Evaluate the following integrals:

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

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

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

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

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

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

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

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

$\Rightarrow I=\int\left(\operatorname{cosec}^{2} x+\operatorname{cosec} x \cot x\right) d x$

$\Rightarrow I=\int \operatorname{cosec}^{2} x d x+\int \operatorname{cosec} x \cot x d x$

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

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

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

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