Question

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

Answer 1

Given to evaluate $\int \frac{x-\sin x}{1-\cos x}dx$.

$\begin{array}{c}\int \frac{x-\sin x}{1-\left(1-2{\sin }^2\frac{x}{2}\right)}dx=\int \frac{x-\sin x}{2{\sin }^2\frac{x}{2}}dx\\ =\frac{1}{2}\int \frac{\mathrm{x}}{{\sin }^2\frac{x}{2}}-\frac{1}{2}\int \frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{{\sin }^2\frac{x}{2}}dx\\ =\frac{1}{2}\int x\cdot {\mathrm{cosec}}^2\frac{x}{2}-\int \cot \frac{x}{2}dx\\ =\frac{1}{2}\left[x\cdot \int {\mathrm{cosec}}^2\frac{x}{2}-\int \left\{\frac{d}{dx}x\cdot \int {\mathrm{cosec}}^2\frac{x}{2}dx\right\}dx\right]-\ln \left|2\sin \frac{x}{2}\right|\\ =\frac{1}{2}\left[-\mathrm{x}\cdot 2\cot \frac{\mathrm{x}}{2}+\int 1\cdot \cot \frac{\mathrm{x}}{2}\cdot 2\mathrm{dx}\right]-\ln \left|2\sin \frac{\mathrm{x}}{2}\right|\\ =\frac{1}{2}\left[-2x\cot \frac{x}{2}+2\ln \left|\sin \frac{x}{2}\cdot 2\right|\right]\\ =\frac{1}{2}\left[-2x\cot \frac{x}{2}+\ln \left|2\cdot \sin \frac{x}{2}\right|\right]+C\end{array}$

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