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?

  • $\begingroup$ I have not worked out entirely, but would multiplying both denominator and numerator by $(1+\cos x)$ work? $\endgroup$ – peterwhy Sep 4 '15 at 22:14
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    $\begingroup$ try the $1-\cos(x)=2 \sin^2 (\frac{x}{2})$ substitution $\endgroup$ – Mihail Sep 4 '15 at 22:18
  • $\begingroup$ try $1-\cos(x)=2\sin^2(x/2)$ after separating into two integrals $\int \frac x{2\sin^2(x/2)}\operatorname d x-\int \frac{\sin x}{1-\cos x}\operatorname d x$ $\endgroup$ – Graham Kemp Sep 4 '15 at 22:21

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

  • $\begingroup$ Downvoted? Did I make any mistake? $\endgroup$ – peterwhy Sep 4 '15 at 22:35
  • $\begingroup$ Apparently there's someone that finds it funny to downvote everything $\endgroup$ – Oussama Boussif Sep 4 '15 at 22:38
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    $\begingroup$ Not only is it correct, but it's a nice method to use on this one. $\endgroup$ – user84413 Sep 4 '15 at 22:43

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