How to integrate following indefinite integal?

Question

The integral is

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

However, the only guess I have is that the denominator is the derivative of the numerator. Probably the integration by substitution will work here?

Answer 1

$$\begin{align*} \int\frac{x-\sin x}{1-\cos x}dx &= \int \frac{(x-\sin x)(1+\cos x)}{(1-\cos x)(1+\cos x)}\ dx \\ &= \int \frac{x + x\cos x - \sin x -\sin x \cos x}{1-\cos^2 x}\ dx\\ &= \int (x\csc ^2 x + x \cot x \csc x - \csc x - \cot x)\ dx\\ &= -\int x\ d(\cot x) - \int x\ d(\csc x) - \int \csc x\ dx - \int \cot x\ dx\\ &= -x \cot x + \int \cot x\ dx - x\csc x + \int \csc x\ dx - \int \csc x\ dx - \int \cot x\ dx\\ &= -x \cot x - x\csc x + C\\ &= -x(\cot x + \csc x) + C \end{align*}$$