Explanation

(c)
Let $I = \int_{}^{}\frac{dx}{1 - \cos{x - \sin x}}$
Put $\cos{x = \frac{1 - \tan^{2}\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}$ and $\sin{x = \frac{2\tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}$
$\therefore I = \int_{}^{}\frac{dx}{1 - \frac{\left( 1 - \tan^{2}\frac{x}{2} \right)}{\left( 1 + \tan^{2}\frac{x}{2} \right)} - \frac{2\tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}$
$= \int_{}^{}\frac{\sec^{2}{\frac{x}{2}dx}}{\left\lbrack 1 + \tan^{2}\frac{x}{2} - 1 + \tan^{2}\frac{x}{2} - 2\tan\frac{x}{2} \right\rbrack}$
$= \int_{}^{}{\frac{\sec^{2}{\frac{x}{2}dx}}{2\tan^{2}\frac{x}{2} - 2\tan\frac{x}{2}} =}\int_{}^{}\frac{\frac{1}{2}\sec^{2}{\frac{x}{2}dx}}{\tan^{2}\frac{x}{2} - \tan\frac{x}{2}}$
Put $\tan{\frac{x}{2} = t\ \Rightarrow \frac{1}{2}\sec^{2}{\frac{x}{2}dx = dt}}$
$\therefore I = \int_{}^{}{\frac{dt}{t^{2} - t} = \int_{}^{}\frac{dt}{(t - 1)}}$
$= \int_{}^{}{\left\lbrack \frac{1}{t - 1} - \frac{1}{t} \right\rbrack dt = \int_{}^{}{\frac{dt}{t - 1} - \int_{}^{}\frac{dt}{t}}}$
$= \log{(t - 1) - \log{t + c = \log{\left| \frac{t - 1}{t} \right| + c}}}$
$= \log{\left| \frac{\tan{\frac{x}{2} - 1}}{\tan\frac{x}{2}} \right| + c = \log{\left| 1 - \cot\frac{x}{2} \right| + c}}$