Question

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

Answer 1

$\int \frac{x-\sin x}{1-(1-2{\sin }^2\frac{x}{2})}dx=\int \frac{x-\sin x}{2{\sin }^2\frac{x}{2}}dx$

$=\frac{1}{2}\int \frac{x}{{\sin }^2\frac{x}{2}}-\frac{1}{2}\int \frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{{\sin }^2\frac{x}{2}}dx$

$=\frac{1}{2}\int x\cdot cose{c}^2\frac{x}{2}-\int \cot \frac{x}{2}dx$

$=\frac{1}{2}[x\cdot \int cose{c}^2\frac{x}{2}-\int \{\frac{d}{dx}x\cdot \int cose{c}^2\frac{x}{2}dx\}dx]-ln\left|2\sin \frac{x}{2}\right|$

$=\frac{1}{2}[-x\cdot 2\cot \frac{x}{2}+\int 1\cdot \cot \frac{x}{2}\cdot 2dx]-ln\left|2\sin \frac{x}{2}\right|$

$=\frac{1}{2}[-2x\cot \frac{x}{2}+2ln|\sin \frac{x}{2}\cdot 2|]$

$=\frac{1}{2}[-2x\cot \frac{x}{2}+ln|2\cdot \sin \frac{x}{2}|]+C$- Answer.