Evaluate the following integrals:

Assume $1+\cos x=t$

$\Rightarrow d(1+\cos x)=d t$

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

Substituting the values oft and dt we get

$\Rightarrow-\int \frac{\mathrm{dt}}{\mathrm{t}^{2}}$

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

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

$\Rightarrow \frac{t^{-1}}{1}+c$

But $t=1+\cos x$

$\Rightarrow \frac{+1}{1+\cos x}+C$