Evaluate the following integrals:

Assume $x-\cos x=t$

$\Rightarrow \mathrm{d}(x-\cos x)=\mathrm{dt}$

$\Rightarrow(1+\sin x) \mathrm{dx}=\mathrm{dt}$

$\therefore$ Substituting $t$ and dt in given equation we get

$\Rightarrow \int \frac{1}{\sqrt{t}} \mathrm{dt}$

$\Rightarrow \int \mathrm{t}^{-1 \backslash 2} \cdot \mathrm{dt}$

$\Rightarrow 2 \mathrm{t}^{1 \backslash 2}+\mathrm{c}$

But $\mathrm{t}=\mathrm{x}-\cos \mathrm{x}$

$\Rightarrow 2(x-\cos x)^{1 / 2}+c$