int_0^{pi /2} frac{x + sin x}{1 + cos x} , dx | vedclass

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(C) Let $I = \int_0^{\pi /2} \frac{x + \sin x}{1 + \cos x} \, dx$. Using the identity $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi /2} \frac{x + \sin x}{1 + \cos x} \, dx$. Using the identity $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get: $I = \int_0^{\pi /2} \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \, dx$ $I = \int_0^{\pi /2} \left( \frac{x}{2} \sec^2 \frac{x}{2} + 	an \frac{x}{2} ight) \, dx$ Consider the integral of the first term using integration by parts: $\int \frac{x}{2} \sec^2 \frac{x}{2} \, dx = x 	an \frac{x}{2} - \int 	an \frac{x}{2} \, dx$ Substituting this back into the expression for $I$: $I = \left[ x 	an \frac{x}{2} - \int 	an \frac{x}{2} \, dx + \int 	an \frac{x}{2} \, dx ight]_0^{\pi /2}$ $I = \left[ x 	an \frac{x}{2} ight]_0^{\pi /2}$ $I = \frac{\pi}{2} 	an \frac{\pi}{4} - 0 \cdot 	an 0 = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}$.

$\int_0^{\pi /2} \frac{x + \sin x}{1 + \cos x} \, dx = $

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