int_{0}^{frac{pi}{2}} frac{dx}{1+cos x} = | vedclass

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(C) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Substituting this into the integral,we get:$I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{2 \cos^2 \frac

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(C) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Substituting this into the integral,we get:$I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{2 \cos^2 \frac{x}{2}} = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sec^2 \frac{x}{2} dx$.Integrating $\sec^2 \frac{x}{2}$ with respect to $x$,we get $2 	an \frac{x}{2}$.$I = \frac{1}{2} \left[ 2 	an \frac{x}{2} ight]_{0}^{\frac{\pi}{2}} = \left[ 	an \frac{x}{2} ight]_{0}^{\frac{\pi}{2}}$.Evaluating the limits:$I = 	an \frac{\pi}{4} - 	an 0 = 1 - 0 = 1$.

$ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\cos x} = $

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