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

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Question

(C) We have the integral $I = \int \frac{x+\sin x}{1+\cos x} \,d x$. Using the trigonometric identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$

Vedclass · Accepted Answer

$x 	an \frac{x}{2}+c$,where $c$ is the constant of integration

Vedclass · Answer

(C) We have the integral $I = \int \frac{x+\sin x}{1+\cos x} \,d x$. Using the trigonometric identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$,we get: $I = \int \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \,d x$ $I = \int \left( \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} ight) \,d x$ $I = \int \left( \frac{1}{2} x \sec^2 \frac{x}{2} + 	an \frac{x}{2} ight) \,d x$ $I = \frac{1}{2} \int x \sec^2 \frac{x}{2} \,d x + \int 	an \frac{x}{2} \,d x$ Using integration by parts for the first term $\int u v' = uv - \int u' v$: Let $u = x$ and $v' = \sec^2 \frac{x}{2}$. Then $u' = 1$ and $v = 2 	an \frac{x}{2}$. $\frac{1}{2} \int x \sec^2 \frac{x}{2} \,d x = \frac{1}{2} \left( x \cdot 2 	an \frac{x}{2} - \int 2 	an \frac{x}{2} \,d x ight) = x 	an \frac{x}{2} - \int 	an \frac{x}{2} \,d x$. Substituting this back into the expression for $I$: $I = x 	an \frac{x}{2} - \int 	an \frac{x}{2} \,d x + \int 	an \frac{x}{2} \,d x = x 	an \frac{x}{2} + c$.

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

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