Evaluate the following integrals:

Assume $x+\cos x=t$

$d(x+\cos x)=d t$

$\Rightarrow 1-\sin x d x=d t$

Put $\mathrm{t}$ and dt in given equation we get

$\Rightarrow \int \frac{d t}{t}$

$=\ln |t|+c$

But $t=x+\cos x$

$=\ln |x+\cos x|+c$