 x - \sec x 	an x) dx$.Now,integrate each term:$I = [	an x - \sec x]_0^\pi$.Evaluate at the limits:$I = (	an \pi - \sec \pi) - (	an 0 - \sec 0)$.Since $	an \pi = 0$,$\sec \pi = -1$,$	an 0 = 0$,and $\sec 0 = 1$:$I = (0 - (-1)) - (0 - 1) = 1 - (-1) = 1 + 1 = 2$.

$\int_0^\pi \frac{dx}{1 + \sin x} = $

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