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 identities $1 + \cos x = 2 \cos^{2} \frac{x}{2}$ and $\sin x = 2 \sin \frac

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 identities $1 + \cos x = 2 \cos^{2} \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have: $I = \int_{0}^{\pi/2} \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} dx$ $I = \int_{0}^{\pi/2} \left( \frac{x}{2} \sec^{2} \frac{x}{2} + 	an \frac{x}{2} ight) dx$ Using integration by parts for the first term $\int \frac{x}{2} \sec^{2} \frac{x}{2} dx$: Let $u = \frac{x}{2}$,$dv = \sec^{2} \frac{x}{2} dx$. Then $du = \frac{1}{2} dx$ and $v = 2 	an \frac{x}{2}$. $\int \frac{x}{2} \sec^{2} \frac{x}{2} dx = \frac{x}{2} (2 	an \frac{x}{2}) - \int (2 	an \frac{x}{2}) \cdot \frac{1}{2} dx = x 	an \frac{x}{2} - \int 	an \frac{x}{2} dx$. Substituting this back into the integral $I$: $I = [x 	an \frac{x}{2} - \int 	an \frac{x}{2} dx] + \int 	an \frac{x}{2} dx$ $I = [x 	an \frac{x}{2}]_{0}^{\pi/2}$ $I = \frac{\pi}{2} 	an \frac{\pi}{4} - 0 \cdot 	an 0 = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2}$.

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

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