int frac{x+sin x}{1+cos x} d x is equal to | vedclass

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(D) Let $I = \int \frac{x+\sin x}{1+\cos x} dx$. Using the identities $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$,we get: $I = \in

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$x 	an \left(\frac{x}{2}ight)+c$

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

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

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