If int frac{x-sin x}{1+cos x} dx = x tan left | vedclass

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(A) Let $I = \int \frac{x-\sin x}{1+\cos x} dx$. Using the identities $1+\cos x = 2\cos^2(\frac{x}{2})$ and $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{

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$-4$

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(A) Let $I = \int \frac{x-\sin x}{1+\cos x} dx$. 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: $I = \int \frac{x}{2\cos^2(\frac{x}{2})} dx - \int \frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{2\cos^2(\frac{x}{2})} dx$ $I = \frac{1}{2} \int x \sec^2(\frac{x}{2}) dx - \int 	an(\frac{x}{2}) dx$. Applying integration by parts to the first term: $\int x \sec^2(\frac{x}{2}) dx = x(2	an(\frac{x}{2})) - \int 1 \cdot 2	an(\frac{x}{2}) dx = 2x	an(\frac{x}{2}) - 2 \int 	an(\frac{x}{2}) dx$. Substituting this back into $I$: $I = \frac{1}{2} [2x	an(\frac{x}{2}) - 2 \int 	an(\frac{x}{2}) dx] - \int 	an(\frac{x}{2}) dx$ $I = x	an(\frac{x}{2}) - \int 	an(\frac{x}{2}) dx - \int 	an(\frac{x}{2}) dx = x	an(\frac{x}{2}) - 2 \int 	an(\frac{x}{2}) dx$. Since $\int 	an(\frac{x}{2}) dx = 2 \log |\sec(\frac{x}{2})|$,we get: $I = x	an(\frac{x}{2}) - 2(2 \log |\sec(\frac{x}{2})|) + C = x	an(\frac{x}{2}) - 4 \log |\sec(\frac{x}{2})| + C$. Comparing this with the given expression $x 	an(\frac{x}{2}) + p \log |\sec(\frac{x}{2})| + C$,we find $p = -4$.

If $\int \frac{x-\sin x}{1+\cos x} dx = x \tan \left(\frac{x}{2}\right) + p \log \left|\sec \left(\frac{x}{2}\right)\right| + C$,then $p$ is equal to

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