Evaluate the following integrals:

Assume $x+\cos ^{2} x=t$

$d\left(x+\cos ^{2} x\right)=d t$

$(1+(-2 \cos x \cdot \sin x)) d x=d t$

$2 \sin x \cdot \cos x=\sin 2 x$

$(1-\sin 2 x) d x=d t$\

Put $t$ and $d t$ in given equation we get

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

$=\ln |\mathrm{t}|+\mathrm{c}$

But $t=x+\cos ^{2} x$

$=\ln \left|x+\cos ^{2} x\right|+c$