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

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(B) We have $I = \int \frac{x - \sin x}{1 - \cos x} dx$. Using the identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \co

Vedclass · Accepted Answer

$-x \cot \frac{x}{2} + c$

Vedclass · Answer

(B) We have $I = \int \frac{x - \sin x}{1 - \cos x} dx$. Using the identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get: $I = \int \frac{x}{2 \sin^2 \frac{x}{2}} dx - \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} dx$ $I = \frac{1}{2} \int x \csc^2 \frac{x}{2} dx - \int \cot \frac{x}{2} dx$. Applying integration by parts to the first term: Let $u = x$ and $dv = \csc^2 \frac{x}{2} dx$. Then $du = dx$ and $v = -2 \cot \frac{x}{2}$. $\int x \csc^2 \frac{x}{2} dx = x(-2 \cot \frac{x}{2}) - \int (-2 \cot \frac{x}{2}) dx = -2x \cot \frac{x}{2} + 2 \int \cot \frac{x}{2} dx$. Substituting this back into the expression for $I$: $I = \frac{1}{2} [-2x \cot \frac{x}{2} + 2 \int \cot \frac{x}{2} dx] - \int \cot \frac{x}{2} dx$ $I = -x \cot \frac{x}{2} + \int \cot \frac{x}{2} dx - \int \cot \frac{x}{2} dx + c$ $I = -x \cot \frac{x}{2} + c$.

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

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