int frac{x + sin x}{1 + cos x} , dx is equal | vedclass

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(B) We have the integral $I = \int \frac{x + \sin x}{1 + \cos x} \, dx$. Using the identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 +

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$x 	an \frac{x}{2} + c$

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(B) We have the integral $I = \int \frac{x + \sin x}{1 + \cos x} \, dx$. Using the identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get: $I = \int \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \, dx$ $I = \int \left( \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} ight) \, dx$ $I = \int \left( \frac{1}{2} x \sec^2 \frac{x}{2} + 	an \frac{x}{2} ight) \, dx$ Now,using integration by parts for the first term $\int \frac{1}{2} x \sec^2 \frac{x}{2} \, dx$: Let $u = x$ and $dv = \frac{1}{2} \sec^2 \frac{x}{2} \, dx$. Then $du = dx$ and $v = 	an \frac{x}{2}$. $\int \frac{1}{2} x \sec^2 \frac{x}{2} \, dx = x 	an \frac{x}{2} - \int 	an \frac{x}{2} \, dx$. Substituting this back into $I$: $I = (x 	an \frac{x}{2} - \int 	an \frac{x}{2} \, dx) + \int 	an \frac{x}{2} \, dx + c$ $I = x 	an \frac{x}{2} + c$.

$\int \frac{x + \sin x}{1 + \cos x} \, dx$ is equal to

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