Evaluate the following integrals:

Let $I=\int \frac{\cos x}{1-\cos x} d x$

On multiplying and dividing $(1+\cos x)$, we can write the integral as

$I=\int \frac{\cos x}{1-\cos x}\left(\frac{1+\cos x}{1+\cos x}\right) d x$

$\Rightarrow I=\int \frac{\cos x(1+\cos x)}{(1-\cos x)(1+\cos x)} d x$

$\Rightarrow I=\int \frac{\cos x+\cos ^{2} x}{1-\cos ^{2} x} d x$

$\Rightarrow I=\int \frac{\cos x+\cos ^{2} x}{\sin ^{2} x} d x\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\Rightarrow I=\int\left(\frac{\cos x}{\sin ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x}\right) d x$

$\Rightarrow I=\int\left(\frac{1}{\sin x} \times \frac{\cos x}{\sin x}+\frac{\cos ^{2} x}{\sin ^{2} x}\right) d x$

$\Rightarrow \mathrm{I}=\int\left(\operatorname{cosecx} \cot \mathrm{x}+\cot ^{2} \mathrm{x}\right) \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int\left(\operatorname{cosec} \mathrm{x} \cot \mathrm{x}+\operatorname{cosec}^{2} \mathrm{x}-1\right) \mathrm{dx}\left[\because \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\right]$

$\Rightarrow \mathrm{I}=\int \operatorname{cosec} \mathrm{x} \cot \mathrm{x} \mathrm{dx}+\int \operatorname{cosec}^{2} \mathrm{x} \mathrm{dx}-\int \mathrm{dx}$

Recall $\int \operatorname{cosec}^{2} x d x=-\cot x+c$ and $\int d x=x+c$

We also have $\int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x+c$

$\therefore I=-\operatorname{cosec} x-\cot x-x+c$

Thus, $\int \frac{\cos \mathrm{x}}{1-\cos \mathrm{x}} \mathrm{dx}=-\operatorname{cosec} \mathrm{x}-\cot \mathrm{x}-\mathrm{x}+\mathrm{c}$