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

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

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$x 	an \left(\frac{x}{2}ight)+c$

Vedclass · Answer

(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 get:$I = \int \frac{x + 2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} dx$$I = \int \frac{x}{2\cos^2 \frac{x}{2}} dx + \int \frac{2\sin \frac{x}{2} \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}} dx$$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 $\int x \sec^2 \frac{x}{2} dx$ (taking $u=x$ and $dv=\sec^2 \frac{x}{2} dx$):$\int x \sec^2 \frac{x}{2} dx = x(2	an \frac{x}{2}) - \int 2	an \frac{x}{2} dx = 2x	an \frac{x}{2} - 2 \int 	an \frac{x}{2} dx$Substituting this back into the expression for $I$:$I = \frac{1}{2} [2x	an \frac{x}{2} - 2 \int 	an \frac{x}{2} dx] + \int 	an \frac{x}{2} dx$$I = x	an \frac{x}{2} - \int 	an \frac{x}{2} dx + \int 	an \frac{x}{2} dx$$I = x	an \frac{x}{2} + c$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

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

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