Evaluate the integral $\int {\dfrac{1}{{1 - \cos x}}dx} $.

Hint: To evaluate the required integral, first we will use the trigonometric formula which is given by $1 - \cos 2\theta = 2{\sin ^2}\theta $. After simplification, we will use the integration formula to integrate the function ${\csc ^2}\theta $.

Complete step-by-step answer:
In this problem, we have to evaluate the integral $\int {\dfrac{1}{{1 - \cos x}}dx} $. Let us say $I = \int {\dfrac{1}{{1 - \cos x}}dx} $. Now we are going to use the trigonometric formula which is given by $1 - \cos x = 2{\sin ^2}\left( {\dfrac{x}{2}} \right)$. So, we can write $I = \int {\left[ {\dfrac{1}{{2{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}} \right]\;} dx$.
In the above integral, $\dfrac{1}{2}$ is constant so we will take it outside the integral and we will use the trigonometric identity which is given by $\dfrac{1}{{\sin \theta }} = \csc \theta $. So, we can write
$I = \dfrac{1}{2}\int {\left[ {{{\csc }^2}\left( {\dfrac{x}{2}} \right)} \right]} \;dx$
Now we will use the integration formula which is given by $\int {{{\csc }^2}\left( {bx} \right)dx = - \dfrac{{\cot \left( {bx} \right)}}{b}} $. Note that when we integrate the function then always include a constant which is called integrating constant. So, we can write
$
 I = \dfrac{1}{2}\left[ { - \dfrac{{\cot \left( {\dfrac{x}{2}} \right)}}{{\dfrac{1}{2}}}} \right] + C \\
   \Rightarrow I = - \cot \left( {\dfrac{x}{2}} \right) + C \\
 $
Hence, we can say that $\int {\dfrac{1}{{1 - \cos x}}dx} = - \cot \left( {\dfrac{x}{2}} \right) + C$ where $C$ is integrating constant.

Note: In this problem, we have evaluated the indefinite integral. If $F\left( x \right)$ is an anti-derivative of the function $f\left( x \right)$ then anti-derivative of $f\left( x \right)$ is called an indefinite integral and it is denoted by $\int {f\left( x \right)dx} = F\left( x \right) + C$ where $C$ is called integrating constant. Note that $f\left( x \right)$ is called the integrand. An integral of the form $\int {f\left( x \right)dx} $ without upper and lower limits is called indefinite integral. In this type of problems of integration, we must remember the trigonometric identities and formulas. Also remember the formulas of integration. There is one useful trigonometric formula which is given by $1 + \cos 2\theta = 2{\cos ^2}\left( {\dfrac{\theta }{2}} \right)$.