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

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

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

Vedclass · Answer

(A) Let $I = \int \frac{x + \sin x}{1 + \cos x} dx$.Using the half-angle formulas,$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Substituting these into the integral:$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{x}{2} \sec^2 \frac{x}{2} + 	an \frac{x}{2} ight) dx$Now,apply integration by parts to the first term $\int \frac{x}{2} \sec^2 \frac{x}{2} dx$ taking $u = \frac{x}{2}$ and $dv = \sec^2 \frac{x}{2} dx$:$\int \frac{x}{2} \sec^2 \frac{x}{2} dx = \frac{x}{2} \cdot 2 	an \frac{x}{2} - \int (1) \cdot 	an \frac{x}{2} dx = x 	an \frac{x}{2} - \int 	an \frac{x}{2} dx$.Substituting this back into the expression for $I$:$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$.

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

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