Solution:

Let $x-\cos x=t$
$\begin{array}{l}
\Rightarrow d(x-\cos x)=d t \\
\Rightarrow(1+\sin x) d x=d t
\end{array}$
$\therefore$ By substituting $\mathrm{t}$ and dt in given equation we obtain
$\begin{array}{l}
\Rightarrow \int \frac{1}{\sqrt{t}} d t \\
\Rightarrow \int t^{-1 \backslash 2} \cdot d t \\
\Rightarrow 2 t^{1 \backslash 2}+c
\end{array}$
But $\mathrm{t}=\mathrm{x}-\cos \mathrm{x}$.
$\Rightarrow 2(x-\cos x)^{1 / 2}+c$