int frac{x + sin x}{1 + cos x} dx = | vedclass

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(D) Consider $I = \int \frac{x + \sin x}{1 + \cos x} dx$ $\Rightarrow I = \int \frac{x}{1 + \cos x} dx + \int \frac{\sin x}{1 + \cos x} dx$ Consider $

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

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(D) Consider $I = \int \frac{x + \sin x}{1 + \cos x} dx$ $\Rightarrow I = \int \frac{x}{1 + \cos x} dx + \int \frac{\sin x}{1 + \cos x} dx$ Consider $I = I_{1} + I_{2}$ Now,$I_{1} = \int \frac{x}{1 + \cos x} dx = \int \frac{x \ dx}{2 \cos^{2} \frac{x}{2}}$ $\left\{ \because 1 + \cos 	heta = 2 \cos^{2} \frac{	heta}{2} ight\}$ $\Rightarrow I_{1} = \int \frac{x}{2} \sec^{2} \frac{x}{2} dx$ Using integration by parts: $\int u v \ dx = u \int v \ dx - \int \left( \frac{du}{dx} \int v \ dx ight) dx$ Let $u = x$ and $v = \frac{1}{2} \sec^{2} \frac{x}{2}$ $\Rightarrow I_{1} = x 	an \frac{x}{2} - \int 1 \cdot 	an \frac{x}{2} dx$ $\Rightarrow I_{1} = x 	an \frac{x}{2} - 2 \log_{e} \left| \sec \frac{x}{2} ight| + c_{1}$ $\Rightarrow I_{1} = x 	an \frac{x}{2} - \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{1}$ Now,$I_{2} = \int \frac{\sin x}{1 + \cos x} dx$ Let $1 + \cos x = t \Rightarrow - \sin x \ dx = dt$ $\Rightarrow I_{2} = - \int \frac{1}{t} dt = - \log_{e} |t| + c_{2} = - \log_{e} |1 + \cos x| + c_{2}$ $\Rightarrow I_{2} = - \log_{e} \left| 2 \cos^{2} \frac{x}{2} ight| + c_{2} = - \log_{e} 2 - \log_{e} \left| \cos^{2} \frac{x}{2} ight| + c_{2} = \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{3}$ Adding $I_{1}$ and $I_{2}$: $I = x 	an \frac{x}{2} - \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{1} + \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{3}$ $\Rightarrow I = x 	an \frac{x}{2} + c$

$\int \frac{x + \sin x}{1 + \cos x} dx =$

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