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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{x+\sin x}{1+\cos x} d x$. 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 get: $I = \int \frac{x}{2 \cos^2 \frac{x}{2}} d x + \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} d x$ $I = \frac{1}{2} \int x \sec^2 \frac{x}{2} d x + \int 	an \frac{x}{2} d x$. Applying integration by parts to the first term: $\int x \sec^2 \frac{x}{2} d x = x \cdot \frac{	an \frac{x}{2}}{1/2} - \int 1 \cdot \frac{	an \frac{x}{2}}{1/2} d x = 2x 	an \frac{x}{2} - 2 \int 	an \frac{x}{2} d x$. Substituting this back into the expression for $I$: $I = \frac{1}{2} (2x 	an \frac{x}{2} - 2 \int 	an \frac{x}{2} d x) + \int 	an \frac{x}{2} d x$ $I = x 	an \frac{x}{2} - \int 	an \frac{x}{2} d x + \int 	an \frac{x}{2} d x$ $I = x 	an \frac{x}{2} + C$.

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

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