We have,
$$
\begin{aligned}
I & =\int \frac{d x}{1-\cos x-\sin x} \\
& =\int \frac{d x}{1-\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}} \\
& =\int \frac{\sec ^2 \frac{x}{2} d x}{1+\tan ^2 \frac{x}{2}-1+\tan ^2 \frac{x}{2}-2 \tan \frac{x}{2}}
\end{aligned}
$$
$$
=\frac{1}{2} \int \frac{\sec ^2 \frac{x}{2} d x}{\tan \frac{x}{2}\left(\tan \frac{x}{2}-1\right)}
$$
Let $\tan \frac{x}{2}=z$
$$
\begin{aligned}
& \Rightarrow \quad \frac{1}{2} \sec ^2 \frac{x}{2} d x=d z \\
& \Rightarrow \quad \sec ^2 \frac{x}{2} d x=2 d z \\
& \therefore I=\frac{1}{2} \int \frac{2 d z}{z(z-1)}=\int\left(\frac{1}{z-1}-\frac{1}{z}\right) d z \\
& \quad=\log (z-1)-\log z+C \\
& =\log \left(\frac{z-1}{z}\right)+C=\log \left[\frac{\tan \frac{x}{2}-1}{\tan \frac{x}{2}}\right]+C \\
& \quad=\log \left[1-\cot \frac{x}{2}\right]+C
\end{aligned}
$$